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Ideal Gas Law - Practice Questions (37)

Question 1: 1. The following statements about the internal energy of an ideal gas are correct

1. The following statements about the internal energy of an ideal gas are correct

  • A. A. Ideal gases have molecular potentials
  • B. B. The internal energy of an ideal gas is the sum of the average potential energy and the average kinetic energy of the molecules
  • C. C. The internal energy of an ideal gas of a given mass is related only to its volume
  • D. D. The internal energy of an ideal gas of a given mass is only dependent on temperature

Answer: D

Solution: A. An ideal gas is conceived as one in which there are no mutual forces between molecules. Molecular potential energy is the energy due to the interaction of molecules. If there are no forces, then there is no molecular potential energy.A Wrong B. Ideal gases do not have molecular potential energy. C. The internal energy of an ideal gas of a certain mass is related to temperature only. D.The internal energy of an ideal gas of a certain mass is related to temperature only. Ideal gases are conceived as molecules without mutual forces.

Question 2: 2. The figure shows a $p - T$ image of the state of a certain mass of an ideal gas that changes cycl...

2. The figure shows a $p - T$ image of the state of a certain mass of an ideal gas that changes cyclically along the sequence $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$. If this cycle is represented by an $V - T$ or $p - V$ image, and $T$ is the thermodynamic temperature, the graphical representation may be correct as ) ![](/images/questions/phys-ideal-gas-law/image-001.jpg)

  • A. A. ![](/images/questions/phys-ideal-gas-law/image-001.jpg)
  • B. B. ![](/images/questions/phys-ideal-gas-law/image-002.jpg)
  • C. C. ![](/images/questions/phys-ideal-gas-law/image-003.jpg)
  • D. D. ![](/images/questions/phys-ideal-gas-law/image-004.jpg)

Answer: A

Solution: By the ideal equation of state $$ \frac { p V } { T } = C $$ It can be seen that $1 \rightarrow 2$ is isovolumetric, $2 \rightarrow 3$ is isovolumetric, and $3 \rightarrow 1$ is isothermal, so only A is correct.

Question 3: 3. A cylindrical vacuum container, a light spring tied to the top of the cylinder under the hanging ...

3. A cylindrical vacuum container, a light spring tied to the top of the cylinder under the hanging of an unaccounted for mass of the piston, the spring is in the natural length, the piston just touch the bottom of the cylinder, as shown in the figure. When a certain mass of ideal gas is injected under the piston, the temperature is $T$, gas The height of the column is $h$, then when the temperature is $T ^ { \prime }$ the height of the column $h ^ { \prime }$ is ![](/images/questions/phys-ideal-gas-law/image-002.jpg)

  • A. A. $\frac { T ^ { \prime } h } { T }$
  • B. B. $\frac { T h } { T ^ { \prime } }$
  • C. C. $h \sqrt { \frac { T ^ { \prime } } { T } }$
  • D. D. $h \sqrt { \frac { T } { T ^ { \prime } } }$

Answer: C

Solution: Let the coefficient of strength of the spring be $k$, the spring force $f = k h$ when the height of the gas column is $h$, the resulting pressure $\frac { f } { s } = \frac { k h } { s }$, $s$ (being the cross-sectional area of the container). Take a closed gas as the object of study: initial state: ($T , h s , \frac { k h } { s }$; end state The equation of state of an ideal gas $\frac { \frac { k h } { s } \cdot h s } { T } = \frac { \frac { k h ^ { \prime } } { s } \cdot h s } { T ^ { \prime } }$ is solved by $h ^ { \prime } = h \sqrt { \frac { T ^ { \prime } } { T } }$, so C is correct and ABD is wrong.

Question 4: 4. A, B, C, D four students in the "study of gas experimental law experiment", respectively, the fol...

4. A, B, C, D four students in the "study of gas experimental law experiment", respectively, the following four images (shown in the figure). Then the following about their statements, is not correct () ![](/images/questions/phys-ideal-gas-law/image-003.jpg) a ![](/images/questions/phys-ideal-gas-law/image-004.jpg) b ![](/images/questions/phys-ideal-gas-law/image-005.jpg) c ![](/images/questions/phys-ideal-gas-law/image-006.jpg) d

  • A. A. If A is studying Charlie's Law, the image he made might be Figure $a$
  • B. B. If B is studying Boyle's law, the image he made might be Figure $b$
  • C. C. If C is studying Charlie's Law, the image he made might be Figure $c$
  • D. D. If Ding is studying Gay-Lussac's law, the image he makes might be the diagram $d$

Answer: C

Solution: AC.Charlie's law studies isovolumetric changes, the pressure is proportional to the temperature, and over the origin of the coordinates, so A is correct, does not meet the meaning of the question, C is wrong, meet the meaning of the question; B. Boyle's law studies isothermal changes, the pressure is inversely proportional to the volume, so B is correct and does not meet the meaning of the question; D. Gay a Lussac's law, the study of isobaric pressure changes, the volume is directly proportional to the temperature, so D is correct, does not meet the meaning of the question.

Question 5: 5. As shown in the figure, a certain mass of a certain gas isobar, isobar on the $a , b$ two states ...

5. As shown in the figure, a certain mass of a certain gas isobar, isobar on the $a , b$ two states compared, the following statements are correct ( ) ![](/images/questions/phys-ideal-gas-law/image-007.jpg)

  • A. A. Number of molecules $b$ hitting a unit area in the same amount of time State more
  • B. B. Number of molecules $a$ hitting a unit area in the same amount of time State more
  • C. C. The number of molecules hitting the same area in the same time is the same in both states
  • D. D. There are as many molecules per unit volume in both states

Answer: B

Solution: ABC. From the ideal gas equation of state $$ V = \frac { n R } { p } T $$ It can be seen that the $V - T$ graph represents isobaric pressure change by the inclined line across the origin, and the volume of the $b$ state is larger than the volume of the $a$ state, so the molecular densification of the $b$ state is less than that of the $a$ state, so the $a$ state has less molecular densification than $a$ state. INLINE_FORMULA_6]] state is less dense than $a$ state, so the $a$ state has more molecules per unit area at the same time as the $a$ state, so B is correct and AC is wrong; D. The total number of molecules remains the same for a given mass of gas. $$ V _ { a } > V _ { b } $$ The number of molecules per unit volume $a$ states more, so D is wrong.

Question 6: 6. As shown in the figure for the internal combustion engine schematic, the principle structure of t...

6. As shown in the figure for the internal combustion engine schematic, the principle structure of the simplified model is a cylinder piston model, the piston is enclosed in the upper part of a certain mass of ideal gas. If the piston moves upward, about the internal gas (ignoring the heat exchange with the outside world) is correct ( ) ![](/images/questions/phys-ideal-gas-law/image-008.jpg)

  • A. A. Volume decreases, pressure stays the same, temperature increases
  • B. B. The outside world does work on the gas and the internal energy increases
  • C. C. Volume decreases, pressure increases, temperature stays the same
  • D. D. The gas does work on the outside world and its internal energy increases

Answer: B

Solution: The piston moves upward, the volume of the enclosed gas decreases, and the outside world does work on the gas, according to the first law of thermodynamics. $$ W + Q = \Delta U , Q = 0 $$ So the internal energy increases, the gas temperature increases, according to the ideal gas equation of state, there are $$ \frac { P V } { T } = C $$ It is known that the pressure of the gas increases.

