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Thermodynamics - Practice Questions (28)

Question 1: Ideal gas in a closed container of fixed volume, when the temperature rises, then the increase in th...

Ideal gas in a closed container of fixed volume, when the temperature rises, then the increase in the

  • A. A. Distance between molecules
  • B. B. The average kinetic energy of a molecule
  • C. C. potential energy of a molecule
  • D. D. Rate of molecules

Answer: B

Solution: Knowledge points] molecular kinetic energy, molecular potential energy [Details] A. The volume of the gas remains constant, so the distance between molecules remains the same. Therefore, A error; B. Temperature is the average kinetic energy of molecules sign, the temperature of the object increases, the average kinetic energy of molecules increase. Therefore, B is correct; C. The distance between the molecules of a gas remains the same, so the kinetic energy of the molecules remains the same. Therefore, C is wrong; D. Temperature is a sign of the average kinetic energy of molecules, is a large number of molecular movement of statistical laws, temperature increases, the average kinetic energy of molecules increases, not every molecule's kinetic energy increases, so the rate of molecules does not necessarily increase. Therefore, D is wrong.

Question 2: To measure Avogadro's constant after the molecular diameter measured by the oil film method, all tha...

To measure Avogadro's constant after the molecular diameter measured by the oil film method, all that is needed is to know the oil droplet's

  • A. A. molar mass
  • B. B. molar volume
  • C. C. volumetric
  • D. D. intensity

Answer: B

Solution: Knowledge points] Avogadro's constant and calculation [Detailed Explanation] Calculate the molecular volume of oil, as long as you know the molar volume of oil can be calculated Avogadro's constant, know the molar volume of oil and density can be calculated its molar volume, Avogadro's constant $$ \mathrm { N } _ { \mathrm { A } } = \frac { V _ { m } } { V _ { 0 } } \cdot \left( V _ { m } \text { —molebodyvolume, } V _ { 0 } \text { —moleculeparticlebodyvolume, } V _ { 0 } = \frac { 1 } { 6 } \pi D ^ { 2 } \right) $$ Avogadro's constant $$ \mathrm { N } _ { \mathrm { A } } = \frac { V _ { m } } { V _ { 0 } } \cdot \left( V _ { m } \text { —molebodyvolume, } V _ { 0 } \text { —moleculeparticlebodyvolume, } V _ { 0 } = \frac { 1 } { 6 } \pi D ^ { 2 } \right) $$ is correct.

Question 3: When the temperature of an object is increased, the temperature in the object is increased by a fact...

When the temperature of an object is increased, the temperature in the object is increased by a factor of 0.5.

  • A. A. The distance between molecules must increase
  • B. B. Intermolecular forces must increase
  • C. C. The average kinetic energy of the molecules must increase
  • D. D. The kinetic energy of each molecule must increase

Answer: C

Solution: Knowledge Points] Molecular kinetic energy [Detailed explanation] temperature is the molecular thermal movement of the average kinetic energy of the sign, the temperature rises, the molecular thermal movement of the average kinetic energy must increase, so ABD error, C correct; choose C.

Question 4: Imagine arranging the molecules one by one. How many molecules would it take to fill a length of 0.1...

Imagine arranging the molecules one by one. How many molecules would it take to fill a length of 0.1 m?

  • A. A. $10 ^ { 8 }$
  • B. B. $10 ^ { 9 }$
  • C. C. $10 ^ { 10 }$
  • D. D. $10 ^ { 11 }$

Answer: B

Solution: Knowledge point] The size of molecules [Detailed Explanation]The order of magnitude of the size of molecules is $10 ^ { - 10 } \mathrm {~m}$,Let n molecules to line up the length of 0.1 m, then $n = \frac { 0.1 } { 10 ^ { - 10 } } = 10 ^ { 9 }$, so B is correct.

Question 5: When two pieces of metal with different temperatures come into contact and reach thermal equilibrium...

