Solved thermodynamics exercise

Work and heat between two states along two paths

Exercise 11 · Lesson 5 — Application to the Ideal Gas

pressure work
heat
isotherm
isochoric transformation
Clapeyron diagram
exact differential
ideal gas

Statement

Consider $n$ moles of a monatomic ideal gas with internal energy $U=\frac{3}{2}nRT$ . State $A(T_1,V_1)$ is connected to state $B(T_2,V_2)$ by two quasi-static paths, with $P_{\mathrm{ext}}=P$ at all times:

$\Gamma_1$ : an isothermal expansion at $T_1$ , from $V_1$ to $V_2$ , followed by an isochoric transformation at $V_2$ , from $T_1$ to $T_2$ ;
$\Gamma_2$ : an isochoric transformation at $V_1$ , from $T_1$ to $T_2$ , followed by an isothermal expansion at $T_2$ , from $V_1$ to $V_2$ .

For $T_2>T_1$ and $V_2>V_1$ , plot paths $\Gamma_1$ and $\Gamma_2$ in the Clapeyron diagram $(P,V)$ . Indicate the intermediate states and the direction of travel.
Calculate the work received, $W_1$ and $W_2$ , along each of the two paths. What is their difference $W_1-W_2$ ?
Without evaluating the two integrals separately, recover $W_1-W_2$ geometrically from the area between the two paths in the Clapeyron diagram.
Calculate $\Delta U$ along each of the two paths, then deduce $Q_1$ and $Q_2$ . Conclude as to the nature of $\delta Q$ .
Numerical application: $n=1$ mol, $T_1=300$ K, $T_2=600$ K, and $V_2=2V_1$ .

Hint

In the Clapeyron diagram, an isochore is vertical, and an ideal-gas isotherm satisfies

P=nRT/V

. The isotherm at temperature

T_2

lies above that at temperature

T_1

. An isochoric transformation makes no contribution to pressure work. For Question 3, the work received is the negative of the signed area under the path in the

(P,V)

diagram.

Detailed solution

Solution

Question 1. Let

C

denote the intermediate state on path

\Gamma_1

and

D

the intermediate state on path

\Gamma_2

. Their pressures and volumes are

\begin{aligned} A&=\left(\frac{nRT_1}{V_1},V_1\right),& C&=\left(\frac{nRT_1}{V_2},V_2\right),\\ D&=\left(\frac{nRT_2}{V_1},V_1\right),& B&=\left(\frac{nRT_2}{V_2},V_2\right). \end{aligned}

Since $T_2>T_1$ , the isotherm $T_2$ lies above the isotherm $T_1$ . Path $\Gamma_1$ runs from $A$ to $C$ , to the right along the lower isotherm $T_1$ , then vertically from $C$ to $B$ along the isochore $V_2$ . Path $\Gamma_2$ first rises vertically from $A$ to $D$ along the isochore $V_1$ , then runs to the right from $D$ to $B$ along the upper isotherm $T_2$ . The arrows on the two paths are therefore

\Gamma_1:A\xrightarrow{\,T=T_1\,}C\xrightarrow{\,V=V_2\,}B, \qquad \Gamma_2:A\xrightarrow{\,V=V_1\,}D\xrightarrow{\,T=T_2\,}B.

Paths _1 and _2 in the Clapeyron diagram. The grey curves extend the isotherms T_1 and T_2. — Figure 1. Paths $\Gamma_1$ and $\Gamma_2$ in the Clapeyron diagram. The grey curves extend the isotherms $T_1$ and $T_2$ .

Question 2. Using the convention that work received by the gas is positive, $\delta W=-P\,\mathrm{d}V$ . An isochoric transformation makes no contribution and, along an isotherm at temperature $T$ ,

W=-\int_{V_1}^{V_2}\frac{nRT}{V}\,\mathrm{d}V =-nRT\ln\frac{V_2}{V_1}.

We therefore obtain

W_1=-nRT_1\ln\frac{V_2}{V_1}, \qquad W_2=-nRT_2\ln\frac{V_2}{V_1},

and then

W_1-W_2 =nR(T_2-T_1)\ln\frac{V_2}{V_1}\neq 0.

The initial and final states are identical, but the works differ: $\delta W$ is not an exact differential.

Question 3. The isochores sweep out no area. Since the work received is the negative of the area under the curve, the difference $W_1-W_2$ is the area between the isotherms $T_2$ and $T_1$ :

W_1-W_2 =\int_{V_1}^{V_2}\left(\frac{nRT_2}{V}-\frac{nRT_1}{V}\right)\mathrm{d}V =nR(T_2-T_1)\ln\frac{V_2}{V_1}>0.

This recovers the result of Question 2.

Question 4. The internal energy of the monatomic ideal gas depends only on temperature. Its change is therefore the same along both paths:

\Delta U=U_B-U_A=\frac{3}{2}nR(T_2-T_1).

The first law, $\Delta U=W+Q$ , gives $Q=\Delta U-W$ , hence

\begin{aligned} Q_1&=\frac{3}{2}nR(T_2-T_1) +nRT_1\ln\frac{V_2}{V_1},\\ Q_2&=\frac{3}{2}nR(T_2-T_1) +nRT_2\ln\frac{V_2}{V_1}. \end{aligned}

Thus,

Q_2-Q_1=nR(T_2-T_1)\ln\frac{V_2}{V_1}\neq 0.

The heat exchanged also depends on the path: $\delta Q$ is not an exact differential, unlike $\mathrm{d}U$ .

Question 5. With $\ln 2\simeq0{,}693$ ,

\begin{aligned} W_1&\simeq-1{,}73\ \mathrm{kJ},& W_2&\simeq-3{,}46\ \mathrm{kJ},\\ \Delta U&\simeq3{,}74\ \mathrm{kJ},& Q_1&\simeq5{,}47\ \mathrm{kJ},& Q_2&\simeq7{,}20\ \mathrm{kJ}. \end{aligned}

The gas does twice as much work during $\Gamma_2$ and receives the corresponding additional heat.