Question 7: 7. In an experiment, as shown in the figure, a certain mass of an ideal gas was enclosed in a test t...

7. In an experiment, as shown in the figure, a certain mass of an ideal gas was enclosed in a test tube with a column of mercury, and after a period of time, it was found that the column of mercury moved down a certain distance. Then about this experiment, the following statements are correct. ![](/images/questions/phys-ideal-gas-law/image-009.jpg)

  • A. A. The average kinetic energy of the molecules of a confined gas decreases and the gas absorbs heat
  • B. B. The impulse of the molecules of a closed gas against the wall per unit area of the vessel is constant in equal times
  • C. C. The number of molecules of a confined gas colliding with a unit area of vessel wall per unit time is constant
  • D. D. Since mercury does not infiltrate the glass, the molecules within the layer of the attacher are denser than those of mercury.

Answer: B

Solution: AB. the gas pressure remains constant and isobaric changes occur $$ \frac { V _ { 0 } } { T _ { 0 } } = \frac { V _ { 1 } } { T _ { 1 } } $$ Also. $$ V _ { 0 } > V _ { 1 } $$ So $$ T _ { 0 } > T _ { 1 } $$ So the temperature decreases, the average molecular kinetic energy decreases, the gas is exothermic, and the impulse of the molecules of the enclosed gas against the wall per unit area of the vessel remains constant for an equal period of time, so A is wrong and B is correct; C. Since the volume decreases and the number of molecules remains unchanged, the number of molecules impinging per unit time should increase, so C is wrong; D. When not infiltrated, the molecules in the attached layer are sparser than the molecules inside the mercury, so D is wrong.

Question 8: 8. An ideal gas of a certain mass absorbs 500 J of heat from the outside world and does 100 J of wor...

8. An ideal gas of a certain mass absorbs 500 J of heat from the outside world and does 100 J of work on the outside world.

  • A. A. The temperature must have risen.
  • B. B. Internal energy may be unchanged
  • C. C. The pressure must be greater.
  • D. D. The pressure must be less.

Answer: A

Solution: From the first law of thermodynamics $Q = W + \Delta E$, substituting the data we get $$ \Delta E = 500 - 100 = 400 \mathrm {~J} > 0 $$ It can be seen that the process increases the internal energy, and because the internal energy of an ideal gas is a single-valued function of temperature, the temperature increases. Work is done on the outside and the volume increases, and again, according to the ideal gas equation of state, it can be seen that in the case of an increase in both volume and temperature, it is not possible to determine how the pressure changes.

Question 9: 9. As shown in the figure. An ideal gas of a certain mass changes from state $A$ to $C$ via $B$. The...

9. As shown in the figure. An ideal gas of a certain mass changes from state $A$ to $C$ via $B$. The volume ![](/images/questions/phys-ideal-gas-law/image-010.jpg)

  • A. A. First constant then decreasing
  • B. B. Decreasing and then unchanged
  • C. C. run
  • D. D. keep on decreasing

Answer: A

Solution: From the image of $V - T$, we know that the volume of an ideal gas of a certain mass changes from the state $A$ to $C$ via $B$, and the volume of the gas first remains unchanged and then decreases.

Question 10: 12. An ideal gas of a certain mass expands isobarically from state $a$ to state $b$, then isobaricly...

12. An ideal gas of a certain mass expands isobarically from state $a$ to state $b$, then isobaricly to state $c$, then isothermally to state $d$, and then finally, through a complex process, it returns to state $a$, where its pressure $p$ and volume $p$ are equalized. Returning to state $a$, its pressure $p$ is related to its volume $V$ as shown in the figure. The following statement is correct. ![](/images/questions/phys-ideal-gas-law/image-011.jpg)

  • A. A. From $a$ to $b$, the kinetic energy of each gas molecule is increased
  • B. B. From $b$ to $c$, the gas temperature decreases.
  • C. C. From $c$ to $d$, the internal energy of the gas remains constant.
  • D. D. From $d$ to $a$, the gas does positive work to the outside world.

Answer: C

Solution: A. From $a$ to $b$ according to $$ \frac { p V } { T } = C $$ It can be seen that the pressure remains the same, the volume increases, then the temperature increases, the average kinetic energy of the molecules increases, but not every gas molecule's kinetic energy increases, so A is wrong; B. According to $$ \frac { p V } { T } = C $$ B. According to $$ \frac { p V } { T } = C $$, from $b$ to $c$, the volume stays the same and the pressure increases, so the temperature of the gas increases, so B is wrong; C. From $c$ to $d$, the temperature of the gas remains unchanged, so the internal energy of the gas remains unchanged, so C is correct; D. From $d$ to $a$, the volume of the gas decreases, so the outside world does positive work on the gas, so D is wrong.

Question 11: 13. For a certain mass of gas enclosed in a cylinder, if the volume of gas is kept constant, when th...

13. For a certain mass of gas enclosed in a cylinder, if the volume of gas is kept constant, when the temperature rises from 300 K to 600 K, the following statements are correct ( )

  • A. A. The density of the gas doubles
  • B. B. The pressure of the gas doubles
  • C. C. The average kinetic energy of a gas molecule is halved
  • D. D. Constant number of molecules per second hitting the wall per unit area of the vessel

Answer: B

Solution: The mass and volume of the gas are unchanged, so the density is unchanged, so A error; according to $\mathrm { PV } / \mathrm { T } = C$, the volume of the gas is unchanged, the temperature is changed to 2 times, so the gas pressure is changed to 2 times, so B is correct; the temperature is the molecular thermal movement of the average kinetic energy of the sign, the molecular thermal movement of the average kinetic energy is directly proportional to the temperature, the temperature is changed to 2 times, so the molecular average kinetic energy is changed to 2 times, so C Wrong; the mass and volume of the gas are unchanged, so the number of molecules in the density is unchanged, but the average kinetic energy increases, so the number of molecules per second to hit the unit area of the wall increases, so D is wrong; Therefore, choose B.

Question 12: 14. A certain mass of ideal gas, the initial state is $p , V , T$, after a series of state changes, ...

14. A certain mass of ideal gas, the initial state is $p , V , T$, after a series of state changes, the pressure is still $p$, then the following process can be realized ( )

  • A. A. Isothermal expansion followed by isovolumetric cooling
  • B. B. Isothermal compression followed by isovolumetric warming
  • C. C. Isothermal warming, then isothermal compression
  • D. D. First isotropic cooling, then isotropic compression

Answer: D

Solution: A. According to the ideal gas equation of state formula $\frac { p V } { T } = C$, it is known that first isothermal expansion pressure decreases, and then isovolumetric cooling pressure decreases, and can not return to the initial value, so A error; B. Similarly, isothermal compression pressure increases, and then isovolumetric warming pressure becomes larger, can not return to the initial value, option B is wrong; C. First isotropic warming pressure increases, then isothermal compression pressure increases, can not return to the initial value, so C error; D. First isovolumetric cooling pressure decreases, and then isothermal compression pressure increases, can return to the initial value, D is correct.