When two pieces of metal with different temperatures come into contact and reach thermal equilibrium, the following physical quantities must be identical

  • A. A. internal energy
  • B. B. average kinetic energy of a molecule
  • C. C. molecular potential energy
  • D. D. average molecular rate

Answer: B

Solution: Knowledge point] Understanding the concept of internal energy [Detailed explanation] internal energy from a microscopic point of view, is the molecular random motion of the statistical average of the sum of energy. In the absence of an external field in the case of molecular random motion of energy including the kinetic energy of molecules, the potential energy of intermolecular interactions and the energy of the internal movement of molecules. When two pieces of metal with different temperatures come into contact and reach thermal equilibrium, the temperatures must be the same, the average kinetic energies of the molecules must be the same, but not the other way around because the volumes are not necessarily the same and the molecules or atoms that make them up are not necessarily the same.

Question 6: As shown in the figure for a certain mass of ideal gas state change $p - \frac { 1 } { V }$ image, t...

As shown in the figure for a certain mass of ideal gas state change $p - \frac { 1 } { V }$ image, the process of change according to the arrow, according to the image can be determined by the gas temperature ![](/images/questions/phys-thermodynamics/f1c059d44c53.jpg)

  • A. A. First constant then rising
  • B. B. First constant, then lower
  • C. C. Reduced and then unchanged
  • D. D. Elevated and then unchanged

Answer: D

Solution: [Knowledge Points] null, isovolumetric changes in the image of the gas, isothermal changes in the image of the gas [Detailed Explanation]According to the image, it can be seen that in the first stage of the isovolumetric process, the pressure increases, and according to the ideal gas equation of state $\frac { P V } { T } = C$, it can be seen that the temperature of the gas increases. The slope of the $p - \frac { 1 } { V }$ image shows a constant related to the temperature, so the slope is constant during the second stage of the process, which means that this stage is isothermal. This means that the temperature of the gas first increases and then remains constant.

Question 7: As shown in the figure, an ideal gas of a certain mass undergoes a change of state as $a \rightarrow...

As shown in the figure, an ideal gas of a certain mass undergoes a change of state as $a \rightarrow b \rightarrow c \rightarrow a$, where the vertical coordinate represents the gas pressure $p$, and the horizontal coordinate represents the gas volume $V , a \rightarrow b$ is a hyperbola asymptotically connected to the $p$ axis and the $V$ axis. $p$ and $V$ axes as asymptotes. The following conclusions are correct ![](/images/questions/phys-thermodynamics/9764df241790.jpg)

  • A. A. State $a \rightarrow b$ ,the internal energy of the ideal gas decreases
  • B. B. State $b \rightarrow c$ ,fewer molecules collide per unit time per unit area of vessel wall
  • C. C. State $b \rightarrow c$ ,Positive external work on ideal gas
  • D. D. State $c \rightarrow a$ ,Ideal gas temperature decrease

Answer: C

Solution: [Knowledge Points] Ideal gas, the microscopic significance of the pressure of the gas A. Because $a \rightarrow b$ is a hyperbola with $p$ and $V$ axes as asymptotes, $a \rightarrow b$ is an isothermal process, and the temperature is constant, so the internal energy of the ideal gas is unchanged, so A is wrong; B. State $b \rightarrow c$, the pressure is constant, the volume becomes smaller, so the number of molecules colliding per unit time per unit area of the wall becomes more, so B is wrong; C. state $b \rightarrow c$, the pressure is unchanged, the volume becomes smaller, so the outside world does positive work on the ideal gas, so C is correct; D. state $c \rightarrow a$, the volume is unchanged, the pressure increases, the temperature of the ideal gas increases, so D is wrong;

Question 8: When an ideal gas of a certain mass undergoes a change of state, the state parameter $p , V , T$ can...

When an ideal gas of a certain mass undergoes a change of state, the state parameter $p , V , T$ cannot be changed in the following way (

  • A. A. $p , V , T$ are increased.
  • B. B. $p$ decreases, $V$ and $T$ increase.
  • C. C. $p$ and $V$ decrease, $T$ increase
  • D. D. $p$ and $T$ increase, $V$ decrease

Answer: C

Solution: [Knowledge Points] Ideal gas equation of state understanding and preliminary applications [Detailed Explanation]According to the ideal gas equation of state, there are $$ \frac { p V } { T } = C $$ Therefore, it is impossible for the pressure and volume to decrease while the temperature increases.

