Solved thermodynamics exercise

Rectangular refrigeration cycle?

Exercise 23 · Lesson 5 — Application to the Ideal Gas

thermodynamic cycle
rectangular cycle
Clapeyron diagram
work
heat
area
ideal gas
refrigeration cycle
refrigerator

Statement

Consider $n$ moles of a monatomic ideal gas, with $U = \frac{3}{2} n R T$ , 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

Δ P = P_{2} - P_{1} > 0, Δ V = V_{2} - V_{1} > 0.

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 $Δ V$ when it varies, then the total work in terms of $Δ P$ and $Δ V$ . Recover its value from the area of the rectangle.
Express $Δ U$ and $Q$ along each branch in terms of $P_{1}$ , $P_{2}$ , $V_{1}$ , $V_{2}$ , $Δ P$ , and $Δ V$ , using $Δ P$ for the isochores and $Δ V$ for the isobars.
Verify that $Δ U_{cycle} = 0$ and $Q_{cycle} = - W_{cycle}$ . Is this a power cycle or a work-consuming cycle? Along which branches does the gas receive heat, and along which does it release heat? Show that the total heat released by the gas, $Q_{out} > 0$ , equals the total heat it receives, $Q_{in} > 0$ , plus the work received $W_{cycle}$ .
Express the temperatures $T_{A}$ , $T_{B}$ , $T_{C}$ and $T_{D}$ of the four vertices. Which is the lowest and which is the highest? How does the temperature of the gas change along each branch? Between which temperatures does the gas receive heat, and between which temperatures does it release heat?
We would like to use this cycle as a refrigerator, extracting heat from a cold compartment and rejecting it to the warmer air of the room. For a body to receive heat, the source it is in contact with must be hotter than the body. What temperatures must the sources that supply heat to the gas have, and those that receive heat from it? Can this cycle work as a refrigerator?
Numerical application: $P_{1} = 1, 0$ bar, $P_{2} = 3, 0$ bar, $V_{1} = 5, 0$ L, and $V_{2} = 15, 0$ L. Calculate $W$ and $Q$ along each branch, then $W_{cycle}$ , $Q_{in}$ and $Q_{out}$ . Finally, calculate the ratios $T_{B} / T_{A}$ , $T_{C} / T_{A}$ and $T_{D} / T_{A}$ , and conclude.

Hint

Along an isochore,

n R Δ T = V Δ P

and the work is zero. Along an isobar,

n R Δ T = P Δ V

and

W = - P Δ 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_{ext} d V .

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

W_{A B} = - P_{1} (V_{2} - V_{1}) = - P_{1} Δ V, W_{C D} = - P_{2} (V_{1} - V_{2}) = P_{2} Δ V .

Thus,

W_{cycle} = (- P_{1} + P_{2}) Δ V = Δ P Δ V > 0

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

Question 3. For the monatomic ideal gas,

Δ U = n c_{V} Δ T = \frac{3}{2} n R Δ T .

Along an isobar, the equation of state gives $n R Δ T = P Δ V_{branch}$ . Consequently,

Δ U_{A B} = \frac{3}{2} P_{1} Δ V, Δ U_{C D} = - \frac{3}{2} P_{2} Δ V .

Along an isochore, it gives $n R Δ T = V Δ P_{branch}$ , hence

Δ U_{B C} = \frac{3}{2} V_{2} Δ P, Δ U_{D A} = - \frac{3}{2} V_{1} Δ P .

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

Branch A \to B B \to C C \to D D \to A W - P_{1} Δ V 0 P_{2} Δ V 0 Δ U \frac{3}{2} P_{1} Δ V \frac{3}{2} V_{2} Δ P - \frac{3}{2} P_{2} Δ V - \frac{3}{2} V_{1} Δ P Q = Δ U - W \frac{5}{2} P_{1} Δ V \frac{3}{2} V_{2} Δ P - \frac{5}{2} P_{2} Δ V - \frac{3}{2} V_{1} Δ P

Question 4. The changes in internal energy add up to zero, as for any state function over a cycle. The first law therefore gives

Q_{cycle} = - W_{cycle} = - Δ P Δ V < 0.