Question 13: 15. As shown in the figure, the cylinder up and down the two sides of the gas by the adiabatic pisto...

15. As shown in the figure, the cylinder up and down the two sides of the gas by the adiabatic piston separation, the piston and the cylinder smooth contact - the initial piston and the two sides of the gas are in equilibrium, because the piston has a mass so the lower side of the gas pressure is the upper side of the gas pressure is twice as much as the upper side of the gas pressure, up and down the volume of the gas ratio $V _ { 1 } : V _ { 2 } = 1 : 2$, the temperature of the ratio ${ } ^ { T _ { 1 } } : T _ { 2 } = 2 :$ 5. FORMULA_1]] 5. Keep the temperature of the gas on the upper side unchanged, change the temperature of the gas on the lower side, so that the volume of the gas on both sides is the same, at this time the ratio of the temperature of the gas on both sides of the upper and lower ( ![](/images/questions/phys-ideal-gas-law/image-012.jpg)

  • A. A. $4 : 5$
  • B. B. 5: 9
  • C. C. 7: 24
  • D. D. $16 : 25$

Answer: D

Solution: Let $V _ { 1 } = V$ , from the question : $V _ { 1 } : V _ { 2 } = 1 : 2$ , then $V _ { 2 } = 2 V$ , let $T _ { 1 } = 2 T$ , it is known that $: T _ { 1 } : T _ { 2 } = 2 : 5$ , then , gas pressure: $P _ { 2 } = P _ { 1 } + \frac { m g } { S } = 2 P _ { 1 }$, then $: \frac { m g } { S } = P _ { 1 }$, the final volume of the two parts of the gas is equal, then: $V _ { 1 } { } ^ { \prime } = V _ { 2 } { } ^ { \prime } = \frac { 3 } { 2 } V$, the temperature of the gas in the upper part of the gas remains unchanged, by Boyle's law: $P _ { 1 } V _ { 1 } = P _ { 1 } { } ^ { \prime } V _ { 1 } { } ^ { \prime }$, solved: [[INLINE_FORMULA_1 9]], solve: $P _ { 1 } { } ^ { \prime } = \frac { 2 } { 3 } P _ { 1 }$, the pressure of the lower part of the gas: $P _ { 2 } ^ { \prime } = P _ { 1 } ^ { \prime } + \frac { m g } { S } = \frac { 5 } { 3 } P _ { 1 }$, for the lower part of the gas, by the ideal gas equation of state: $\frac { P _ { 2 } V _ { 2 } } { T _ { 2 } } = \frac { P _ { 2 } { } ^ { \prime } V _ { 2 } { } ^ { \prime } } { T _ { 2 } { } ^ { \prime } }$, solve $: T _ { 2 } { } ^ { \prime } = \frac { 25 } { 8 } T$, the upper and lower sides of the gas The ratio of the temperatures of the upper and lower gases is $\frac { T _ { 1 } { } ^ { \prime } } { T _ { 2 } { } ^ { \prime } } = \frac { 2 T } { \frac { 25 } { 8 } T } = \frac { 16 } { 25 }$, so D is correct and A, B and C are wrong;

Question 14: 16. As shown in the figure, a certain amount of ideal gas changes from the state $A$ to $C$ via $B$,...

16. As shown in the figure, a certain amount of ideal gas changes from the state $A$ to $C$ via $B$, and its pressure ( ) ![](/images/questions/phys-ideal-gas-law/image-013.jpg)

  • A. A. run
  • B. B. keep on decreasing
  • C. C. increase and then decrease
  • D. D. decreasing then increasing

Answer: A

Solution: According to the ideal gas equation of state $\frac { P V } { T } = C$, from $A$ to $B$, the volume remains unchanged, the temperature rises, and the pressure increases; from $B$ to $C$, the temperature remains unchanged, the volume decreases and the pressure increases. From $B$ to $C$, the temperature remains unchanged, the volume decreases and the pressure increases.

Question 15: 17. As shown in the figure for an ideal gas $p - T$ image, the following statements are correct () !...

17. As shown in the figure for an ideal gas $p - T$ image, the following statements are correct () ![](/images/questions/phys-ideal-gas-law/image-014.jpg)

  • A. A. From A to B, the gas does work on the outside and increases its internal energy.
  • B. B. From B to C, work is done on the gas and heat is released.
  • C. C. As we move from B to C, the temperature of the gas increases and the kinetic energy of all the gas molecules increases.
  • D. D. From A to B is isovolumetric, from B to C is isobaric, and the volume of the gas in the C state is greater than the volume in the B state.

Answer: D

Solution: A. The gas undergoes isovolumetric change from $A$ to $B$ state, the gas does not do any work on the outside, the temperature rises and the internal energy increases, so A is wrong; B. From $B$ to $C$ process, the volume of the gas increases, the gas external work, according to the first law of thermodynamics $$ \Delta U = Q + W $$ The temperature rises, the internal energy increases, the gas absorbs heat, so B is wrong; C. By $B$ to $C$ process, the gas temperature increases, the average kinetic energy of the gas molecules increase, but not every molecule of kinetic energy are increased, so C error; D. The gas from $B$ to $C$ occurs isobaric change, according to the Gelusak $$ \frac { V _ { B } } { T _ { B } } = \frac { V _ { C } } { T _ { C } } $$ As a result of $$ T _ { C } > T _ { B } $$ Therefore $$ V _ { C } > V _ { B } $$ Therefore D is correct.

Question 16: 18. Traditional Chinese medicine cupping therapy has a long history in China, as early as written in...

18. Traditional Chinese medicine cupping therapy has a long history in China, as early as written in the Western Han Dynasty in the silk book "fifty-two disease prescription". There are similar to the later fire cupping therapy. The method is to use a can as a tool, put a piece of lit paper into a small can, when the piece of paper burns out, quickly press the open end of the can tightly on the skin, the can will be tightly "sucked" in the skin, resulting in localized blood stasis, in order to achieve the effect of therapy to activate the meridians and collaterals, activate the blood circulation, reduce swelling and relieve pain, eliminate the wind and dispel the cold. In the first short period of time, the main reason why the fire pot "sucks" on the skin is that

  • A. A. The temperature of the gas inside the fire can remains constant, the volume decreases and the pressure increases
  • B. B. The pressure of the gas inside the fire can remains constant, the temperature decreases, and the volume decreases
  • C. C. The volume of gas in the fire can remains constant, the temperature decreases, and the pressure decreases
  • D. D. The volume of gas in the fire can remains constant, the temperature decreases and the pressure increases

Answer: C

Solution: In the beginning of a short period of time, the volume of the gas inside the fire pot is unchanged, because of the good thermal conductivity of the fire pot, so the temperature of the gas inside the fire pot decreases rapidly, according to $\frac { p V } { T } = C$ can be known, the gas pressure decreases, in the outside world of the atmospheric pressure under the action of the fire pot "sucked" in the skin, ABD wrong, C ABD is wrong, C is correct.