Question 9: The pressure of an ideal gas of a certain mass, the change in internal energy and the temperature of...

The pressure of an ideal gas of a certain mass, the change in internal energy and the temperature of the volume of the gas in relation to ()

  • A. A. If the volume is kept constant and the temperature is increased, the pressure of the gas increases and the internal energy increases.
  • B. B. If the volume is kept constant and the temperature is increased, the pressure of the gas increases and the internal energy decreases.
  • C. C. If its temperature is kept constant and its volume is increased, the pressure of the gas decreases and its internal energy increases.
  • D. D. If its temperature is kept constant and its volume is increased, the pressure of the gas decreases and its internal energy decreases.

Answer: A

Solution: [Knowledge Points] Application of the First Law of Thermodynamics,null [Detailed explanation] (1) If the volume remains unchanged and the temperature rises, the pressure of the gas increases, as can be seen from $\frac { p V } { T } = c$. The internal energy of an ideal gas of a certain mass is only related to the temperature, and when the temperature increases, the internal energy increases, so A is correct and B is wrong; (2) If the temperature remains unchanged and the volume increases, then $\frac { p V } { T } = c$ can be known that the pressure of the gas decreases, and the internal energy of a certain mass of an ideal gas is only related to the temperature, and if the temperature remains unchanged, the internal energy remains unchanged, so the CD is wrong; therefore, the correct answer to the question is A.

Question 10: As shown in the figure, A, B two containers of the same volume, with a long straight conduit connect...

As shown in the figure, A, B two containers of the same volume, with a long straight conduit connected, both sealed into the pressure of P, the temperature is T of a certain mass of ideal gas, now that the gas temperature in the A temperature rise to $T ^ { \prime }$, the gas temperature in the B to remain unchanged, set the stable After stabilization, the pressure in container A is $P ^ { \prime }$,then ![](/images/questions/phys-thermodynamics/fa8ed5dba1f7.jpg)

  • A. A. $P ^ { \prime } = \sqrt { \frac { T ^ { \prime } } { T } } P$
  • B. B. $P ^ { \prime } = \frac { 2 T ^ { \prime } } { T + T ^ { \prime } } P$
  • C. C. $P ^ { \prime } = P$
  • D. D. $P ^ { \prime } = \sqrt { \frac { T + T } { 2 T } } P$

Answer: B

Solution: [Knowledge Point]null [Detailed Explanation]Let the volume of $A , B$ be $V$ respectively, the pressure of the initial state is $P$, the temperature is $T$, and the temperature of the $A$ part of the final state is $T ^ { \prime }$, after stabilization $A$ the vessel is $A$. Warming $T ^ { \prime }$ , after stabilization $A$ the pressure of the vessel is ${ } _ { P ^ { \prime } }$ , then the pressure of the $B$ is also ${ } _ { P ^ { \prime } }$ , by the ideal gas equation of state: $\frac { P V } { T } = C$: $2 \frac { P V } { T } = \frac { P ^ { \prime } V } { T ^ { \prime } } + \frac { P ^ { \prime } V } { T }$ ,Solve: $P ^ { \prime } = \frac { 2 T ^ { \prime } } { T + T ^ { \prime } } P$ ,so B is correct,ACD is wrong.

Question 11: The ideal gas constant $R$ is a constant that characterizes the properties of an ideal gas, which ca...

The ideal gas constant $R$ is a constant that characterizes the properties of an ideal gas, which can be deduced from the ideal gas equation of state $R = \frac { p V _ { \mathrm { m } } } { T }$, where $p$ is the pressure of the gas, $V _ { \mathrm { m } }$ is the molar volume of the gas (the volume of one mole of a substance in $0 ^ { \circ } \mathrm { C } , 1 \mathrm {~atm}$), $T$ is the thermodynamic temperature. thermodynamic temperature, the ideal gas constant $R$ is expressed in the basic units of the International System of Units.

