Solved thermodynamics exercise

Textbook exercise: elementary transformations

Exercise 17 · Lesson 5 — Application to the Ideal Gas

textbook exercise
monatomic ideal gas
isochoric transformation
isothermal transformation
isobaric transformation
internal energy
work
heat
first law

Statement

Consider $n$ moles of a monatomic ideal gas, initially in state $A (P_{A}, V_{A}, T_{i})$ . We study separately three quasi-static transformations, all starting from $A$ , with $T_{f} > T_{i}$ :

$A \to B$ : isochoric heating from $T_{i}$ to $T_{f}$ ;
$A \to C$ : isothermal compression at $T_{i}$ , from $V_{A}$ to $V_{A} /2$ ;
$A \to D$ : isobaric heating from $T_{i}$ to $T_{f}$ .

We use the banker's convention: transfers received by the gas are positive.

Plot the three transformations on the same Clapeyron diagram $(P, V)$ and formally determine states $B$ , $C$ , and $D$ .
Formally calculate $W$ , $Δ U$ , and $Q$ for each of the three transformations.
Comment on the signs of the energy transfers.
Numerical application: $n = 1 mol$ , $T_{i} = 300 K$ , $T_{f} = 500 K$ , and $R = 8, 314 J mo l^{- 1} K^{- 1}$ .

Hint

For a monatomic ideal gas,

Δ U = \frac{3}{2} n R Δ T

. Use

W = - \int P d V

and the first law

Δ U = Q + W

Detailed solution

Solution

Question 1. The ideal-gas equation gives

State A B C D P P_{A} P_{A} \frac{T _{f}}{T _{i}} 2 P_{A} P_{A} V V_{A} V_{A} \frac{V _{A}}{2} V_{A} \frac{T _{f}}{T _{i}} T T_{i} T_{f} T_{i} T_{f}

In the $(P, V)$ diagram, $A \to B$ is vertical, $A \to C$ follows the isotherm $T_{i}$ to the left, and $A \to D$ is horizontal to the right.

Question 2. Isochoric heating

A \to B

. Since

d V = 0

W_{A B} = 0, Δ U_{A B} = \frac{3}{2} n R (T_{f} - T_{i}), Q_{A B} = \frac{3}{2} n R (T_{f} - T_{i}) .

Isothermal compression $A \to C$ . The temperature is constant, so $Δ U_{A C} = 0$ . Moreover,

W_{A C} = - \int_{V_{A}}^{V_{A} /2} \frac{n R T _{i}}{V} d V = n R T_{i} ln 2, Q_{A C} = - n R T_{i} ln 2.

Isobaric heating $A \to D$ . At constant pressure, $P_{A} (V_{D} - V_{A}) = n R (T_{f} - T_{i})$ , so

W_{A D} = - n R (T_{f} - T_{i}), Δ U_{A D} = \frac{3}{2} n R (T_{f} - T_{i}), Q_{A D} = \frac{5}{2} n R (T_{f} - T_{i}) .

Question 3. Along the isochoric transformation, all the heat received increases the internal energy. During the isothermal compression, the gas receives work and releases exactly the same amount of energy as heat. Along the isobaric transformation, the gas receives heat, increases its internal energy, and does work as it expands.

Question 4. With

T_{f} - T_{i} = 200 K

Transformation Isochoric A \to B Isothermal A \to C Isobaric A \to D W (kJ) 0 + 1, 73 - 1, 66 Δ U (kJ) + 2, 49 0 + 2, 49 Q (kJ) + 2, 49 - 1, 73 + 4, 16