The gas receives net work: this is a work-consuming cycle. According to the table, it receives heat along $A \to B$ and $B \to C$ , and releases heat along $C \to D$ and $D \to A$ :

Q_{in} = Q_{A B} + Q_{B C} = \frac{5}{2} P_{1} Δ V + \frac{3}{2} V_{2} Δ P, Q_{out} = - (Q_{C D} + Q_{D A}) = \frac{5}{2} P_{2} Δ V + \frac{3}{2} V_{1} Δ P .

Subtracting, with $P_{2} - P_{1} = Δ P$ and $V_{2} - V_{1} = Δ V$ ,

Q_{out} - Q_{in} = \frac{5}{2} Δ P Δ V - \frac{3}{2} Δ P Δ V = Δ P Δ V = W_{cycle},

hence

Q_{out} = Q_{in} + W_{cycle}

Over a cycle, the gas releases as heat all the energy it has received, heat and work included.

Question 5. The equation of state gives

T = P V / (n R)

at each vertex:

T_{A} = \frac{P _{1} V _{1}}{n R}, T_{B} = \frac{P _{1} V _{2}}{n R}, T_{C} = \frac{P _{2} V _{2}}{n R}, T_{D} = \frac{P _{2} V _{1}}{n R} .

Since $P_{1} < P_{2}$ and $V_{1} < V_{2}$ , the product $P V$ is smallest at $A$ and largest at $C$ : $T_{A}$ is the lowest temperature of the cycle, $T_{C}$ the highest, and $T_{B}$ and $T_{D}$ lie in between.

Along the isobar

A \to B

V

increases at constant pressure, so

T

rises from

T_{A}

T_{B}

. Along the isochore

B \to C

P

increases at constant volume, so

T

rises from

T_{B}

T_{C}

. Likewise,

T

falls from

T_{C}

T_{D}

along

C \to D

, then from

T_{D}

T_{A}

along

D \to A

. The gas therefore receives heat while its temperature climbs from the lowest value

T_{A}

to the highest value

T_{C}

, and releases heat while it falls back from

T_{C}

T_{A}

. Question 6. To receive heat, the gas must be in contact with sources at least as hot as itself; in the quasi-static limit, their temperature follows that of the gas. The sources that supply heat to the gas therefore have temperatures ranging from

T_{A}

up to

T_{C}

, the highest temperature of the cycle. Conversely, the sources that receive its heat have temperatures ranging from

T_{C}

down to

T_{A}

, the lowest temperature of the cycle. There is thus no separate cold side and hot side: the gas takes heat from very hot sources and gives heat to very cold ones. This cycle cannot serve as a refrigerator, which must extract heat at the lowest temperatures and reject it at the highest. It receives work and returns it entirely as heat (

Q_{cycle} = - W_{cycle}

), while redistributing heat among sources at various temperatures. A refrigerator must, on the contrary, absorb heat on the coldest parts of its cycle and reject it on the hottest. This is the case for the Carnot cycle run in reverse, studied in Lesson 6: the gas receives heat along the cold isotherm and releases it along the hot isotherm. A cycle being work-consuming is therefore not enough for it to be a refrigeration cycle. Question 7. Here

Δ P = 2, 0 \times 1 0^{5}

Pa and

Δ V = 1, 0 \times 1 0^{- 2}

^{3}

Branch A \to B B \to C C \to D D \to A W (J) - 1000 030000 Δ U (J) 15004500 - 4500 - 1500 Q (J) 25004500 - 7500 - 1500

Thus $W_{cycle} = + 2000 J$ , $Q_{cycle} = - 2000 J$ and $Δ U_{cycle} = 0$ . The heat received is $Q_{in} = 2500 + 4500 = 7000 J$ and the heat released is $Q_{out} = 7500 + 1500 = 9000 J = Q_{in} + W_{cycle}$ . The temperature ratios do not depend on $n$ :

\frac{T _{B}}{T _{A}} = \frac{V _{2}}{V _{1}} = 3, \frac{T _{C}}{T _{A}} = \frac{P _{2} V _{2}}{P _{1} V _{1}} = 9, \frac{T _{D}}{T _{A}} = \frac{P _{2}}{P _{1}} = 3.

If the lowest temperature of the cycle were that of a cold compartment, say $T_{A} = 270 K$ , the gas would have to receive heat from sources reaching up to $T_{C} = 9 T_{A} ≃ 2400 K$ . Such a cycle obviously does not cool the compartment.