  • A. A. J/mol ⋅ K
  • B. B. $\mathrm { kg } \cdot \mathrm { m } ^ { 2 } / \left( \mathrm { s } ^ { 2 } \cdot \mathrm {~K} \right)$
  • C. C. $\mathrm { kg } \cdot \mathrm { m } ^ { 2 } / \left( \mathrm { s } ^ { 2 } \cdot \mathrm {~mol} \cdot \mathrm {~K} \right)$
  • D. D. $\mathrm { kg } \cdot \mathrm { m } ^ { 3 } / \left( \mathrm { s } ^ { 2 } \cdot \mathrm {~mol} \cdot \mathrm {~K} \right)$

Answer: C

Solution: [Knowledge Point]Derivation of units [Detailed Explanation]From $$ p = \frac { F } { S } , F = m a $$ The unit of pressure can be expressed as $\mathrm { kg } / \mathrm { m } \cdot \mathrm { s } ^ { 2 }$ . The unit of molar volume ${ } ^ { V _ { \mathrm { m } } }$ is $\mathrm { m } ^ { 3 } / \mathrm { mol }$ , $T$ is K . Then the ideal gas constant $R$ has units of $\mathrm { kg } \cdot \mathrm { m } ^ { 2 } / \left( \mathrm { s } ^ { 2 } \cdot \mathrm {~mol} \cdot \mathrm {~K} \right)$ .

Question 12: An ideal gas of a certain mass undergoes a $a \rightarrow b \rightarrow c \rightarrow d$ change duri...

An ideal gas of a certain mass undergoes a $a \rightarrow b \rightarrow c \rightarrow d$ change during an experiment as shown in the figure. The following statement is correct. ![](/images/questions/phys-thermodynamics/8d4c5e50718b.jpg)

  • A. A. The process from $a$ to $b$ is isothermal.
  • B. B. The process from $b$ to $c$ is an isobaric change.
  • C. C. During the process from $c$ to $d$, the gas does external work.
  • D. D. During the whole process, the point $c$ has the highest temperature.

Answer: C

Solution: [Knowledge Points] Application of the first law of thermodynamics, understanding and preliminary application of the ideal gas equation of state A. According to the ideal gas equation of state, $a$ to $b$ process, due to the product of pressure and volume increases, the gas temperature increases, so A error; B. According to the image, $b$ to $c$ process, the volume of the gas is unchanged, the pressure of the gas increases, isovolumetric change, so B error; C. According to the image, $c$ to $d$ process, the volume of the gas increases, then the gas external work, so C is correct; D. According to the ideal gas equation of state, in the whole process, since the $d$ point has the largest product of pressure and volume, then in the whole process, the $d$ point has the highest temperature, so D is wrong.

Question 13: A certain mass of an ideal gas undergoes a series of changes of state, during which the internal ene...

A certain mass of an ideal gas undergoes a series of changes of state, during which the internal energy of the gas increases by 135 J, and the outside world does 90 J of work on the gas. In this process, the heat exchanged between the gas and the outside world is

  • A. A. The gas gives off heat to the outside world 45J
  • B. B. Heat absorbed by a gas from the outside world 45 J
  • C. C. The gas gives off 225 J of heat to the outside world.
  • D. D. A gas absorbs heat from the outside world 225J

Answer: B

Solution: [Knowledge Point]null [Detailed Explanation]According to the first law of thermodynamics $$ \Delta U = W + Q $$ where $$ \begin{aligned} \Delta U & = 135 \mathrm {~J} \\ W & = 90 \mathrm {~J} \end{aligned} $$ has to be $$ Q = 45 \mathrm {~J} $$ i.e., the gas absorbs heat from the outside world 45J.

Question 14: An ideal gas of a certain mass changes from state $A$ to state $B , p - V$ as shown in the image. Th...

