Solved thermodynamics exercise

Textbook exercise: Laplace's laws

Exercise 10 · Lesson 5 — Application to the Ideal Gas

textbook exercise
Laplace's law
adiabatic index
gamma
adiabatic expansion
differential form of the first law

Statement

Consider $n$ moles of an ideal gas with constant adiabatic index $\gamma = c_P/c_V$ . The system is assumed to be closed. Mayer's relation $c_P - c_V = R$ is taken as given (see the exercise on the derivation of $c_P$ ).

Show that $c_V = \frac{R}{\gamma - 1}, \qquad c_P = \frac{\gamma R}{\gamma - 1}.$
The gas undergoes a quasi-static adiabatic transformation for which $P_{\text{ext}} = P$ is assumed. Starting from the differential form of the first law, show that $\frac{\mathrm{d}T}{T} + (\gamma - 1)\frac{\mathrm{d}V}{V} = 0,$ then deduce that $TV^{\gamma-1} = \text{const}$ along the transformation.
Deduce the other two forms of Laplace's law: $PV^\gamma = \text{const}$ and $T^\gamma P^{1-\gamma} = \text{const}$ .
Consider such an expansion that doubles the volume of a monatomic gas ( $\gamma = 5/3$ ), with $n = 1$ mol and initial temperature $T_i = 300$ K. Calculate $T_f$ , then $\Delta T$ and $W$ , and comment on the signs of $\Delta T$ and $W$ .
In the Clapeyron diagram $(P,V)$ , compare the slope of such an adiabat with that of an isotherm at the same point.

Hint

For Question 2:

\mathrm{d}U = n c_V\,\mathrm{d}T

and

\delta W = -P\,\mathrm{d}V

with

P = nRT/V

, then separate the variables. For Question 5, differentiate

PV^\gamma = \text{const}

and

PV = \text{const}

Detailed solution

Solution

Question 1. From

c_P = \gamma c_V

and

c_P - c_V = R

, we obtain

c_V(\gamma - 1) = R

, which gives the two formulas.

Question 2. For an adiabatic transformation, $\delta Q = 0$ , so $\mathrm{d}U = \delta W$ , that is, $n c_V\,\mathrm{d}T = -P\,\mathrm{d}V = -\dfrac{nRT}{V}\mathrm{d}V$ . With $c_V = R/(\gamma-1)$ :

\frac{\mathrm{d}T}{T} = -(\gamma-1)\frac{\mathrm{d}V}{V}.

Integrating gives $\ln T + (\gamma-1)\ln V = \text{const}$ , that is, $TV^{\gamma-1} = \text{const}$ .

Question 3. Substituting $T = PV/(nR)$ gives $PV^\gamma = \text{const}$ because $n$ is constant. Substituting $V = nRT/P$ gives $T \left(\frac{T}{P}\right)^{\gamma-1} = \text{const}$ , and therefore $T^\gamma P^{1-\gamma} = \text{const}$ .

Question 4. The simplest approach is to use Laplace's law $T V^{\gamma - 1} = \text{const}$ , which gives $T_i V_i^{\gamma - 1} = T_f V_f^{\gamma - 1}$ , hence $T_f = T_i\,2^{-(\gamma-1)} = 300 \times 2^{-2/3} \approx 189$ K. Note that the temperature must be expressed in kelvins in this calculation to avoid an error! It follows that $\Delta T = T_f-T_i \approx -111$ K. For the work, since $Q = 0$ :

W = \Delta U = \tfrac{3}{2}R\Delta T \approx \tfrac{3}{2}\times 8{,}314 \times (-111) \approx -1{,}4\ \text{kJ}.

$W < 0$ : the gas does work on the surroundings during expansion. Since it cannot draw this energy from any heat reservoir ( $Q=0$ ), it draws on its internal energy, hence the cooling $\Delta T < 0$ .

Question 5. Differentiating $PV = \text{const}$ gives $\left(\frac{\mathrm{d}P}{\mathrm{d}V}\right)_{\text{isoT}} = -\dfrac{P}{V}$ . Differentiating $PV^\gamma = \text{const}$ gives $\left(\frac{\mathrm{d}P}{\mathrm{d}V}\right)_{\text{adiab}} = -\gamma\dfrac{P}{V}$ . The adiabat is $\gamma$ times steeper than the isotherm: compressing the gas without removing heat makes the pressure rise more rapidly because the temperature also increases.