Question 17: 19 . The following statements are correct

19 . The following statements are correct

  • A. A. Knowing the molar volume of hydrogen and the volume of each hydrogen molecule at the standard conditions does not allow one to calculate Avogadro's constant
  • B. B. The irregular motion of pollen particle molecules suspended in a liquid is Brownian motion
  • C. C. Dust suspended in the air can be observed in sunlight in constant motion, the motion of this dust is Brownian motion
  • D. D. When you compress a gas, the pressure of the gas must increase.

Answer: A

Solution: A. Because gas molecules are not tightly packed, Avogadro's constant cannot be calculated by knowing the molar volume of the gas and the volume of each molecule, so A is correct; B. Brownian motion is the irregular motion of small particles suspended in liquids or gases, and the motion of pollen particle molecules belongs to molecular irregular thermal motion, not Brownian motion, so B is wrong; C. In the sunlight to see the dust constantly moving, is due to the air caused by the macro-motion, does not belong to the Brownian motion, so C error; D. By the ideal gas equation of state $$ \frac { p V } { T } = C $$ It can be seen that when the volume decreases, the change in pressure cannot be judged because the change in temperature is unknown, so D is wrong. D. D is wrong.

Question 18: 20. As shown in the figure is a gun reset device schematic, open the gun, the gun barrel recoil driv...

20. As shown in the figure is a gun reset device schematic, open the gun, the gun barrel recoil drive rod piston oil compressed air, the process of air and the outside world without heat transfer, recoil after the end of the compressed air to promote the piston so that the gun barrel reset, set the pressure of the closed air before the opening of the gun is $p _ { 1 }$, the thermodynamic temperature of , the volume is $V _ { 1 }$, the thermodynamic temperature of the air is ${ } ^ { T }$ and the volume is compressed to ${ } ^ { V 2 }$ by recoil of the gun barrel, then the pressure of the air after recoil is ![](/images/questions/phys-ideal-gas-law/image-015.jpg)

  • A. A. $\frac { p _ { 1 } T _ { 1 } } { V _ { 2 } }$
  • B. B. $\frac { p _ { 1 } T _ { 1 } } { V _ { 1 } }$
  • C. C. $\frac { p _ { 1 } V _ { 1 } T _ { 2 } } { V _ { 2 } T _ { 1 } }$
  • D. D. $\frac { p _ { 1 } V _ { 2 } T _ { 2 } } { V _ { 1 } T _ { 1 } }$

Answer: C

Solution: According to the ideal gas equation of state $$ \frac { p _ { 1 } V _ { 1 } } { T _ { 1 } } = \frac { p _ { 2 } V _ { 2 } } { T _ { 2 } } $$ the pressure of the air after recoil is solved as $$ p _ { 2 } = \frac { p _ { 1 } V _ { 1 } T _ { 2 } } { V _ { 2 } T _ { 1 } } $$

Question 19: 21. A certain mass of an ideal gas, the pressure becomes 2 times the original pressure at the same t...

21. A certain mass of an ideal gas, the pressure becomes 2 times the original pressure at the same temperature, the volume becomes 2 times the original volume.

  • A. A. 4 times
  • B. B. 2 times
  • C. C. $\frac { 1 } { 2 }$
  • D. D. $\frac { 1 } { 4 }$

Answer: C

Solution: According to the ideal gas equation of state we have $$ \frac { p V } { T } = C $$ At constant temperature the pressure becomes twice the original and the volume becomes $\frac { 1 } { 2 }$. Option C is correct and ABC is incorrect.

Question 20: 22. A test tube of sufficient length has its opening firmly downward and a certain mass of an ideal ...

22. A test tube of sufficient length has its opening firmly downward and a certain mass of an ideal gas enclosed in the center by mercury, as shown in the figure. The test tube is now turned slowly to the right around the fixed point to the dotted line, then the following image may be correct () ![](/images/questions/phys-ideal-gas-law/image-016.jpg)

  • A. A. ![](/images/questions/phys-ideal-gas-law/image-005.jpg)
  • B. B. ![](/images/questions/phys-ideal-gas-law/image-006.jpg)
  • C. C. ![](/images/questions/phys-ideal-gas-law/image-007.jpg)
  • D. D. ![](/images/questions/phys-ideal-gas-law/image-008.jpg)

Answer: D

Solution: Let the pressure of the gas in the tube be $p$, the volume $V$, the length of the column of mercury be $h$, and the angle of rotation be $\theta$. Then $$ p = p _ { 0 } - p _ { h } \cos \theta $$ When $\theta$ increases, $\cos \theta$ decreases, and the pressure of the closed gas increases, the volume decreases, and the temperature remains constant. A. According to the ideal gas equation of state $\frac { p V } { T } = C$, it can be obtained that $$ V = \frac { C } { p } \cdot T $$ A. According to the ideal gas equation of state $\frac { p V } { T } = C$, the $$ V = \frac { C } { p } \cdot T $$ pressure increases, and the slope of the $V - T$ image decreases; B. According to the ideal gas equation of state $\frac { p V } { T } = C$, it can be obtained that $$ p = \frac { C } { V } T $$ B. According to the ideal gas equation $$ p = \frac { C } { V } T $$, the volume of the gas decreases and the slope of the $p - T$ image increases, so B is wrong; C. According to the ideal gas equation of state $\frac { p V } { T } = C$, it can be obtained as follows $$ p = \frac { C } { V } T $$ is the ideal gas. C. According to the ideal gas equation of state $\frac { p V } { T } = C$, $$ p = \frac { C } { V } T $$ can be obtained; D. The isotherm of the $p - V$ image is a branch of the hyperbola, and the temperature is unchanged because the pressure of the closed gas increases, the volume decreases and the temperature remains unchanged, so D is correct.

Question 21: 24. Regarding ideal gases, the correct statement is ( )

24. Regarding ideal gases, the correct statement is ( )

  • A. A. Real gases can be treated as ideal gases only when the temperature is very low
  • B. B. A real gas can be treated as an ideal gas only if the pressure is very high
  • C. C. At room temperature and pressure, many real gases can be treated as ideal gases
  • D. D. All real gases can be treated as ideal gases in any case

Answer: C

Solution: As long as the pressure of the actual gas is not very high and the temperature is not very large, it can be approximated to be treated as an ideal gas. Ideal gas is an idealized model introduced in physics for the purpose of simplification, and it does not exist in real life; gases that strictly obey the gaseous equations under normal conditions are called ideal gases, and many actual gases can be treated as ideal gases at room temperature and pressure.

Question 22: 25. As shown in the figure, the cylinder with smooth inner wall is placed straight on the ground, th...

25. As shown in the figure, the cylinder with smooth inner wall is placed straight on the ground, the mass of the T-shaped piston is $M$, the area of the lower bottom is $S$, and the area of the upper bottom is $4 S$, and the atmospheric pressure is $p _ { 0 }$, then the pressure of the enclosed gas $p$ is equal to $p$. ], then the pressure of the enclosed gas $p$ is equal to ( ).