An ideal gas of a certain mass changes from state $A$ to state $B , p - V$ as shown in the image. The gas changes from state [[INLINE_FORMULA_0 ![](/images/questions/phys-thermodynamics/43a2c1a393c8.jpg)

  • A. A. constant temperature
  • B. B. internal energy reduction
  • C. C. Negative work done on the outside
  • D. D. draw heat from outside

Answer: D

Solution: [Knowledge Points] Application of the first law of thermodynamics, isobaric pressure changes in the image of gas [Detailed Explanation]AB . From the figure, the gas from the state $A$ to the state $B$ in the process, the pressure is unchanged, according to the $\frac { V } { T } = C$ can be understood, the temperature increases, the internal energy increases, so AB error; C. Because the volume of the gas increases, the gas does positive work on the outside, so C is wrong; D. According to the first law of thermodynamics $\Delta U = W + Q$, where $$ W < 0 $$ $$ \Delta U > 0 $$ So $$ Q > 0 $$ That is, the gas absorbs heat from the outside world, so D is correct.

Question 15: In physics, some physical quantities are related to the state of an object, while others are related...

In physics, some physical quantities are related to the state of an object, while others are related to the physical process, the following physical quantities are related to the state ( )

  • A. A. potential energy
  • B. B. average speed
  • C. C. accomplishment
  • D. D. calorimetric

Answer: A

Solution: Knowledge Points] Understand the formulation and expression of the first law of thermodynamics, the definition and properties of gravitational potential energy, the definition of work (formula), rate and speed. [Detailed explanation] average speed, work and heat are process quantities, corresponding to a process; and potential energy is a state quantity, corresponding to a moment, related to the state.

Question 16: Compress the air in the cylinder with the piston, 900J of work done on the air, while the cylinder d...

Compress the air in the cylinder with the piston, 900J of work done on the air, while the cylinder dissipates 210J of heat to the outside, the internal energy of the air in the cylinder changed ().

  • A. A. 900J
  • B. B. 210J
  • C. C. 690J
  • D. D. 1110J

Answer: C

Solution: [Knowledge Point]null [Detailed Explanation]From the first law of thermodynamics $\Delta U = W + Q$ we get $$ \Delta U = 900 \mathrm {~J} - 210 \mathrm {~J} = 690 \mathrm {~J} $$

Question 17: A certain mass of ideal gas, from the outside of the heat absorbed 500 J, at the same time the exter...

A certain mass of ideal gas, from the outside of the heat absorbed 500 J, at the same time the external work done 100 J, then the gas ()

  • 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: [Knowledge Points] Application of the first law of thermodynamics,null [Detailed explanation] by the first law of thermodynamics $Q = W + \Delta E$ ,substituting the data can be obtained $$ \Delta E = 500 - 100 = 400 \mathrm {~J} > 0 $$ It can be seen that the process, the internal energy increases, and because the ideal gas internal energy is a single-valued function of temperature, so the temperature increases. Work is done on the outside, the volume increases, and according to the ideal gas equation of state can be obtained, in the case of volume and temperature are increased, how the pressure changes can not be judged.

Question 18: A certain mass of an ideal gas, in a state change process, the gas to the outside world to do work 4...

A certain mass of an ideal gas, in a state change process, the gas to the outside world to do work 4 J, the gas internal energy decreased by 12 J, then in the process ( )

  • A. A. Gas exothermic 8J
  • B. B. Gas heat absorption 8 J
  • C. C. Gas heat absorption 16 J
  • D. D. Gas exothermic 16 J

Answer: A

Solution: [Knowledge Point]null [Detailed Explanation]According to the first law of thermodynamics, it is known that $$ \Delta U = Q + W $$ So $$ Q = \Delta U - W = - 12 - ( - 4 ) = - 8 \mathrm {~J} $$ A is correct .

Question 19: The initial state of a system has an internal energy of 1 J, in the outside world to do 0.5 J of wor...

The initial state of a system has an internal energy of 1 J, in the outside world to do 0.5 J of work on it, it gives off 0.2 J of heat, the system in the process of the increment of internal energy is ()

