Quiz 3, with solutions

Useful equations provided with the quiz included $H = U + PV$, $\Delta S = \int_i^f \frac{dq_\text{rev}}{T}$, $dq = nC_v\,dT$, and $\Delta U = q + w$ (the last carrying a correction from the original quiz).

A trip around the PV graph

The figure shows the pressure-volume state of an ideal gas along three different paths I, II, III. Express your answers with the given variables and/or universal constants.

(a) Suppose $n$ moles of an ideal gas undergo a reversible isothermal expansion at temperature $T_1$ along path I. Determine the heat released or absorbed by the gas.

Solution. We begin with the first law of thermodynamics, $\Delta U = q + w$. For a reversible isothermal expansion $\Delta U = 0$, and therefore $q = -w$. For an ideal gas, the pressure-volume work for a reversible process is given by

$$w = -\int_{V_1}^{V_2}\frac{nRT}{V}\,dV$$

Solving the integral, we get

$$q = -w = nRT\ln\frac{V_2}{V_1}$$

(b) Determine the change in internal energy $\Delta U$ and change in enthalpy $\Delta H$ along path I.

Solution. We know that for an isothermal process $\Delta U = 0$. Starting from the definition of enthalpy, $H = U + PV$, and for an ideal gas $H = U + nRT$. Consider the change $\Delta H = \Delta U + \Delta(nRT)$. Since all quantities on the right hand side remain constant, $\Delta H = 0$.

(c) Determine the change in entropy $\Delta S$ along path I.

Solution. To obtain the change in entropy we begin with $\Delta S = q_\text{rev}/T$, where $q_\text{rev}$ means we need a reversible heat for the desired process. From part (a) we found the heat along path I, and it is indeed reversible. Combining the equations, we get

$$\Delta S = nR\ln\frac{V_2}{V_1}$$

(d) Consider path II from point B to C, where the gas is cooled reversibly at constant volume from $T_1$ to $T_2$. Determine the heat released or absorbed during the process. Let the heat capacity at constant volume of the gas be $C_v$.

Solution. For the heat of a constant-volume process we use the definition of heat capacity, $dq = nC_v\,dT$, which integrates to $q = nC_v(T_2 - T_1)$.

(e) Consider path III from point C to A, where the gas goes through an irreversible adiabatic process (i.e. no heat exchange). Determine $\Delta U$ from point C to A.

Solution. Recall that $U$ is a state variable, which means it does not matter how we go from C to A. Since we have not learned much about irreversible, let alone adiabatic, processes, it is best we choose a path that is familiar to us. We can combine our answers from previous parts, writing

$$\Delta U_{C\to A} = \Delta U_{C\to B} + \Delta U_{B\to A}$$

For $\Delta U_{C\to B}$ we determined the heat in part (d), and the work for a constant-volume process is 0, hence $\Delta U_{C\to B} = -nC_v(T_2 - T_1)$, with the negative sign because we reverse the process of part (d). From B to A, $\Delta U = 0$. Therefore

$$\Delta U_{C\to A} = -nC_v(T_2 - T_1)$$

(f) Determine $\Delta S$ from point C to A.

Solution. Similar to part (e), $S$ is a state function, so we disregard the complicated irreversible adiabatic path and instead choose the path $C \to B \to A$ that we have already solved. From C to B we have $dq = nC_v\,dT$. Dividing both sides by $T$ and integrating (the process is reversible, so $q$ is $q_\text{rev}$),

$$\Delta S_{C\to B} = nC_v\ln\frac{T_1}{T_2}$$

From B to A, part (c) gives $\Delta S_{A\to B} = nR\ln\frac{V_2}{V_1}$, so $\Delta S_{B\to A} = nR\ln\frac{V_1}{V_2}$. Altogether

$$\Delta S_{C\to A} = nC_v\ln\frac{T_1}{T_2} + nR\ln\frac{V_1}{V_2}$$

Remark. To determine entropy you must find a reversible path. The adiabatic ($q=0$) path does not imply $\Delta S = 0$, because that $q$ is irreversible.

Bonus

Interpret the change in entropy from point A to B above using the concept of microstates.

Solution. The fundamental definition of entropy relates to microstates according to $S = k_B \ln\Omega$. From A to B we increase the volume, hence the number of available configurations (microstates) the gas can arrange itself in increases, and entropy increases.