Solved thermodynamics exercise

Heating at constant pressure: deriving $c_{P}$

Exercise 21 · Lesson 5 — Application to the Ideal Gas

heat capacity
constant pressure
Mayer's relation
isobaric heating
ideal gas

Statement

One mole of an ideal gas with molar heat capacity $c_{V}$ is enclosed in a cylinder with initial volume $V_{0}$ . The cylinder is closed by a frictionless movable adiabatic piston subjected to a constant external pressure $P_{0}$ . An electrical resistor inside the gas supplies it with heat $Q > 0$ quasi-statically.

Draw a simple diagram of the apparatus. Indicate the gas, the piston, the resistor, the external pressure $P_{0}$ , and the energy transfers.
Without using numerical values, express the initial temperature $T_{0}$ , then determine the final state $(P_{f}, V_{f}, T_{f})$ , the work $W$ received by the gas, and its change in internal energy $Δ U$ .
Deduce the molar heat capacity at constant pressure $c_{P} = (δ Q / d T)_{P}$ in terms of $c_{V}$ and $R$ .
Numerical application for a monatomic ideal gas: $c_{V} = \frac{3}{2} R$ , $V_{0} = 25 L$ , $P_{0} = 1, 0 \times 1 0^{5} Pa$ , $Q = 1000 J$ , and $R = 8, 314 J mo l^{- 1} K^{- 1}$ .

Hint

The transformation is isobaric:

W = - P_{0} Δ V

. For one mole of an ideal gas,

P_{0} Δ V = R Δ T

. Substitute these relations into the first law.

Detailed solution

Solution

Question 1. The diagram should show the gas in a cylinder closed by a movable piston. An arrow inside the cylinder represents the heat

Q

supplied by the resistor. The external pressure

P_{0}

acts on the piston, which moves upward when the gas expands. Question 2. The initial equation of state gives

T_{0} = \frac{P _{0} V _{0}}{R} .

The pressure remains constant, so $P_{f} = P_{0}$ . Moreover, $W = - P_{0} Δ V = - R Δ T$ . The first law then gives

Δ U = c_{V} Δ T = Q + W = Q - R Δ T,

hence

Δ T = \frac{Q}{c _{V} + R} .

The final state is therefore

P_{f} = P_{0}, T_{f} = T_{0} + \frac{Q}{c _{V} + R}, V_{f} = V_{0} + \frac{R Q}{P _{0} ( c _{V} + R )} .

Finally,

W = - \frac{R Q}{c _{V} + R}, Δ U = \frac{c _{V} Q}{c _{V} + R} .

Question 3. Since $Q = (c_{V} + R) Δ T$ at constant pressure,

c_{P} = c_{V} + R .

This is Mayer's relation.

Question 4. For a monatomic gas,

c_{P} = \frac{5}{2} R ≃ 20, 8 J mo l^{- 1} K^{- 1} .

We obtain

T_{0} ≃ 301 K, Δ T ≃ 48, 1 K, T_{f} ≃ 349 K, V_{f} = 29, 0 L .

Finally,

W = - 400 J, Δ U = 600 J .