AP Physics 2- 9.1 Kinetic Theory of Temperature and Pressure- Exam Style questions - FRQs- New Syllabus
Kinetic Theory of Temperature and Pressure AP Physics 2 FRQ
Unit 9: Thermodynamics
Weightage : 15–18%
Question

A sample of ideal gas is taken through the thermodynamic cycle shown above. Process \(C\) is isothermal.

Most-appropriate topic codes (AP Physics 2):
• Topic \(9.3\) — Thermal Energy Transfer and Equilibrium (Part \( \mathrm{(c)} \))
• Topic \(9.4\) — The First Law of Thermodynamics (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)(i)} \))
▶️ Answer/Explanation
(a)
From the graph, state \(1\) and state \(3\) lie on the same isothermal curve \(C\). For an ideal gas, internal energy depends only on temperature. Since the initial and final temperatures are the same, the change in internal energy is
\(\Delta U=0\,\text{J}\)
During process \(A\), the gas expands from \(V_1=1.0\times 10^{-3}\,\text{m}^3\) to \(V_2=4.0\times 10^{-3}\,\text{m}^3\) at constant pressure \(P=100\times 10^3\,\text{Pa}\).
The work done by the gas during process \(A\) is \(W_{\text{by}}=P\Delta V\).
\(W_{\text{by}}=\left(100\times 10^3\,\text{Pa}\right)\left[\left(4.0\times 10^{-3}\,\text{m}^3\right)-\left(1.0\times 10^{-3}\,\text{m}^3\right)\right]\)
\(W_{\text{by}}=\left(100\times 10^3\,\text{Pa}\right)\left(3.0\times 10^{-3}\,\text{m}^3\right)=300\,\text{J}\)
The question asks for the work \(W\) done on the gas. Since the gas expands, the gas does positive work on the surroundings, so the work done on the gas is negative.
\(W=-300\,\text{J}\)
During process \(B\), the volume is constant, so no work is done.
\(W_B=0\,\text{J}\)
Therefore, the total work done on the gas from state \(1\) to state \(3\) by processes \(A\) and \(B\) is
\(\boxed{W=-300\,\text{J}}\)
Use the first law of thermodynamics with the sign convention \( \Delta U=Q+W \), where \(W\) is work done on the gas.
\(0=Q+\left(-300\,\text{J}\right)\)
\(Q=+300\,\text{J}\)
Therefore,
\(\boxed{\Delta U=0\,\text{J}}\), \(\boxed{W=-300\,\text{J}}\), and \(\boxed{Q=+300\,\text{J}}\).
Energy must be transferred to the gas by heating so that the internal energy remains constant while the gas does work during expansion.
(b)(i)
In process \(C\), the gas goes from state \(3\) back to state \(1\). The volume decreases from \(4.0\times 10^{-3}\,\text{m}^3\) to \(1.0\times 10^{-3}\,\text{m}^3\), so the gas is compressed.
Since the volume decreases, work is done on the gas, so the work in process \(C\) is positive.
The magnitude of the work equals the area under the curve on a \(PV\) graph. Process \(C\) is below the horizontal line for process \(A\), so the area under process \(C\) is less than the rectangular area for process \(A\).
Therefore, the work done on the gas in process \(C\) is positive, while the work in part (a) is negative.
Also, the magnitude of the work in process \(C\) is less than \(300\,\text{J}\).
\(\boxed{0<W_C<300\,\text{J}}\)
So compared with part (a), the work in process \(C\) has the opposite sign and smaller magnitude.
(b)(ii)
Process \(C\) is isothermal, so the temperature of the gas remains constant.
For an ideal gas, average molecular kinetic energy depends only on temperature. Since the temperature does not change, the average kinetic energy and average speed of the gas particles do not change.
However, during process \(C\), the volume decreases. The gas particles are confined to a smaller space, so they collide with the container walls more frequently.
Since the particles have the same average speed but collide with the walls more often, the force per unit area on the walls increases. Therefore, the pressure increases from state \(3\) to state \(1\).
This agrees with the graph because pressure is lower at state \(3\) and higher at state \(1\).
(c)
For an ideal gas with the same number of particles, the temperature is proportional to the product \(PV\), since \(PV=nRT\).
At state \(2\), the pressure is \(P_2=100\times 10^3\,\text{Pa}\), and the volume is \(V_2=4.0\times 10^{-3}\,\text{m}^3\).
\(P_2V_2=\left(100\times 10^3\right)\left(4.0\times 10^{-3}\right)=400\,\text{J}\)
At state \(3\), the pressure is \(P_3=25\times 10^3\,\text{Pa}\), and the volume is \(V_3=4.0\times 10^{-3}\,\text{m}^3\).
\(P_3V_3=\left(25\times 10^3\right)\left(4.0\times 10^{-3}\right)=100\,\text{J}\)
Since \(P_2V_2>P_3V_3\), sample \(2\) has a higher temperature than sample \(3\).
Thermal energy flows spontaneously from the hotter sample to the colder sample.
Therefore, energy transfers from sample \(2\) to sample \(3\).
\(\boxed{\text{Energy transfers from sample }2\text{ to sample }3.}\)
