Statement

Consider $n$ moles of a monatomic ideal gas, with $U=\frac{3}{2}nRT$ , traversing quasi-statically the counterclockwise rectangular cycle

A(P_1,V_1)\to B(P_1,V_2)\to C(P_2,V_2) \to D(P_2,V_1)\to A,

where $V_2>V_1$ and $P_2>P_1$ . We write

\Delta P=P_2-P_1>0, \qquad \Delta V=V_2-V_1>0.

This refrigeration cycle is used as a model of a refrigerator. Its refrigeration coefficient of performance (COP) is defined by

\mathrm{COP}=\frac{Q_f}{W_{\mathrm{in}}} =\frac{Q_f}{W_{\mathrm{cycle}}},

where $Q_f>0$ is the total heat extracted by the gas along the branches where it receives heat, and $W_{\mathrm{cycle}}>0$ is the net work it receives.

Plot the cycle in the Clapeyron diagram $(P,V)$ and specify the nature of each branch.
Calculate the work received along each of the four branches in terms of $\Delta V$ when it varies, then the total work in terms of $\Delta P$ and $\Delta V$ . Recover its value from the area of the rectangle.
Express $\Delta U$ and $Q$ along each branch in terms of $P_1$ , $P_2$ , $V_1$ , $V_2$ , $\Delta P$ , and $\Delta V$ , using $\Delta P$ for the isochores and $\Delta V$ for the isobars.
Verify that $\Delta U_{\mathrm{cycle}}=0$ and $Q_{\mathrm{cycle}}=-W_{\mathrm{cycle}}$ . Is the cycle a power or refrigeration cycle? Along which branches does the gas receive heat, and along which does it release heat? Interpret these exchanges by identifying the cold and hot reservoirs: what is being cooled here?
Deduce the refrigeration COP of the cycle in terms of $P_1$ , $P_2$ , $V_1$ , and $V_2$ .
Interpret the COP physically. Why can it be greater than 1, unlike the efficiency of a heat engine?
Numerical application: $P_1=1{,}0$ bar, $P_2=3{,}0$ bar, $V_1=5{,}0$ L, and $V_2=15{,}0$ L. Give $W$ and $Q$ along each branch, then their balances over the cycle and the COP.

Hint

Along an isochore,

nR\,\Delta T=V\,\Delta P

and the work is zero. Along an isobar,

nR\,\Delta T=P\,\Delta V

and

W=-P\Delta V

. Every state function returns to its initial value at the end of a cycle.

Detailed solution

Solution

Question 1. Points

A

,

B

,

C

, and

D

are at the bottom left, bottom right, top right, and top left, respectively. Branches

A\to B

and

C\to D

are isobaric;

B\to C

and

D\to A

are isochoric. The path

A\to B\to C\to D\to A

is counterclockwise.

Figure 1. The rectangular isobaric—isochoric cycle in the Clapeyron diagram.

Question 2. The work received by the gas is

W=-\int P_{\mathrm{ext}}\,\mathrm dV.

It is zero along the isochores $B\to C$ and $D\to A$ . Along the two isobars,

W_{AB}=-P_1(V_2-V_1)=-P_1\Delta V, \qquad W_{CD}=-P_2(V_1-V_2)=P_2\Delta V.

Thus,

\boxed{W_{\mathrm{cycle}} =(-P_1+P_2)\Delta V =\Delta P\,\Delta V>0}.

This value is indeed the area of the rectangle in the Clapeyron diagram.

Question 3. For the monatomic ideal gas,

\Delta U=nc_V\Delta T=\frac32nR\Delta T.

Along an isobar, the equation of state gives $nR\Delta T=P\Delta V_{\mathrm{branche}}$ . Consequently,

\Delta U_{AB}=\frac32P_1\Delta V, \qquad \Delta U_{CD}=-\frac32P_2\Delta V.

Along an isochore, it gives $nR\Delta T=V\Delta P_{\mathrm{branche}}$ , hence

\Delta U_{BC}=\frac32V_2\Delta P, \qquad \Delta U_{DA}=-\frac32V_1\Delta P.