  • A. A. $4 p _ { 0 } + \frac { M g } { S }$
  • B. B. $3 p _ { 0 } + \frac { M g } { S }$
  • C. C. $p _ { 0 } + \frac { M g } { S }$
  • D. D. Insufficient conditions to judge

Answer: C

Solution: The piston is supported by the force of gravity, atmospheric pressure, and the enclosed gas, and is balanced by the $$ M g + p _ { 0 } S = p S $$ Solve for $$ p = p _ { 0 } + \frac { M g } { S } $$ Therefore, ABD is wrong and C is correct;

Question 23: 26. The P--T diagram of the change of state of an ideal gas of a certain mass is shown in the figure...

26. The P--T diagram of the change of state of an ideal gas of a certain mass is shown in the figure. If $\rho _ { a } , \rho _ { b } , \rho _ { c }$ and $\mathrm { Va } , \mathrm { Vb } , V c$ are used to represent the density and volume of the gas in the $a , b , c$ states respectively, then ([INLINE_FORMULA_2]) is used to represent the density and volume of the gas. The density and volume of a gas in three states of $a , b , c$ are then () ![](/images/questions/phys-ideal-gas-law/image-017.jpg)

  • A. A. $\rho _ { a } > \rho _ { b } > \rho _ { c } \quad V _ { a } > V _ { b } > V _ { c }$
  • B. B. $\rho _ { a } = \rho _ { b } < \rho _ { c } \quad V _ { a } = V _ { b } > V _ { c }$
  • C. C. $\rho _ { a } = \rho _ { b } > \rho _ { c } \quad V _ { a } < V _ { b } = V _ { c }$
  • D. D. $\rho _ { a } < \rho _ { b } < \rho _ { c } \quad V _ { a } < V _ { b } < V _ { c }$

Answer: B

Solution: According to the ideal gas equation of state: $\frac { P V } { T } = C$, it can be seen that from $a$ to $b$ the volume is unchanged i.e., $V _ { a } = V _ { b }$, according to From $\rho = \frac { M } { V }$ to $\rho _ { a } = \rho _ { b }$ the density is constant i.e. $\rho _ { a } = \rho _ { b }$ and from $b$ to c the temperature is constant and the pressure increases according to [INLINE_FORMULA_6]] and according to [INLINE_FORMULA_6]], the density is constant. Program $: \frac { P V } { T } = C$, it can be seen that the volume decreases, so there is $: V _ { c } < V _ { b }$, according to $\rho = \frac { M } { V }$ it can be seen that $\rho _ { c } > \rho _ { b }$, which gives : $\rho _ { c } > \rho _ { b } = \rho _ { a } , V _ { c } < V _ { b } = V _ { a }$, so B is correct and ACD is wrong.

Question 24: 27. The theory of molecular dynamics better explains the macroscopic thermodynamic properties of mat...

27. The theory of molecular dynamics better explains the macroscopic thermodynamic properties of matter. Accordingly, the following statements can be judged to be correct (

  • A. A. Brownian motion reflects the irregular motion of solid molecules
  • B. B. The potential energy of an ideal gas molecule may decrease or increase as the distance between molecules increases
  • C. C. The number of molecules in an object with a high rate of thermal motion as a proportion of the total number of molecules is temperature dependent
  • D. D. A drop of an alcoholic solution of oleic acid of volume $V$ forms an area of $S$ on the water surface, and the diameter of the molecules of the oil film is $\frac { V } { S }$

Answer: C

Solution: A. Brownian motion is the motion of small solid particles suspended in a liquid, Brownian motion shows that the liquid molecules keep doing irregular motion, A error; B. Ideal gas intermolecular force is zero, there is no molecular potential energy, B error; C. According to Maxwell's statistical law can be known, the object of thermal movement rate of the number of molecules in the total number of molecules proportion and temperature, C is correct; D. The volume is the volume of oleic acid alcohol solution, not the volume of oleic acid, then the diameter of the oleic acid molecule is not $\frac { V } { S }$, D error.

Question 25: 28. "Early in the morning to wear a coat afternoon to wear a yarn, holding the stove to eat watermel...

28. "Early in the morning to wear a coat afternoon to wear a yarn, holding the stove to eat watermelon", describing the Turpan region of Xinjiang, China, the natural phenomenon of large temperature differences between day and night, the use of this feature can be made of good quality raisins. Existing a grape drying room four wall openings, as shown in the figure, the room at night temperature $7 ^ { \circ } \mathrm { C }$, the temperature rises at noon $37 ^ { \circ } \mathrm { C }$, assuming that the atmospheric pressure at noon than the night to reduce $7 \%$, then at noon in the room escaped the air mass and the night of the room the ratio of the air mass ( ) ![](/images/questions/phys-ideal-gas-law/image-018.jpg)

  • A. A. $\frac { 4 } { 25 }$
  • B. B. $\frac { 4 } { 21 }$
  • C. C. $\frac { 3 } { 31 }$
  • D. D. $\frac { 3 } { 28 }$

Answer: A

Solution: Let the volume of the room be $V$ and the pressure at night be $p$, then at noon and at night $$ \frac { p V } { 273 + 7 } = \frac { 0.93 p ( V + \Delta V ) } { 273 + 37 } $$ The solution is $$ \Delta V = \frac { 4 V } { 21 } $$ Ratio of the mass of air escaping from the room at noon to the mass of air in the room at night $$ \frac { \Delta V } { \Delta V + V } = \frac { 4 } { 25 } $$

Question 26: 29. As shown in the figure, a thin under the thick, thick end and thin end of the glass tube are pan...

29. As shown in the figure, a thin under the thick, thick end and thin end of the glass tube are panned on the upper end of the opening, the lower end (thick end) closed, the upper end is long enough, the lower end (thick end) in the middle of a section of mercury closed a certain mass of ideal gas. Now the gas is slowly heated, the gas temperature continues to rise, the mercury column rises, then the volume of the gas is enclosed with the thermodynamic temperature of the relationship closest to the figure below ( )

  • A. A. ![](/images/questions/phys-ideal-gas-law/image-009.jpg)
  • B. B. ![](/images/questions/phys-ideal-gas-law/image-010.jpg)
  • C. C. ![](/images/questions/phys-ideal-gas-law/image-011.jpg)
  • D. D. ![](/images/questions/phys-ideal-gas-law/image-012.jpg)

Answer: A

Solution: According to $\frac { P V } { T } = C$: $V = \frac { C } { P } T$, the slope of the $V - T$ graph line is $\frac { C } { P }$, before the mercury column rises into the thin tube, the closed gas does isobaric pressure change first, the slope remains unchanged, and the graph line is a straight line; after part of the mercury column After the mercury column into the tube, the gas pressure increases, the slope decreases; when all the mercury column into the tube, the pressure of the gas and unchanged, $V - T$ line and straight line, just the slope is smaller than the original, so the A is correct, BCD is wrong.

Question 27: An ideal gas of a certain mass goes from the initial state $A$ through the states $B$, $C , D$ and b...

An ideal gas of a certain mass goes from the initial state $A$ through the states $B$, $C , D$ and back to the state $A$, where the volume $V$ is related to the temperature $T$ as shown in the figure. temperature $T$ is shown in the figure. The following statement is correct ( ) ![](/images/questions/phys-ideal-gas-law/image-019.jpg)

  • A. A. The internal energy of the gas decreases from state $A$ to state $B$.
  • B. B. During the transition from state $B$ to state $C$, the gas emits heat.
  • C. C. During the process from state $C$ to state $D$, the gas does work on the outside.
  • D. D. The gas pressure increases from state $D$ to state $A$.

