Chapter 12 Practice Problems [Thermal & Statistical Mechanics I Studied #19]

Prob 12.1

In adiabatic expansion of an ideal gas, $pV^\gamma$ is a constant. Show that the following equation holds.

The two relations to show: T V^(γ−1) = constant and T = constant × p^((γ−1)/γ)

Proof: constant = pV^γ = (n_mRT/V)V^γ = n_mRT V^(γ−1), so T V^(γ−1) = constant; then T = constant × V^(1−γ) = constant × (n_mRT/p)^(1−γ) = constant × T^(1−γ) p^(γ−1), which boils down to T = constant × p^((γ−1)/γ)

Prob 12.2

Suppose a gas behaves according to the law given by $pV = f(T)$. Here $f(T)$ is a function of temperature. From this, show the following equation.

Proof: p = f(T)/V, hold V and differentiate by T to get (∂p/∂T)_V = (1/V) df/dT; V = f(T)/p, hold p to get (∂V/∂T)_p = (1/p) df/dT; ‘What, that’s simple!’

Also show that the following equation holds.

Proof: all we know is differentiating heat by T, so use the chain rule: C_p = (∂Q/∂T)_p = (∂Q/∂V)_p (∂V/∂T)_p, hence (∂Q/∂V)_p = C_p/(∂V/∂T)_p = C_p (∂T/∂V)_p; the same with C_V and p gives (∂Q/∂p)_V = C_V (∂T/∂p)_V

In an adiabatic change, the following equation is satisfied.

Therefore, show that $pV^\gamma$ is a constant.

The equation above is the one where $Q$ is thought of as a function defined by the two variables $p$ and $V$.

Earlier we proved

and now I’ll substitute it into the equation above.

Substituting into dQ = 0: C_V (∂T/∂p)_V dp + C_p (∂T/∂V)_p dV = 0, so C_V (∂T/∂p)_V dp = −C_p (∂T/∂V)_p dV

And I’ll substitute in what we proved before:

Substituting: after the cancelling (‘can to the cel’) V C_V dp = −C_p p dV, so (1/p) dp = −γ (1/V) dV; ‘indefinite integral of both sides, gogo’: ln p = −γ ln V + C, so ln(pV^γ) = C and pV^γ is constant

Prob 12.3

Explain why the following can be written as such.

Here $A$ and $B$ are just coefficients (they can change from state to state of the gas, but each is a plain number at a given state). From these equations,

show that this holds.

If Q is a function of T and p: dQ = (∂Q/∂T)_p dT + (∂Q/∂p)_T dp = C_p dT + (∂Q/∂p)_T dp

The author’s reasoning that (∂Q/∂p)_T ‘can be said to be a constant’ since Q = cmT = cm pV/(n_mR)

So dQ = C_p dT + A dp

If Q is a function of T and V: dQ = (∂Q/∂T)_V dT + (∂Q/∂V)_T dV = C_V dT + (∂Q/∂V)_T dV

The author’s reasoning that (∂Q/∂V)_T ‘can be said to be a constant’ since Q = cmT = cm pV/(n_mR)

So dQ = C_V dT + B dV

Setting the two dQ equal: C_p dT + A dp = C_V dT + B dV (the picture writes C_p instead of C_V on the right of the first line), so (C_p − C_V) dT = B dV − A dp, and ‘Huh? Done already!’

When the temperature is a constant, show that the following is satisfied.

Above we had

but $T$ being constant means $dT = 0$, i.e.

With dT = 0: 0 = B dV − A dp, so dp/dV = B/A, and since T is constant, (∂p/∂V)_T = B/A

Show that in an adiabatic change the following equation holds.

For an adiabatic change: dp = −(C_p/A) dT and dV = −(C_V/B) dT

This one is easy.

For an adiabatic change dQ = C_p dT + A dp = 0, so −A dp = C_p dT and dp = −(C_p/A) dT

The one below too:

Likewise dQ = C_V dT + B dV = 0 gives dV = −(C_V/B) dT

Therefore, show that in an adiabatic change the following equation is satisfied.

For an adiabatic change: (∂p/∂V)_adiabatic = γ (∂p/∂V)_T, (∂V/∂T)_adiabatic = 1/(1−γ) (∂V/∂T)_p and (∂p/∂T)_adiabatic = γ/(γ−1) (∂p/∂T)_V

First,

First relation: with dp = −(C_p/A) dT and dV = −(C_V/B) dT, (∂p/∂V)_adiabatic = (C_p/C_V)(B/A) = γ (∂Q/∂V)_T/(∂Q/∂p)_T = γ (∂p/∂V)_T

Diving into the second.

Second relation: C_V dT + B dV = 0 gives (∂V/∂T)_adiabatic = −C_V/B; with dp = 0, B = (C_p − C_V)(∂T/∂V)_p; so (∂V/∂T)_adiabatic = −1/(γ−1) (∂V/∂T)_p = 1/(1−γ) (∂V/∂T)_p

Let’s cook up the last equation too.

Third relation: (∂p/∂T)_adiabatic = −C_p/A; with dV = 0, A = −(C_p − C_V)(∂T/∂p)_V; so (∂p/∂T)_adiabatic = 1/(1 − 1/γ) (∂p/∂T)_V = γ/(γ−1) (∂p/∂T)_V

Prob 12.4

Using this, explain the slopes of adiabatic and isothermal on the $p$-$V$ graph in relation to each other.

The ratio of the adiabatic and isothermal slopes is γ; for a monatomic ideal gas γ = 5/3, so the adiabatic line’s slope turns out steeper

Prob 12.5

Two thermally insulated cylinders A and B of equal volume, each having a piston, are connected by a valve. Initially A has its piston completely pulled out and contains a monatomic gas inside the cylinder at temperature $T$; B has its piston pushed completely in, and the valve is closed.

After the operation described below is carried out, calculate the final temperature of the gas in each case.

In each operation the initial arrangement starts from the same state, and the heat capacity of the cylinder is neglected.

a) Fully open the valve and pull B’s piston all the way out so that the gas slowly enters into B. A’s piston is fixed so that it doesn’t move.

Hand sketch of the two connected cylinders A and B before and after: B’s piston is pulled out (arrows) and the gas spreads into B

Since it was said to be completely thermally insulated, $\delta Q = 0$.

The reason they emphasize that it moves slowly is that the change is reversible, which I take to mean that we can write $\delta W = -p\,dV$.

Derivation for a): δQ = 0 and reversible δW = −p dV give dU = −p dV = −(n_mRT/V) dV; dU = C_V dT for an ideal gas, so C_V (1/T) dT = −(n_mR/V) dV; integrating between the initial and final states (V to 2V) gives C_V ln(T₂/T₁) = −n_mR ln 2, so T₂ = T₁ 2^(−n_mR/C_V)

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

The two pictures that call $\left(\frac{\partial Q}{\partial p}\right)_T$ and $\left(\frac{\partial Q}{\partial V}\right)_T$ constants are wrong there: heat is not a function of state, so $Q = cmT$ does not apply. For an ideal gas at constant $T$, $\delta Q = p\,dV = -V\,dp$, so $A = -V$ and $B = p$. Nothing later needs them to be constant.
In the picture that sets the two $dQ$ equal, the right side should be $C_V\,dT + B\,dV$; the next line, $(C_p - C_V)\,dT = B\,dV - A\,dp$, has it right.

Prob 12.1

Prob 12.2

Prob 12.3

Prob 12.4

Prob 12.5

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