Finally, the first law $\Delta U=Q+W$ , that is, $Q=\Delta U-W$ , gives the summary table:

\begin{aligned} \begin{array}{c|c|c|c} \text{Branch} & W & \Delta U & Q=\Delta U-W\\ \hline A\to B & -P_1\Delta V & \frac32P_1\Delta V & \frac52P_1\Delta V\\[2pt] B\to C & 0 & \frac32V_2\Delta P & \frac32V_2\Delta P\\[2pt] C\to D & P_2\Delta V & -\frac32P_2\Delta V & -\frac52P_2\Delta V\\[2pt] D\to A & 0 & -\frac32V_1\Delta P & -\frac32V_1\Delta P \end{array} \end{aligned}

Question 4. The sum of the changes in internal energy is zero. The first law therefore requires

Q_{\mathrm{cycle}}=-W_{\mathrm{cycle}}=-\Delta P\,\Delta V<0.

Overall, the gas receives work: this is a refrigeration cycle. It receives heat along $A\to B$ and $B\to C$ . Along these two branches, it extracts heat from the cold reservoirs with which it is placed in contact: it is therefore these external reservoirs (for example, the compartment and food in a refrigerator), rather than the working gas itself, that are cooled.

Along $C\to D$ and $D\to A$ , by contrast, the gas releases heat to the hot reservoirs, for example the air in the room. Denoting by $Q_{\mathrm{chaud}}>0$ the total heat received by these reservoirs,

Q_{\mathrm{chaud}} =-(Q_{CD}+Q_{DA}) =\frac52P_2\Delta V+\frac32V_1\Delta P.

The cycle thus transfers heat from the cold reservoirs to the hot reservoirs by means of the work it receives.

Question 5. The heat extracted by the gas is

Q_f =Q_{AB}+Q_{BC} =\frac52P_1\Delta V+\frac32V_2\Delta P.

It follows that

\boxed{ \mathrm{COP} =\frac{\frac52P_1\Delta V+\frac32V_2\Delta P} {\Delta P\,\Delta V}. }

Question 6. The COP measures the refrigeration benefit obtained relative to the work input: a COP equal to $k$ means that each joule of work supplied allows $k$ joules of heat to be extracted from the cold reservoirs.

Unlike the efficiency of a heat engine, the COP is not the fraction of received energy that is converted into work. The gas receives both the work $W_{\mathrm{in}}$ and the heat $Q_f$ extracted from the cold reservoirs, then rejects their sum to the hot reservoirs:

\boxed{Q_{\mathrm{chaud}}=Q_f+W_{\mathrm{in}}}.

There is therefore no reason for $Q_f$ to be less than $W_{\mathrm{in}}$ , and the COP can readily be greater than 1 without violating energy conservation.

Question 7. Here $\Delta P=2{,}0\times10^5$ Pa and $\Delta V=1{,}0\times10^{-2}$ m $^3$ :

\begin{aligned} \begin{array}{c|r|r|r} \text{Branch} & W\ (\mathrm J) & \Delta U\ (\mathrm J) & Q\ (\mathrm J)\\ \hline A\to B & -1000 & 1500 & 2500\\ B\to C & 0 & 4500 & 4500\\ C\to D & 3000 & -4500 & -7500\\ D\to A & 0 & -1500 & -1500 \end{array} \end{aligned}

Thus $W_{\mathrm{cycle}}=+2000$ J, $Q_{\mathrm{cycle}}=-2000$ J, and $\Delta U_{\mathrm{cycle}}=0$ . The heat extracted is

Q_f=2500+4500=7000\ \mathrm{J},

whereas the heat released to the hot reservoirs is

Q_{\mathrm{chaud}}=7500+1500=9000\ \mathrm{J} =Q_f+W_{\mathrm{cycle}}.

The refrigeration coefficient of performance is therefore

\mathrm{COP}=\frac{7000}{2000}=3{,}5.