  • A. A. 0.7 J
  • B. B. 0.3 J
  • C. C. 1.3 J
  • D. D. - 0.3 J

Answer: B

Solution: [Knowledge Points] null [Detailed Explanation]According to the first law of thermodynamics: $\Delta \mathrm { U } = \mathrm { W } + \mathrm { Q }$ , because the outside world on it to do 0.5 J of work, it puts out 0.2 J of heat, then $\Delta \mathrm { U } = 0.5 \mathrm {~J} - 0.2 \mathrm {~J} = 0.3 \mathrm {~J}$ , $\Delta \mathrm { U } = 0.5 \mathrm {~J} - 0.2 \mathrm {~J} = 0.3 \mathrm {~J}$ , $\Delta \mathrm { U } = 0.5 \mathrm {~J} - 0.2 \mathrm {~J} = 0.3 \mathrm {~J}$ , $\Delta \mathrm { U } = 0.5 \mathrm {~J} - 0.2 \mathrm {~J} = 0.3 \mathrm {~J}$ , $\Delta \mathrm { U } = 0.5 \mathrm {~J} - 0.2 \mathrm {~J} = 0.3 \mathrm {~J}$

Question 20: A gas of a certain mass expands from 20 L to 30 L at a constant pressure equal to $1.0 \times 10 ^ {...

A gas of a certain mass expands from 20 L to 30 L at a constant pressure equal to $1.0 \times 10 ^ { 5 } \mathrm {~Pa}$. During this process the gas absorbs a total of $4 \times 10 ^ { 3 } \mathrm {~J}$ heat from the outside world, the change in internal energy of the gas is ( ).

  • A. A. Added $5 \times 10 ^ { 3 } \mathrm {~J}$
  • B. B. Reduced $5 \times 10 ^ { 3 } \mathrm {~J}$
  • C. C. Added $3 \times 10 ^ { 3 } \mathrm {~J}$
  • D. D. Reduced $3 \times 10 ^ { 3 } \mathrm {~J}$

Answer: C

Solution: [Knowledge Points] null [Detailed explanation] by the first law of thermodynamics, the increase in internal energy of an object is equal to the work done by the outside world on the object and the object from the outside world to absorb the heat of the sum, and the object's internal energy is equal to the work done by the outside world on the object. That is $$ \Delta U = W + Q $$ then $$ W = P \cdot \Delta V = 1.0 \times 10 ^ { 5 } \mathrm { pa } \times 10 \times 10 ^ { - 3 } \mathrm {~m} ^ { 3 } = 1.0 \times 10 ^ { 3 } \mathrm {~J} , Q = 4 \times 10 ^ { 3 } \mathrm {~J} , \Delta U = 5 \times 10 ^ { 3 } \mathrm {~J} $$

Question 21: An object absorbs 58 J of heat from the outside world and does 42 J of work to the outside world.

An object absorbs 58 J of heat from the outside world and does 42 J of work to the outside world.

  • A. A. Increased by 16 J
  • B. B. Reduced by 16 J
  • C. C. Increased by 100 J
  • D. D. Reduced by 100 J

Answer: A

Solution: [Knowledge Point]null [Detailed Explanation]According to the first law of thermodynamics can be obtained $$ \Delta U = W + Q = - 42 + 58 \mathbf { J } = 16 \mathbf { J } $$ Therefore, the internal energy increased by 16 J, so A correct, BCD error.

Question 22: The gas in the cylinder absorbs $4.2 \times 10 ^ { 3 } \mathrm {~J}$ of heat from the outside world ...

The gas in the cylinder absorbs $4.2 \times 10 ^ { 3 } \mathrm {~J}$ of heat from the outside world while the gas pushes the piston to do $2.2 \times 10 ^ { 3 } \mathrm {~J}$ of work on the outside world. Then the internal energy of the gas (

  • A. A. Added $2.0 \times 10 ^ { 3 } \mathrm {~J}$
  • B. B. Reduced $2.0 \times 10 ^ { 3 } \mathrm {~J}$
  • C. C. Added $6.4 \times 10 ^ { 3 } \mathrm {~J}$
  • D. D. Reduced $6.4 \times 10 ^ { 3 } \mathrm {~J}$

Answer: A

Solution: Knowledge point] null [Detailed explanation] for a certain mass of gas heating, the gas absorbed 4200 J of heat, so the $$ Q = 4200 \mathrm {~J} $$ At the same time, the work done on the outside is 2200 J, therefore $$ W = - 2200 \mathrm {~J} $$ According to the first law of thermodynamics we have $$ \Delta U = W + Q = 4200 \mathrm {~J} - 2200 \mathrm {~J} = 2000 \mathrm {~J} $$ That is, the internal energy has increased by $2.0 \times 10 ^ { 3 } \mathrm {~J}$, so A is correct, and BCD is wrong.

