Solved thermodynamics exercise

Textbook exercise: Laplace's laws

Exercise 18 · 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 $γ = c_{P} / c_{V}$ . 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 $c_{V} = \frac{R}{γ - 1}, c_{P} = \frac{γ R}{γ - 1} .$
The gas undergoes a quasi-static adiabatic transformation for which $P_{ext} = P$ is assumed. Starting from the differential form of the first law, show that $\frac{d T}{T} + (γ - 1) \frac{d V}{V} = 0,$ then deduce that $T V^{γ - 1} = const$ along the transformation.
Deduce the other two forms of Laplace's law: $P V^{γ} = const$ and $T^{γ} P^{1 - γ} = const$ .
Consider such an expansion that doubles the volume of a monatomic gas ( $γ = 5/3$ ), with $n = 1$ mol and initial temperature $T_{i} = 300$ K. Calculate $T_{f}$ , then $Δ T$ and $W$ , and comment on the signs of $Δ T$ 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:

d U = n c_{V} d T

and

δ W = - P d V

with

P = n R T / V

, then separate the variables. For Question 5, differentiate

P V^{γ} = const

and

P V = const

Detailed solution

Solution

Question 1. From

c_{P} = γ c_{V}

and

c_{P} - c_{V} = R

, we obtain

c_{V} (γ - 1) = R

, which gives the two formulas. Question 2. For an adiabatic transformation,

δ Q = 0

, so

d U = δ W

, that is,

n c_{V} d T = - P d V = - \frac{n R T}{V} d V

. With

c_{V} = R / (γ - 1)

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

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

Question 3. Substituting

T = P V / (n R)

gives

P V^{γ} = const

because

n

is constant. Substituting

V = n R T / P

gives

T (\frac{T}{P})^{γ - 1} = const

, and therefore

T^{γ} P^{1 - γ} = const

. Question 4. The simplest approach is to use Laplace's law

T V^{γ - 1} = const

, which gives

T_{i} V_{i}^{γ - 1} = T_{f} V_{f}^{γ - 1}

, hence

T_{f} = T_{i} 2^{- (γ - 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

Δ T = T_{f} - T_{i} \approx - 111

K. For the work, since

Q = 0

W = Δ U = \frac{3}{2} R Δ T \approx \frac{3}{2} \times 8, 314 \times (- 111) \approx - 1, 4 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 $Δ T < 0$ .

Question 5. Differentiating

P V = const

gives

(\frac{d P}{d V})_{isoT} = - \frac{P}{V}

. Differentiating

P V^{γ} = const

gives

(\frac{d P}{d V})_{adiab} = - γ \frac{P}{V}

. The adiabat is

γ

times steeper than the isotherm: compressing the gas without removing heat makes the pressure rise more rapidly because the temperature also increases.