Problem Statement
Express the Joule-Thomson coefficient $\mu_{JT} = (\partial T/\partial p)_H$ in terms of measurable quantities using a Maxwell relation.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: From $dH = TdS + Vdp$ and the Maxwell relation from $G$: $(\partial S/\partial p)_T = -(\partial V/\partial T)_p$.
Step 2 — Apply the relevant physical law or equation: At constant $H$: $0 = T(\partial S/\partial p)_H + V$, and using the triple product rule:
Step 3 — Solve algebraically for the unknown: $$\mu_{JT} = \left(\frac{\partial T}{\partial p}\right)_H = \frac{1}{C_p}\left[T\left(\frac{\partial V}{\partial T}\right)_p – V\right]$$
Step 4 — Substitute numerical values with units: For an ideal gas, $V = \nu RT/p$ so $(\partial V/\partial T)_p = \nu R/p = V/T$, giving $\mu_{JT}=0$. ✓
Step 5 — Compute and check the result: For a van der Waals gas to first order: $\mu_{JT} = (2a/RT – b)/C_p$, which is positive (cooling on expansion) below the inversion temperature.
Worked Calculation
$$\mu_{JT} = \left(\frac{\partial T}{\partial p}\right)_H = \frac{1}{C_p}\left[T\left(\frac{\partial V}{\partial T}\right)_p – V\right]$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{\mu_{JT} = \left(\frac{\partial T}{\partial p}\right)_H = \frac{1}{C_p}\left[T\left(\frac{\partial V}{\partial T}\right)_p – V\right]}$$
From $dH = TdS + Vdp$ and the Maxwell relation from $G$: $(\partial S/\partial p)_T = -(\partial V/\partial T)_p$.
At constant $H$: $0 = T(\partial S/\partial p)_H + V$, and using the triple product rule:
$$\mu_{JT} = \left(\frac{\partial T}{\partial p}\right)_H = \frac{1}{C_p}\left[T\left(\frac{\partial V}{\partial T}\right)_p – V\right]$$
For an ideal gas, $V = \nu RT/p$ so $(\partial V/\partial T)_p = \nu R/p = V/T$, giving $\mu_{JT}=0$. ✓
For a van der Waals gas to first order: $\mu_{JT} = (2a/RT – b)/C_p$, which is positive (cooling on expansion) below the inversion temperature.
Answer
$$\boxed{\mu_{JT} = \left(\frac{\partial T}{\partial p}\right)_H = \frac{1}{C_p}\left[T\left(\frac{\partial V}{\partial T}\right)_p – V\right]}$$
Physical Interpretation
Maxwell’s thermodynamic relations connect seemingly unrelated thermodynamic derivatives, allowing quantities that are hard to measure directly (like entropy changes at constant pressure) to be computed from measurable ones.
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