Answer: B

Solution: A. From state $A$ to state $B$, the temperature of the gas increases, then the internal energy of the ideal gas increases, so A is wrong; B. From state $B$ to state $C$, the temperature is unchanged, the internal energy is unchanged, and from the figure, the volume decreases, and the outside world does work on the gas. $$ \Delta U = Q + W $$ By the first law of thermodynamics $$ \Delta U = Q + W $$, it can be seen that the gas exothermic, so B is correct; C. From state $C$ to state $D$, the volume decreases, and the outside world does work on the gas, so C is wrong; D. In the process from state $D$ to state $A$, the gas makes isobaric change, so D is wrong.

Question 28: 31. The figure for Galileo designed a temperature measurement device schematic, the upper end of the...

31. The figure for Galileo designed a temperature measurement device schematic, the upper end of the glass tube and good thermal conductivity of the glass bubble connected to the lower end of the inserted into the water, the glass bubble closed with a certain amount of air. If the water column rises in the glass tube, the change in the external atmosphere may be ( ) ![](/images/questions/phys-ideal-gas-law/image-020.jpg)

  • A. A. Temperature decreases, pressure increases
  • B. B. Temperature increases, pressure stays the same
  • C. C. Temperature increases, pressure decreases
  • D. D. Temperature remains constant, pressure decreases

Answer: A

Solution: Let the pressure of the gas in the glass bubble is p, the outside atmospheric pressure is $\mathrm { p } ^ { \prime }$, then $\mathrm { p } ^ { \prime } = \mathrm { p } + \mathrm { pgh }$, and the temperature of the gas in the glass bubble is the same as that of the outside atmosphere, the liquid column rises, and the volume of the gas decreases; by the equation of state of an ideal gas $\frac { p V } { T } = C$, it can be known that, when V decreases, then T may increase or remain the same. By the ideal gas equation of state $\frac { p V } { T } = C$, when V decreases, T may increase, decrease or remain unchanged if p increases, so A is correct; if p does not change, T decreases, so BD is wrong; if p decreases, T decreases, so C is wrong.

Question 29: 32 . The following statements are correct

32 . The following statements are correct

  • A. A. Heat cannot be transferred from a low temperature object to a high temperature object
  • B. B. Brownian motion of pollen suspended in water reflects the thermal motion of pollen molecules
  • C. C. In complete weightlessness, the pressure of the gas against the walls of the container is zero
  • D. D. Ideal gases don't actually exist, they're just an ideal model.

Answer: D

Solution: Heat may be transferred from a low-temperature object to a high-temperature object, so A is wrong; the Brownian motion of pollen suspended in water reflects the thermal motion of liquid molecules, so B is wrong; gas pressure is caused by the collision of molecules on the container, and the density of molecules and the average kinetic energy of molecules in thermal motion, so C is wrong; an ideal gas does not actually exist, but only an idealized physical model, so D is correct. Therefore, D is correct and ABC is wrong. $33 . \mathrm { B }$ [INLINE_FORMULA_6]] [Detailed Explanation] AB . The pressure, temperature, and volume of the gas in the cylinder at the beginning are $$ p _ { 1 } = p _ { 0 } + 50 \mathrm { cmHg } = 125 \mathrm { cmHg } \quad , T _ { 1 } = 300 \mathrm {~K} \quad , V _ { 1 } = 50 \mathrm {~S} $$ The pressure, volume of the gas when the mercury is exactly all the way into the upper cylinder are $$ p _ { 2 } = p _ { 0 } + 75 \mathrm { cmHg } = 150 \mathrm { cmHg } , \quad V _ { 2 } = 100 \mathrm {~S} $$ According to the ideal gas equation of state we have $$ \frac { p _ { 1 } V _ { 1 } } { T _ { 1 } } = \frac { p _ { 2 } V _ { 2 } } { T _ { 2 } } $$ Substituting the data to solve for $$ T _ { 2 } = 720 \mathrm {~K} $$ So A is wrong; B is correct; CD. When the temperature is 900 K, the gas in the cylinder changes isovolumically, then there are $$ \frac { p _ { 2 } } { T _ { 2 } } = \frac { p _ { 3 } } { T _ { 3 } } $$ The solution is $$ p _ { 3 } = 187.5 \mathrm { cmHg } $$ So CD is wrong ;

Question 30: 33. As shown in the figure, an adiabatic cylinder with an opening at the top is placed in a straight...

33. As shown in the figure, an adiabatic cylinder with an opening at the top is placed in a straight position, the upper part of the cylinder is 100 cm high, the lower part of the cylinder is 50 cm high, and the internal cross-sectional areas of the upper and lower parts are $S$ and ${ } ^ { 2 S }$, respectively, and the lower part of the cylinder is closed to a certain mass of gas with an adiabatic piston, and at the bottom of the cylinder, there is a heater ([INLINE_FORMULA_2]]) to heat the gas. volume and mass) can be heated on the gas, there is mercury above the piston, when the gas temperature is $27 ^ { \circ } \mathrm { C }$ when the lower mercury column is 25 cm high, the upper mercury column is 25 cm high, known as the atmospheric pressure is 75 cmHg, the thickness of the piston is not counted, then ![](/images/questions/phys-ideal-gas-law/image-021.jpg) electric heating wire

  • A. A. Heating the gas so that all of the mercury is in the upper cylinder when the temperature is 360 K.
  • B. B. Heating the gas so that all of the mercury enters the upper cylinder when the temperature is 720 K.
  • C. C. The pressure of the gas in the cylinder at 900 K is 180 cmHg.
  • D. D. The pressure of the gas in the cylinder at 900 K is 90 cmHg.

Answer: B

Solution:

Question 31: 34. As shown in the figure, straight placed uniform thickness U-shaped test tube, cross-sectional ar...

34. As shown in the figure, straight placed uniform thickness U-shaped test tube, cross-sectional area $S = 5 \mathrm {~cm} ^ { 2 }$, the left side of the tube length $l = 25 \mathrm {~cm}$ and closed a certain mass of the ideal gas, the right tube is long enough and the mouth of the tube opening, initially the left and right two tubes of the mercury level is equal to the height of the height $h = 10 \mathrm {~cm}$, known the ambient temperature $T _ { 0 } = 300 \mathrm {~K}$, the atmospheric pressure $p _ { 0 } = 75 \mathrm { cmHg }$. _2]], the ambient temperature $T _ { 0 } = 300 \mathrm {~K}$ and the atmospheric pressure $p _ { 0 } = 75 \mathrm { cmHg }$ are known. Now on the left side of the closed gas slowly heated until the two tubes of mercury surface formation of 10 cm height difference, and then keep this temperature unchanged under the same circumstances, from the mouth of the right tube slowly added mercury, so that the gas in the left tube to restore the initial volume, then add the volume of mercury is about ( ) (results are retained in three significant figures) ![](/images/questions/phys-ideal-gas-law/image-022.jpg)

