Enthalpy and Entropy Change of a Real Gas Across a Valve: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2013, Q7

Question 7: Enthalpy and Entropy Change of a Real Gas Across a Valve (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Isothermal expansion at $T = 300$ K (heat supplied to hold $T$ constant); real gas $Z = 1 + (B/RT)P$ with $B = 10$ cm³/mol $= 1.0\times10^{-5}$ m³/mol, constant. $P_1 = 10$ bar, $P_2 = 1$ bar; $C_v^{ig}=1.5R$.

Find. (a) $\Delta H$ across the valve; (b) $\Delta S$.

Figure 4 — Slow leak across a partially-open valve held isothermal (300 K) by rapid heat transfer; 10 bar → 1 bar.

Approach. Since $T$ is held constant, the ideal-gas contributions to $\Delta H$ and $\Delta S$ reduce to the isothermal pressure terms; add the residual (departure) properties obtained from the given EOS. The volume-explicit form $V = RT/P + B$ makes both departures elementary.

Residual properties from the EOS. Writing $V = RT/P + B$ with $B$ constant gives $(\partial V/\partial T)_P = R/P$. The standard departure integrals then collapse: $$H^R = \int_0^P\!\!\Big[V - T\big(\tfrac{\partial V}{\partial T}\big)_P\Big]dP = \int_0^P\! B\,dP = BP,\qquad S^R = -\int_0^P\!\!\Big[\big(\tfrac{\partial V}{\partial T}\big)_P-\tfrac{R}{P}\Big]dP = 0.$$ The residual entropy vanishes because $B$ does not depend on temperature.
Enthalpy change (part a). At constant $T$ the ideal-gas enthalpy change is zero, so only the residual term survives: $$\Delta H = \underbrace{\Delta H^{ig}}_{=0} + \Delta H^R = B(P_2-P_1) = (1.0\times10^{-5})(1\times10^{5}-10\times10^{5}) = \boxed{-9\ \text{J/mol}}.$$
Entropy change (part b). With $S^R=0$ at both states, $\Delta S$ is the ideal-gas isothermal pressure term: $$\Delta S = \underbrace{-R\ln\frac{P_2}{P_1}}_{\Delta S^{ig}} + \underbrace{\Delta S^R}_{=0} = -8.314\ln\!\frac{1}{10} = \boxed{+19.1\ \text{J/mol}\cdot\text{K}}.$$
Interpretation. Because heat is supplied to keep $T$ fixed, this is not an isenthalpic throttle: the enthalpy falls slightly ($-9$ J/mol, entirely the pressure dependence of the real-gas enthalpy), while the entropy rises with the pressure drop. The given $C_v^{ig}=1.5R$ is not needed here — it would only matter if the temperature changed (an adiabatic valve).

Quantity	Result
Residual enthalpy $H^R = BP$	at 10 bar: +10 J/mol; at 1 bar: +1 J/mol
(a) $\Delta H$ across valve	−9 J/mol
(b) $\Delta S$ across valve	+19.1 J/mol·K

Concept & Theory

Topic. Residual (departure) enthalpy and entropy for a pressure-explicit virial EOS; isothermal real-gas property changes.

Key relations. $H^R = \int_0^P[V-T(\partial V/\partial T)_P]\,dP$ and $S^R = -\int_0^P[(\partial V/\partial T)_P - R/P]\,dP$; for $Z=1+BP/RT$ these give $H^R = P[B - T\,dB/dT]$ and $S^R = -P\,dB/dT$. Ideal isothermal: $\Delta H^{ig}=0$, $\Delta S^{ig}=-R\ln(P_2/P_1)$.

Why this works. The two-term virial in pressure makes $V$ linear in $P$, so the departure integrals are trivial. The key physical insight is that a temperature-independent $B$ carries an enthalpy departure but no entropy departure — the gas's non-ideality shows up in energy, not in the number of accessible microstates at fixed $T$ and $P$.

Common pitfall. Assuming the valve is isenthalpic (Joule–Thomson). That holds only for an adiabatic throttle; here the problem explicitly keeps $T$ constant by heat transfer, so $\Delta H\ne0$ in general. Also watch units: $B$ in m³/mol times $P$ in Pa yields J/mol directly.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §6.2–6.3 (residual properties; property changes from a virial EOS).

Question 7 of 7: Enthalpy and Entropy Change of a Real Gas Across a Valve

Question 7: Enthalpy and Entropy Change of a Real Gas Across a Valve (equal value)