Question 23: The intermolecular force $F$ varies with the intermolecular distance $r$ as shown in the figure, the...

The intermolecular force $F$ varies with the intermolecular distance $r$ as shown in the figure, then the molecular potential energy is related to the magnitude of (). ) ![](/images/questions/phys-thermodynamics/06b3a11a128b.jpg)

  • A. A. $E _ { 1 } < E _ { 3 } < E _ { 2 }$
  • B. B. $E _ { 1 } < E _ { 2 } < E _ { 3 }$
  • C. C. $E _ { 3 } < E _ { 2 } < E _ { 1 }$
  • D. D. $E _ { 2 } < E _ { 3 } < E _ { 1 }$

Answer: B

Solution: Knowledge point] Molecular potential energy [Detailed Explanation]According to the figure, ${ } ^ { r = r _ { 1 } }$, the molecules are in equilibrium, ${ } ^ { r > r _ { 1 } }$, the molecular force is gravitational, with the fraction of the molecular potential energy, the molecular potential energy of the molecular force is the gravitational force. As the distance between the molecules increases, the molecular force does negative work and the molecular potential energy increases, i.e. $E _ { 1 } < E _ { 2 } < E _ { 3 }$

Question 24: A gas bubble slowly floats from the bottom of a thermostatic tank, the gas inside the bubble is rega...

A gas bubble slowly floats from the bottom of a thermostatic tank, the gas inside the bubble is regarded as an ideal gas, and the number of gas molecules is constant, and the external atmospheric pressure is constant. In the process of floating the gas inside the bubble

  • A. A. Absorption of heat from water
  • B. B. constant volume
  • C. C. pressure is constant
  • D. D. The average kinetic energy of the molecules becomes larger

Answer: A

Solution: [Knowledge Points] Molecular kinetic energy, understanding and preliminary application of the equation of state of an ideal gas, application of the first law of thermodynamics [Detailed Explanation]C.Pressure inside a bubble $p = p _ { 0 } + \rho _ { \text {water } } g h$ $h _ { \text {decreasesmall } }$ , $p _ { \text {decreasesmall } }$ , so C is wrong; B. When the bubble floats, the temperature remains the same, according to Boyle's law $p V = C$, the pressure $p$ decreases, and the volume $V$ increases, so B is wrong; A. When a gas bubble floats, the temperature remains the same and the internal energy remains the same. A. When the bubble floats, the temperature remains the same and the internal energy remains the same. When the volume increases, the gas does work on the outside ($\mathrm { W } < 0$), and the thermodynamic principle is the same. The first law of thermodynamics $\Delta U = Q + W$ is that a bubble can float at a constant temperature. By the first law of thermodynamics $\Delta U = Q + W$, $Q > 0$ is $Q > 0$, i.e., heat is absorbed; D. The average kinetic energy of the molecules remains the same at the same temperature, so D is wrong.

Question 25: A drop of water from a chopstick has a volume of about $0.1 ^ { c m ^ { 3 } }$ . Avogadro's constant...

A drop of water from a chopstick has a volume of about $0.1 ^ { c m ^ { 3 } }$ . Avogadro's constant is known to be $N _ { \mathrm { A } } = 6 \times 10 ^ { 23 } \mathrm {~mol} ^ { - 1 }$ . The molar volume of water is $V _ { \mathrm { mol } } = 18 \mathrm {~cm} ^ { 3 } / \mathrm { mol }$ , then the number of water molecules in this drop of water is approximately

  • A. A. $6 \times 10 ^ { 2 }$
  • B. B. $3 \times 10 ^ { 17 }$
  • C. C. $6 \times 10 ^ { 19 }$
  • D. D. $3 \times 10 ^ { 21 }$

Answer: D

Solution: [Knowledge Point] Avogadro's constant and calculation [Details] The amount of substance in this drop of water is $n = \frac { V } { V _ { \mathrm { mol } } } = \frac { 0.1 \mathrm {~cm} ^ { 3 } } { 18 \mathrm {~cm} ^ { 3 } / \mathrm { mol } } = \frac { 1 } { 180 } \mathrm {~mol}$ The number of molecules is $N = n N _ { \mathrm { A } } = \frac { 1 } { 180 } \times 6 \times 10 ^ { 23 }$ $= 3.3 \times 10 ^ { 21 }$

Question 26: The $p - t$ image of an ideal gas of a certain mass is shown in the figure, during the change of sta...

