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AP Physics 2- 9.3 Thermal Energy Transfer and Equilibrium- Exam Style questions - FRQs- New Syllabus

Thermal Energy Transfer and Equilibrium AP  Physics 2 FRQ

Unit 9: Thermodynamics

Weightage : 15–18%

AP Physics 2 Exam Style Questions – All Topics

Question

A sample of a monatomic ideal gas is sealed in a thermally conducting container by a movable piston of mass \( M \) and area \( A \). The container is in a large water bath that is held at a constant temperature \( T_0 \). The piston is free to move with negligible friction. At the instant shown, the gas is in thermal equilibrium with the water bath, the piston is at rest, and the gas occupies volume \( V_0 \). The pressure of the air above the piston is \( P_{\mathrm{atm}} \).
A. On the dot shown, representing the piston, draw and label the forces that are exerted on the piston. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
 
B. Derive an expression for the internal energy of the gas in terms of \( M \), \( A \), \( V_0 \), \( P_{\mathrm{atm}} \), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
C. A block, also of mass \( M \), is placed on the piston at time \( t=t_0 \) and is slowly lowered. The piston comes to rest at time \( t=t_f \) when the block is completely released. On the axes provided, sketch the expected relationship between the pressure \( P \) and volume \( V \) of the gas for the thermodynamic process that the gas undergoes during time interval \( t_0 \le t \le t_f \). Draw an arrow on your sketch to represent the direction of the thermodynamic process.
D. With the block still on the piston, the temperature of the water bath is changed to a new constant temperature \( T_{\mathrm{new}} \). The gas occupies the original volume \( V_0 \) when the sample of gas and the water bath come to thermal equilibrium.
Indicate whether \( T_{\mathrm{new}} \) is greater than \( T_0 \), less than \( T_0 \), or equal to \( T_0 \).
_____ \( T_{\mathrm{new}} > T_0 \)
_____ \( T_{\mathrm{new}} < T_0 \)
_____ \( T_{\mathrm{new}} = T_0 \)
Briefly justify your answer by referencing at least one feature of your answers to parts \( \mathrm{A} \), \( \mathrm{B} \), or \( \mathrm{C} \).

Most-appropriate topic codes (AP Physics 2):

• Topic \( 9.2 \) — The Ideal Gas Law (Part \( \mathrm{B} \), Part \( \mathrm{D} \))
• Topic \( 9.3 \) — Thermal Energy Transfer and Equilibrium (Part \( \mathrm{A} \), Part \( \mathrm{C} \), Part \( \mathrm{D} \))
• Topic \( 9.4 \) — The First Law of Thermodynamics (Part \( \mathrm{B} \), Part \( \mathrm{C} \), Part \( \mathrm{D} \))
▶️ Answer/Explanation

A.
The piston has three forces acting on it:

\( \bullet \) Upward force exerted by the gas, \( F_{\mathrm{gas}} \)
\( \bullet \) Downward weight of the piston, \( F_g = Mg \)
\( \bullet \) Downward force from the atmosphere, \( F_{\mathrm{atm}} = P_{\mathrm{atm}}A \)

A correct force diagram is shown below.

Because the piston is at rest, these forces balance.

B.
For a monatomic ideal gas,

\( U = \dfrac{3}{2}nRT \)

and from the ideal gas law,

\( PV = nRT \)

so

\( U = \dfrac{3}{2}PV \)

Now find the gas pressure using force balance on the piston:

Upward force from the gas: \( PA \)

Downward forces: \( P_{\mathrm{atm}}A + Mg \)

Since the piston is at rest,

\( PA – P_{\mathrm{atm}}A – Mg = 0 \)

\( P = P_{\mathrm{atm}} + \dfrac{Mg}{A} \)

Therefore,

\( U = \dfrac{3}{2}PV_0 \)

\( \boxed{U = \dfrac{3}{2}\left(P_{\mathrm{atm}} + \dfrac{Mg}{A}\right)V_0} \)

This works nicely because the gas is ideal and monatomic, so internal energy depends only on temperature, and \( PV \) gives the same result through the ideal gas law.

C.
Because the container is thermally conducting and the water bath remains at constant temperature, the process is approximately isothermal.

When the extra block is slowly added, the external pressure on the gas increases, so the gas is compressed. For an isothermal compression, \( PV = \text{constant} \), so the \( P\text{-}V \) curve is a hyperbola.

The process moves from a point at lower pressure and larger volume to a point at higher pressure and smaller volume, so the arrow points up and to the left along a concave-up curve.

D.
\( \boxed{T_{\mathrm{new}} > T_0} \)

With the extra block still on the piston, the gas must support a larger downward force than it did initially. Therefore, at equilibrium the gas pressure is larger than before.

In part \( \mathrm{B} \), the initial pressure was \( P_{\mathrm{atm}} + \dfrac{Mg}{A} \). With an added block of the same mass \( M \), the final equilibrium pressure at volume \( V_0 \) is

\( P_{\mathrm{final}} = P_{\mathrm{atm}} + \dfrac{2Mg}{A} \)

Since the gas ends at the same volume \( V_0 \) but at a larger pressure, the ideal gas law implies a larger temperature:

\( PV = nRT \)

With \( n \) constant and \( V \) unchanged, larger \( P \) means larger \( T \). Therefore, \( T_{\mathrm{new}} > T_0 \).

This is also consistent with the \( P\text{-}V \) picture in part \( \mathrm{C} \): to return to the original volume while supporting the added block, the gas must end at a higher pressure than it had initially.

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