Chapter 11 Practice Problems [Thermal & Statistical Mechanics I Studied #17]

Prob 11.1

1 mole of monatomic ideal gas is trapped in a cylinder by a piston, and is kept at constant temperature $T_0$ by being in contact with a heat reservoir. This gas expands slowly while being kept at the same temperature $T_0$, and its volume changes from $V_1$ to $V_2$.

Why does the internal energy of the gas not change? Calculate the work done by the gas and the heat that flowed into the gas.

First lines of the answer: for the ideal gas U = N·(3/2)k_BT and the temperature is held at T₀, so U does not change; dU = 0 = δQ + δW, hence δQ = −δW

Here $\delta W$ means work done on the gas, so it is negative during expansion. For the logarithm below, we also assume a frictionless, reversible expansion; slow by itself is not enough. Then the work done by the gas and the heat supplied are both $n_mRT_0\ln(V_2/V_1)$.

Heat for the slow, reversible expansion from V₁ to V₂: ΔQ = ∫δQ = −∫δW = ∫p dV = n_mRT ∫dV/V = n_mRT ln(V₂/V₁)

So the gas does n_mRT ln(V₂/V₁) of work while the same amount, n_mRT ln(V₂/V₁), of heat flows in

Prob 11.2

Show that for an ideal gas, the following holds.

For these two identities, use molar heat capacities, so $C_p-C_V=R$ and $\gamma=C_p/C_V$. The notes below also show total capacities for $n_m$ moles; those instead satisfy $C_p-C_V=n_mR$. The $3/2$ line uses the monatomic example; the heat-capacity difference above holds for any ideal gas. Here’s how I wrapped it up in the last post.

From the last post: C_V = (∂U/∂T)_V = ∂U/∂T = (3/2)n_mR, so dU = C_V dT

C_p from the first law: the red term (∂U/∂V)_T is 0 for an ideal gas because U does not depend on V, and the ideal gas law gives C_p = C_V + n_mR

Showing R/C_V = γ − 1: γ − 1 = (C_p − C_V)/C_V = n_mR/C_V, and per 1 mol this is R/C_V

Showing R/C_p = (γ − 1)/γ: (γ − 1)/γ = (C_p/C_V − 1)/(C_p/C_V) = (R/C_V)/(C_p/C_V) = R/C_p

Prob 11.3

Consider the following differential.

The differential dz = 2xy dx + (x² + 2y) dy and two paths made of straight segments from (x₁, y₁) to (x₂, y₂): i) along x to (x₂, y₁), then along y; ii) along y to (x₁, y₂), then along x

Path i): the two straight pieces give y₁(x₂² − x₁²) and x₂²(y₂ − y₁) + (y₂² − y₁²), in total x₂²y₂ − x₁²y₁ + y₂² − y₁²

Path-ii picture mistakenly repeats the path-i integrals in the other order; the final total is nevertheless right, x₂²y₂ − x₁²y₁ + y₂² − y₁²

Huh, it doesn’t depend on the path.

Prob 11.4

Polar coordinates x = r cosθ, y = r sinθ. From the definition of x, (∂x/∂r) = cosθ = x/r. Differentiating x² + y² = r² by r instead gives ∂x/∂r = r/x. So ∂x/∂r = ∂r/∂x???? What went wrong?

The answer: (∂x/∂r)_θ = cosθ holds θ constant, while differentiating x² + y² = r² by r holds y constant, so 2x(∂x/∂r)_y = 2r and (∂x/∂r)_y = r/x; the two partial derivatives are strictly different things

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

For path ii), the picture repeats the two pieces of path i) in the other order. Path ii) has its own pieces: along $x = x_1$, $\int_{y_1}^{y_2}(x_1^2 + 2y)\,dy = x_1^2(y_2 - y_1) + (y_2^2 - y_1^2)$, and along $y = y_2$, $\int_{x_1}^{x_2} 2xy_2\,dx = y_2(x_2^2 - x_1^2)$. They add up to the same $x_2^2y_2 - x_1^2y_1 + y_2^2 - y_1^2$, because $dz$ is exact.
The isothermal logarithm assumes a reversible expansion; slow alone does not guarantee it. Here $\delta W$ is work on the gas. The identities with $R/C_V$ and $R/C_p$ use molar capacities; the $3/2$ heat capacity is the monatomic example.

Prob 11.1

Prob 11.2

Prob 11.3

Prob 11.4

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