AP Physics 2 Exam Question

A container contains an ideal gas at an initial temperature of $T_i=300$ K and an initial volume of $V_i=2$ m$^3$. The gas undergoes a series of processes as described below:

1. The gas is compressed at constant pressure until its volume is reduced to $V_1=1$ m$^3$.
2. The gas is then heated at constant volume until its pressure reaches $P_2=3$ atm.
3. Finally, the gas is allowed to expand adiabatically until its temperature returns to the initial value.

Assuming the gas behaves ideally and the specific heat capacity at constant pressure is $C_p=20$ J/(mol K), respond to the following:

a) Calculate the final pressure of the gas after the compression in part 1.

b) Calculate the amount of heat transferred to the gas during the heating in part 2.

c) Determine the final volume the gas has after the adiabatic expansion in part 3.

---

Answer and Explanation

a) To calculate the final pressure of the gas after the compression in part 1, we will use the relationship $P_1V_1 = P_2V_2$ for an ideal gas.

Given: $V_1 = 1$ m$^3$ $V_2 = 2$ m$^3$ $P_2 = 3$ atm

Substituting the given values into the equation, we can solve for $P_1$:

$P_1V_1 = P_2V_2$

$P_1(1\, \text{m}^3) = (3\, \text{atm})(2\, \text{m}^3)$

$P_1 = \frac{6}{1} \text{ atm}$

Therefore, the final pressure after compression is $P_1 = 6$ atm.

---

b) To calculate the amount of heat transferred to the gas during the heating in part 2, we will use the equation for specific heat at constant volume:

$Q = nC_v\Delta T$

Given: $T_i = 300$ K $P_2 = 3$ atm $C_p = 20$ J/(mol K)

Since the process occurs at constant volume, the temperature change can be calculated using the ideal gas law:

$P_1V_1 = nRT_1$

$T_1 = \frac{P_1V_1}{nR} = \frac{(6\, \text{atm})(1\, \text{m}^3)}{nR}$

Substitute this equation into the specific heat equation:

$Q = nC_v(T_2 - T_i) = nC_p(T_2 - \frac{(6\, \text{atm})(1\, \text{m}^3)}{nR})$

We are given the final pressure $P_2 = 3$ atm, which we can substitute after solving the equation:

$Q = nC_p(T_2 - \frac{(P_1V_1)}{nR}) = nC_p(T_2 - \frac{(6\, \text{atm})(1\, \text{m}^3)}{nR})$

The value of $nR$ cancels out in the equation. Rearranging terms and substituting values:

$Q = nC_pT_2 - 6C_p$ J

$c = \frac{Q}{nC_pT_2 - 6C_p}$

Therefore, the amount of heat transferred to the gas during the heating in part 2 is given by $Q = nC_pT_2 - 6C_p$ J.

---

c) To determine the final volume the gas has after the adiabatic expansion in part 3, we will use the relationship for adiabatic expansion:

$T_f = T_i\left(\frac{V_i}{V_f}\right)^{(\gamma-1)}$

Given: $T_i = 300$ K $V_i = 2$ m$^3$ $T_f = T_i = 300$ K, since the final temperature is the same as the initial temperature.

Substituting the values and solving for $V_f$:

$T_i\left(\frac{V_i}{V_f}\right)^{(\gamma-1)} = T_f$

$300\left(\frac{2}{V_f}\right)^{(\gamma-1)} = 300$

Simplifying:

$\left(\frac{2}{V_f}\right)^{(\gamma-1)} = 1$

$(\gamma-1) \ln\left(\frac{2}{V_f}\right) = 0$

Since the process is adiabatic and the value of $\gamma$ for an ideal monoatomic gas is $\gamma = \frac{5}{3}$, the equation simplifies to:

$\ln\left(\frac{2}{V_f}\right) = 0$

$\frac{2}{V_f} = e^0$

$V_f = 2$ m$^3$

Therefore, the final volume the gas has after the adiabatic expansion in part 3 is $V_f = 2$ m$^3$.