  • A. A. $176 \mathrm {~cm} ^ { 3 }$
  • B. B. $192 \mathrm {~cm} ^ { 3 }$
  • C. C. $203 \mathrm {~cm} ^ { 3 }$
  • D. D. $212 \mathrm {~cm} ^ { 3 }$

Answer: B

Solution: Taking the initial state after warming, the pressure of the closed gas at this time is $$ p _ { 1 } = 75 \mathrm { cmHg } + 10 \mathrm { cmHg } = 85 \mathrm { cmHg } $$ Volume of the gas after warming $$ V _ { 1 } = S \left( l - \frac { h } { 2 } \right) $$ Let the height of mercury added be $\Delta h$ and the pressure of the enclosed gas at this time $$ p _ { 2 } = p _ { 0 } + \rho g \Delta h $$ During the addition of mercury, the closed gas undergoes an isothermal process with $$ p _ { 1 } V _ { 1 } = p _ { 2 } V _ { 0 } $$ Substituting the data to solve for $$ \Delta h \approx 38.3 \mathrm {~cm} $$ Volume of mercury added $$ V = \Delta h \cdot S \approx 192 \mathrm {~cm} ^ { 3 } $$

Question 32: 35. As shown in the figure, a certain mass of hydrogen (can be regarded as an ideal gas) from the st...

35. As shown in the figure, a certain mass of hydrogen (can be regarded as an ideal gas) from the state $A$ by the state $B$ to the state $C$, the $A$ to $B$, from $B$ to $C$ process outside the work done on the gas were $B$ to $C$, respectively. FORMULA_4]] and $B$ to $C$, the work done by the outside world on the gas is $W _ { 1 } , W _ { 2 }$, and the heat absorbed by the gas from the outside world is $W _ { 1 } , W _ { 2 }$ and $Q _ { 1 } , Q _ { 2 }$ respectively. $Q _ { 1 } , Q _ { 2 }$, then () ![](/images/questions/phys-ideal-gas-law/image-023.jpg)

  • A. A. $W _ { 1 } > 0 , W _ { 2 } > 0$
  • B. B. $Q _ { 1 } > 0 , Q _ { 2 } > 0$
  • C. C. $\left| W _ { 1 } \right| + \left| W _ { 2 } \right| = \left| Q _ { 1 } \right| + \left| Q _ { 2 } \right|$
  • D. D. $\left| W _ { 1 } \right| + \left| W _ { 2 } \right| > \left| Q _ { 1 } \right| + \left| Q _ { 2 } \right|$

Answer: B

Solution: FORMULA_5]] know, the volume of the gas is increasing, the gas to the outside world to do work, so $W _ { 1 } < 0 , W _ { 2 } <$ 0 , so A error; ![](/images/questions/phys-ideal-gas-law/image-024.jpg) B. In the $A B$ process, the temperature of the gas is increasing, while doing work to the outside world, to absorb heat, so $Q _ { 1 } > 0$ , in the $B C$ process, the temperature of the gas is unchanged, the internal energy is unchanged, but at the same time, the work done to the outside world, to absorb heat, so [$B$, so B is correct; CD. in the whole process, the temperature rises, the internal energy increases, according to the first law of thermodynamics $\triangle U = Q + W$ know, the heat absorbed by the gas is greater than the gas external work, that is, there are $$ \left| W _ { 1 } \right| + \left| W _ { 2 } \right| < \left| Q _ { 1 } \right| + \left| Q _ { 2 } \right| . $$ So CD is wrong.

Question 33: 36. The following statements are correct ()

36. The following statements are correct ()

  • A. A. Temperature is an indication of the average kinetic energy of a molecule
  • B. B. An increase in temperature increases the kinetic energy of every molecule in an object
  • C. C. For a given mass of an ideal gas, the higher the temperature and the higher the pressure, the lower the intensity.
  • D. D. The higher the temperature, the smaller the volume of an ideal gas of a given mass.

Answer: A

Solution: A. Temperature is a reflection of the intensity of molecular heat movement and is a sign of the average kinetic energy of molecules, so A is correct; B. object temperature increases, the object of the molecular average kinetic energy, but because the molecular movement is irregular, not every molecule's kinetic energy are increased, so B error; C. A certain mass of ideal gas, according to the gas equation $$ \frac { p V } { T } = c $$ According to the gas equation $$ \frac { p V } { T } = c $$, the higher the temperature, the pressure is not necessarily smaller, but also with the change of volume, so C is wrong; D. Similarly, according to the gas equation $$ \frac { p V } { T } = c $$ D. Similarly, according to the gas equation $$ \frac { p V } { T } = c $$, the higher the temperature, the volume is not necessarily smaller, but also with the change of pressure, so D error.

Question 34: 37. As shown in the figure, the adiabatic container with a piston closed to a certain mass of ideal ...

37. As shown in the figure, the adiabatic container with a piston closed to a certain mass of ideal gas (piston adiabatic), now compressed gas, the process ( ) ![](/images/questions/phys-ideal-gas-law/image-025.jpg)

  • A. A. The gas does work on the outside world and increases its internal energy
  • B. B. The outside world does work on the gas and the internal energy decreases
  • C. C. Gas temperature increases, pressure increases
  • D. D. Gas temperature decreases, pressure increases

Answer: C

Solution: AB . Compressing a gas, the outside world does work on the gas, in an adiabatic container, the internal energy increases and the temperature rises, option AB is wrong; CD. When the gas is compressed with a piston, the volume of the gas decreases; the outside world does work on the gas, the temperature rises, according to the equation of state of an ideal gas there are $$ \frac { p V } { T } = C $$ $V$ decreases, $T$ increases, so the pressure $p$ must increase. Therefore, the pressure $p$ must increase. Option C is correct and D is wrong.

Question 35: 38. As shown in the figure, one end of the opening, the other end of the closed glass tube with a ce...

38. As shown in the figure, one end of the opening, the other end of the closed glass tube with a certain mass of air sealed with a column of mercury, when the opening is placed down straight, the lower mercury surface is exactly level with the mouth of the tube. The glass tube is now turned around the opening by an angle of $30 ^ { \circ }$, and the enclosed gas () ![](/images/questions/phys-ideal-gas-law/image-026.jpg)

  • A. A. Pressure increases and the length of the air column shortens
  • B. B. Pressure increases and some of the mercury flows out of the tube
  • C. C. The pressure decreases and some of the mercury flows out of the tube
  • D. D. Pressure decreases and the length of the air column shortens

Answer: A

Solution: Let the atmospheric pressure be $p _ { 0 }$, the pressure of the enclosed gas be $p$, the length of the mercury column be $L$, and for the mercury column according to the condition of equilibrium it can be seen $p s + \rho g L s = p _ { 0 } s$ that when the glass tube is turned $30 ^ { \circ }$, the vertical pressure becomes smaller. INLINE_FORMULA_4]], along the direction of the glass tube $p s + \rho g L s \cos 30 ^ { \circ } < p _ { 0 } s$, the vertical downward pressure becomes smaller, then the atmospheric pressure pushes the mercury column upward, so the volume of the closed gas becomes smaller, after rebalancing there is $p ^ { \prime } s + \rho g L s \cos 30 ^ { \circ } = p _ { 0 } s$, it can be known [[INLINE_FORMULA_7 ]], then the pressure of the closed gas becomes larger; so A is correct, BCD is wrong.