The $p - t$ image of an ideal gas of a certain mass is shown in the figure, during the change of state $A$ to state $B$, the volume ( ) ![](/images/questions/phys-thermodynamics/342bac656352.jpg)

  • A. A. invariant (math.)
  • B. B. must decrease
  • C. C. must increase
  • D. D. possibly unchanged

Answer: D

Solution: [Knowledge Points] The graph of isovolumetric changes of gases, the understanding and preliminary application of the equation of state of ideal gases [Detailed Explanation]A certain mass of ideal gas has $\frac { p V } { T } = n R$ can be obtained $p V = n R ( t + 273 )$ The collation has $p = \frac { n R } { V } t + \frac { 273 n R } { V }$ If the change of state $A$ to state $B$ is isotropic, then $A B$ is a straight line, and the inverse prolongation intersects with the time axis at the intersection of $- 273 ^ { \circ } \mathrm { C }$, but there is no specific data in the graph, so it is impossible to determine the volume change.

Question 27: The temperature of the gas (which can be regarded as an ideal gas) in a vegetable greenhouse in the ...

The temperature of the gas (which can be regarded as an ideal gas) in a vegetable greenhouse in the morning is $7 ^ { \circ } \mathrm { C }$ , and the temperature rises at noon due to sunlight $18 ^ { \circ } \mathrm { C }$ . If the atmospheric pressure remains constant, the mass of the gas in the greenhouse at noon is $7 ^ { \circ } \mathrm { C }$ , and the temperature of the gas in the greenhouse at noon is $18 ^ { \circ } \mathrm { C }$ .

  • A. A. $\frac { 280 } { 298 }$ times
  • B. B. $\frac { 298 } { 280 }$ times
  • C. C. $\frac { 7 } { 18 }$ times
  • D. D. $\frac { 7 } { 25 }$ times

Answer: A

Solution: [Knowledge Points] null [Detailed Explanation]Atmospheric pressure is unchanged, according to the ideal gas equation of state can be $\frac { p V } { T } = n R$ The ratio of the mass of the gas in the shed at noon to the mass of the gas in the shed in the morning can be obtained as follows $\frac { m _ { 2 } } { m _ { 1 } } = \frac { n _ { 2 } } { n _ { 1 } } = \frac { T _ { 1 } } { T _ { 2 } } = \frac { ( 273 + 7 ) \mathrm { K } } { ( 273 + 7 + 18 ) \mathrm { K } } = \frac { 280 } { 298 }$ The mass of gas in the greenhouse at noon is $\frac { 280 } { 298 }$ times the mass of gas in the greenhouse in the morning.

Question 28: A certain mass of ideal gas, experienced as shown in the figure $1 \rightarrow 2 \rightarrow 3$ stat...

A certain mass of ideal gas, experienced as shown in the figure $1 \rightarrow 2 \rightarrow 3$ state change process, then the thermodynamic temperature of the three states is the ratio ( ) ![](/images/questions/phys-thermodynamics/797a1cb54446.jpg)

  • A. A. $1 : 3 : 5$
  • B. B. 3:2:1
  • C. C. $3 : 6 : 5$
  • D. D. 5:6:3

Answer: C

Solution: [Knowledge Point]null [Detailed Explanation]According to the ideal gas equation of state $\frac { p V } { T } = C$ to organize there are $$ T \propto p V $$ So the ratio of the temperatures of the three states 1,2 and 3 is $$ T _ { 1 } : T _ { 2 } : T _ { 3 } = p _ { 1 } V _ { 1 } : p _ { 2 } V _ { 2 } : p _ { 3 } V _ { 3 } = 3 : 6 : 5 $$
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