Solved thermodynamics exercise

Triangular cycle

Exercise 20 · Lesson 5 — Application to the Ideal Gas

power cycle
Clapeyron diagram
cycle area
first law
efficiency

Statement

Consider $n$ moles of an ideal gas with constant molar heat capacity $c_{V}$ undergoing the quasi-static cycle $A \to B \to C \to A$ , with

A = (P_{0}, V_{0}), B = (2 P_{0}, V_{0}), C = (P_{0}, 2 V_{0}) .

Branch $A \to B$ is isochoric, branch $B \to C$ is a straight-line segment in the $(P, V)$ plane, and branch $C \to A$ is isobaric.

For a power cycle, the efficiency is defined by

η = \frac{W _{out}}{Q _{in}} = \frac{- W _{cycle}}{Q _{in}},

where $Q_{in}$ is the sum of the elementary heat transfers $δ Q > 0$ received by the gas during the cycle.

Plot the cycle in the Clapeyron diagram and specify its direction of travel. Is it a power or refrigeration cycle?
Determine the equation $P (V)$ of the line connecting $B$ to $C$ .
Letting $T_{0} = P_{0} V_{0} / (n R)$ , determine the temperatures $T_{A}$ , $T_{B}$ , and $T_{C}$ . Why is $Δ U_{B \to C}$ zero even though this branch is not isothermal?
Calculate the work received along each of the three branches, then over the complete cycle. Recover the absolute value of the total work from the area of the triangle.
Calculate $Δ U$ and $Q$ along each branch in terms of $P_{0}$ , $V_{0}$ , and $c_{V}$ only, then verify that $Δ U_{cycle} = 0$ and $Q_{cycle} = - W_{cycle}$ .
Letting $a = \frac{c _{V}}{R} > 1$ and $x = V / V_{0}$ , show that the sign of $δ Q$ changes at a point on branch $B \to C$ . Deduce $Q_{in}$ , then the efficiency $η$ of this cycle. Give its value for a monatomic ideal gas, for which $c_{V} = \frac{3}{2} R$ .

Hint

Along

B \to C

, seek

P (V) = aV + b

by requiring the line to pass through

B

and

C

. To calculate a change in internal energy, only the temperatures at the ends of the branch are required. For Question 6, use the differential form of the first law.

Detailed solution

Solution

Questions 1–2. The cycle runs clockwise: it is a power cycle. The equation of the line

B \to C

P (V) = P_{0} (3 - \frac{V}{V _{0}}) .

The triangular cycle A B C A in the Clapeyron diagram. — Figure 1. The triangular cycle $A \to B \to C \to A$ in the Clapeyron diagram.

Question 3. The equation of state gives

T_{A} = T_{0}, T_{B} = T_{C} = 2 T_{0} .

The temperature varies along $B \to C$ , but returns to the same value at both endpoints. Since the internal energy of an ideal gas depends only on temperature for a fixed amount of substance, $Δ U_{B \to C} = 0$ .

Question 4. The works received are

W_{A B} = 0, W_{B C} = - \int_{V_{0}}^{2 V_{0}} P (V) d V = - \frac{3}{2} P_{0} V_{0}, W_{C A} = - P_{0} Δ V = P_{0} V_{0} .

Thus,

W_{cycle} = - \frac{1}{2} P_{0} V_{0}

Its sign is negative: the gas does work over the cycle. Its absolute value is indeed the area of the triangle, with base $V_{0}$ and height $P_{0}$ , from the formula $A = \frac{1}{2} base \times height$ for a right triangle.

Question 5. Since

n R T_{0} = P_{0} V_{0}

Branch A \to B B \to C C \to A W 0 - \frac{3}{2} P_{0} V_{0} P_{0} V_{0} Δ U \frac{c _{V}}{R} P_{0} V_{0} 0 - \frac{c _{V}}{R} P_{0} V_{0} Q = Δ U - W \frac{c _{V}}{R} P_{0} V_{0} \frac{3}{2} P_{0} V_{0} - (\frac{c _{V}}{R} + 1) P_{0} V_{0}

Summing the three branches gives

Δ U_{cycle} = 0, Q_{cycle} = \frac{1}{2} P_{0} V_{0} = - W_{cycle}

Question 6. Let $a = c_{V} / R$ and $x = V / V_{0}$ . Along branch $B \to C$ , $P = P_{0} (3 - x)$ and $T = T_{0} (3 x - x^{2})$ . The first law gives

δ Q = d U - δ W = P_{0} V_{0} [3 (a + 1) - (2 a + 1) x] d x .

This elementary heat transfer vanishes at

x_{⋆} = \frac{3 ( a + 1 )}{2 a + 1} .

When $a > 1$ , we have $1 < x_{⋆} < 2$ : the gas receives heat from $B$ to the point with abscissa $x_{⋆}$ , then releases heat up to $C$ . It also receives the heat $Q_{A B} = a P_{0} V_{0}$ along $A \to B$ . Consequently,

Q_{in} = P_{0} V_{0} [a + \int_{1}^{x_{⋆}} (3 (a + 1) - (2 a + 1) x) d x] = P_{0} V_{0} [a + \frac{( a + 2 ) ^{2}}{2 ( 2 a + 1 )}] .

The efficiency is therefore

η = \frac{\frac{1}{2} P _{0} V _{0}}{Q _{in}} = \frac{2 a + 1}{5 a ^{2} + 6 a + 4}, a = \frac{c _{V}}{R}

For a monatomic ideal gas, $a = \frac{3}{2}$ , hence

η = \frac{16}{97} ≃ 16, 5 %.