Question 36: 39. As shown in the figure, $20 ^ { \circ } \mathrm { C }$ of oxygen and $10 ^ { \circ } \mathrm { C...

39. As shown in the figure, $20 ^ { \circ } \mathrm { C }$ of oxygen and $10 ^ { \circ } \mathrm { C }$ of hydrogen are of the same volume, and the column of mercury is in the center of a sufficiently long tube connecting the two containers, when the temperatures of the oxygen and the hydrogen are both increased by $10 ^ { \circ } \mathrm { C }$, the column of mercury ( ) ![](/images/questions/phys-ideal-gas-law/image-027.jpg)

  • A. A. immobile
  • B. B. pan
  • C. C. right-hand side
  • D. D. Move right then left.

Answer: B

Solution: According to the ideal gas equation of state $$ \frac { p V } { T } = C $$ Assuming a constant volume of gas, there is $$ \frac { p _ { 1 } } { T _ { 1 } } = \frac { p _ { 2 } } { T _ { 2 } } = \frac { \Delta p } { \Delta T } \Rightarrow \Delta p = \frac { p _ { 1 } } { T _ { 1 } } \Delta T $$ From the question, we know that at the beginning moment, the pressure is equal on both sides of the gas and that $$ T _ { \text {氧气 } 1 } > T _ { \text {氢气 } 1 } $$ It can be obtained that when both sides are raised to the same temperature, there are $$ \Delta p _ { \text {氧气 } } < \Delta p _ { \text {氢气 } } $$ Then the pressure of hydrogen on the right will be greater than the pressure of oxygen on the left, and the mercury column will move to the left, so choose B. $40 . \mathrm { C }$ [KNOWLEDGE POINT] null [Detailed Explanation]A. When the temperature of the gas in the cylinder is $12 ^ { \circ } \mathrm { C }$, the piston is analyzed, by the equilibrium condition have $$ p _ { 1 } S + m _ { 0 } g = p _ { 0 } S $$ Substitute the data to solve $$ p _ { 1 } = 9 \times 10 ^ { 4 } \mathrm {~Pa} $$ Therefore, A is correct; B. The spring is compressed when the length of the cylinder ${ } ^ { L _ { 2 } } = 22 \mathrm {~cm}$ is ${ } ^ { L _ { 2 } } = 22 \mathrm {~cm}$. [BLOCK_FORMULA_6]] Analyze the piston, according to the equilibrium conditions have $$ p _ { 0 } S + k \Delta x = m g + p _ { 2 } S $$ According to the gas equation of an ideal gas of a certain mass there is $$ \frac { p _ { 1 } L _ { 1 } S } { T _ { 1 } } = \frac { p _ { 2 } L _ { 2 } S } { T _ { 2 } } $$ where $$ T _ { 1 } = ( 273 + 12 ) \mathrm { K } = 285 \mathrm {~K} $$ Solving by association yields $$ T _ { 2 } = 383 \mathrm {~K} $$ Therefore, B is correct and inconsistent with the meaning of the question; CD. According to the meaning of the question can be known, when the cylinder pressure on the ground is zero, and then raise the temperature of the gas, the gas pressure remains unchanged, set the temperature of the gas at this time for ${ } ^ { T _ { 3 } }$, on the cylinder, by the equilibrium conditions $$ p _ { 0 } S + M g = p _ { 3 } S $$ For the piston, according to the equilibrium condition $$ p _ { 3 } S + m g = p _ { 0 } S + k \Delta x _ { 1 } $$ According to the gas equation for an ideal gas of a certain mass we have $$ \frac { p _ { 1 } L _ { 1 } S } { T _ { 1 } } = \frac { p _ { 3 } \left( L _ { 1 } + \Delta x _ { 1 } \right) S } { T _ { 3 } } $$ The solution by association is $$ T _ { 3 } = 437 \mathrm {~K} $$ That is, the temperature of the gas in the cylinder rises above 437 K, and the gas expands isobaric, so C is incorrect and D is incorrect; Therefore, C is incorrect and D is incorrect.

Question 37: 40. The cylinder is placed vertically on a horizontal floor, the mass of the cylinder $M = 4 \mathrm...

40. The cylinder is placed vertically on a horizontal floor, the mass of the cylinder $M = 4 \mathrm {~kg}$, the mass of the piston $m _ { 0 } = 2 \mathrm {~kg}$, the cross-sectional area of the piston $S = 2 \times 10 ^ { - 3 } \mathrm {~m} ^ { 2 }$. The cylinder above the piston encloses a certain mass of ideal gas, and the lower part is connected to the outside world through a gas port $P$. The atmospheric pressure is $p _ { 0 } = 1.0 \times 10 ^ { 5 } \mathrm {~Pa}$. The piston is connected to a light spring with a coefficient of strength $k = 2 \times 10 ^ { 3 } \mathrm {~N} / \mathrm { m }$ below the piston, and when the gas is released from the cylinder, the piston is connected to the spring. When the temperature of the gas in the cylinder is $12 ^ { \circ } \mathrm { C }$, the spring is at its natural length and the length of the gas column in the cylinder is $L _ { 1 } = 20 \mathrm {~cm}$. The $T = ( 273 + t ) \mathrm { K } , g$ is known. Taking $10 \mathrm {~m} / \mathrm { s } ^ { 2 }$, the piston does not leak, has no friction with the cylinder wall, and never reaches the $P$ hole. The closed gas is now heated, and the following statement is false ( ) ![](/images/questions/phys-ideal-gas-law/image-028.jpg) High School Physics Assignment, October 30, 2025

  • A. A. The gas pressure in the cylinder is $9 \times 10 ^ { 4 } \mathrm {~Pa}$ at cylinder column length $L _ { 1 } = 20 \mathrm {~cm}$.
  • B. B. At the cylinder column length ${ } ^ { L _ { 2 } } = 22 \mathrm {~cm}$, the cylinder gas temperature is approximately 383 K.
  • C. C. The temperature of the gas in the cylinder rises above 522 K, and the gas expands isobaric.
  • D. D. The temperature of the gas in the cylinder rises above 437 K, and the gas expands isobaric.

Answer: C

Solution:
Back to Topics

Ideal Gas Law

理想气体状态方程

37 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

The ideal gas equation of state is the central formula describing the relationship between pressure, volume, and temperature of an ideal gas, usually expressed as PV = nRT.In the CSCA Physics exam, this point is often examined in conjunction with thermodynamic processes, image transformations, and real-world situations (e.g., piston cylinders), and candidates are required to be flexible in their use of the equation to analyze the change in state of a gas.

Questions:37

Key Points

  • 1Understand the meaning and units of the physical quantities in the equation PV=nRT
  • 2Master isothermal, isobaric, isovolumic process in the gas state change rule
  • 3Ability to convert p-T, V-T and p-V images to each other for analysis
  • 4Analyze changes in the state of a gas in the context of mechanical equilibrium (e.g., force on a piston)

Study Tips

It is recommended that the analysis be aided by plotting the change of state and that care be taken to differentiate between thermodynamic and Celsius temperatures in the scenarios in which they are used.

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