Which statement about c_p − c_v = T v β²/κ_T is correct for a stable pure substance?

c_p − c_v is never negative, and it vanishes whenever β = 0, as for liquid water at its density maximum near 4 °C

c_p − c_v can be negative for liquids that contract on heating, because β is then negative

c_p − c_v equals R for every substance, since the relation is general and R is a universal constant

c_p − c_v is zero only for an ideal gas, because only an ideal gas has β = 1/T

The Clausius-Clapeyron equation ln(p₂/p₁) = (h_fg/R)(1/T₁ − 1/T₂) for a liquid-vapour equilibrium rests on which set of assumptions?

v_f ≪ v_g, the vapour behaves as an ideal gas, and h_fg is constant over the temperature range

v_f = v_g at the critical point, the liquid is incompressible, and s_fg is constant over the range

the process is isenthalpic, the vapour is incompressible, and the pressure is constant

the vapour is ideal, the liquid and vapour have equal specific volumes, and T is constant

For ordinary ice melting to liquid water at 0 °C, the density of ice is lower than that of water. The melting curve on the p-T diagram therefore has a slope dp/dT that is

negative, because v_sf = v_liquid − v_solid is negative while h_sf and T are positive

positive, because h_sf is positive and the Clapeyron equation then always gives a positive slope

zero, because melting occurs at constant temperature and pressure

positive, because the Clausius-Clapeyron equation applies and v_g is large

negative, because v_sf = v_liquid − v_solid is negative while h_sf and T are positive

A hypothetical gas obeys p(v − b) = RT, with b > 0 and constant c_p. Which statement about its behaviour on throttling is correct?

μ_JT = −b/c_p, so the gas warms on throttling at every temperature

μ_JT = 0, because u of this gas is independent of volume, as for an ideal gas

μ_JT = +b/c_p, so the gas cools on throttling at every temperature

μ_JT = 2b/c_p at low temperature and −b/c_p at high temperature

Which of the following statements are correct for a simple compressible substance?

Saturated liquid and saturated vapour at the same T and p have equal specific Gibbs functions

A spontaneous change at constant T and v has da ≤ 0

Saturated liquid and saturated vapour at the same T and p have equal specific Gibbs functions

A spontaneous change at constant T and v has da ≤ 0

A spontaneous change at constant T and p has da ≤ 0

Which of the following statements about throttling and the Joule-Thomson effect are correct?

An ideal gas has μ_JT = 0 at all temperatures and pressures

A real gas throttled at a state where μ_JT > 0 is cooled as its pressure falls

The inversion curve is the locus where T(∂v/∂T)p = v

An ideal gas has μ_JT = 0 at all temperatures and pressures

A real gas throttled at a state where μ_JT > 0 is cooled as its pressure falls

Steady adiabatic throttling is an isentropic process

The inversion curve is the locus where T(∂v/∂T)p = v

Which of the following relations are valid for a pure simple compressible substance?

(∂u/∂v)T = T(∂p/∂T)v − p

Thermodynamic Relations: T-ds Equations, Maxwell Relations, Joule-Thomson Effect and the Clapeyron Equation

This chapter belongs to the Engineering Sciences (XE) paper. It answers Section XE4.6 of XE4 Thermodynamics, one of the nine optional sections, any two of which a candidate chooses in the exam: “T-ds relations, Helmholtz and Gibbs functions, Gibbs relations, Maxwell relations, Joule-Thomson coefficient, coefficient of volume expansion, adiabatic and isothermal compressibilities, Clapeyron and Clausius-Clapeyron equations.” It teaches the T-ds and Gibbs relations, the four thermodynamic potentials, the Maxwell relations and how to use them, the expansion coefficient and the two compressibilities with c_p − c_v, the Joule-Thomson coefficient with its inversion curve, and the Clapeyron and Clausius-Clapeyron equations. The numericals are c_p − c_v of a liquid from β and κ_T, the entropy change of a van der Waals gas, a Joule-Thomson coefficient, a vapour-pressure estimate and the pressure needed to shift a melting point. All constants are stated in each stem.

1. T-ds relations, Helmholtz and Gibbs functions, the Gibbs relations

For a simple compressible substance in a reversible process, δq = T ds and δw = p dv. The first law du = δq − δw therefore gives the first T-ds relation, and substituting h = u + pv gives the second:
T ds = du + p dv and T ds = dh − v dp.
These relations contain only properties, so they hold for every process, reversible or not, between neighbouring equilibrium states of a simple compressible substance; only the identification of T ds with heat needs reversibility. Rearranged they are the energy forms of the Gibbs relations: du = T ds − p dv and dh = T ds + v dp.

Two more potentials remove s as the natural variable. The Helmholtz function a = u − Ts and the Gibbs function g = h − Ts are properties. Differentiating and using the relations above:
da = −s dT − p dv and dg = −s dT + v dp.
Hence (∂a/∂T)v = −s, (∂a/∂v)T = −p, (∂g/∂T)p = −s and (∂g/∂p)T = v. Each potential is a function of its own natural pair of variables, u(s,v), h(s,p), a(T,v), g(T,p), and from one of them all other properties follow by differentiation.

Potential	Definition	Differential	Spontaneous change at fixed …
u	u	du = T ds − p dv	s, v: du ≤ 0
h	u + pv	dh = T ds + v dp	s, p: dh ≤ 0
Helmholtz a	u − Ts	da = −s dT − p dv	T, v: da ≤ 0
Gibbs g	h − Ts	dg = −s dT + v dp	T, p: dg ≤ 0

🎯 Why g matters for phase change

At fixed T and p a spontaneous change lowers g, so two phases in equilibrium have equal g. For saturated liquid and vapour, g_f = g_g, i.e. h_f − T s_f = h_g − T s_g, which gives s_fg = h_fg/T_sat. This one equality is the starting point of the Clapeyron equation in section 5.

2. Maxwell relations and their use

Each of the four differentials is exact, so the mixed second derivatives are equal. Applying this to du, dh, da and dg gives the four Maxwell relations:

From	Relation
du	(∂T/∂v)s = −(∂p/∂s)v
dh	(∂T/∂p)s = +(∂v/∂s)p
da	(∂s/∂v)T = +(∂p/∂T)v
dg	(∂s/∂p)T = −(∂v/∂T)p

🧠 Remembering the signs

The last two matter most because they replace entropy, which cannot be measured, by p, v and T, which can. Derive them on the spot: from da = −s dT − p dv, differentiate −s with respect to v and −p with respect to T, which gives −(∂s/∂v)T = −(∂p/∂T)v. From dg = −s dT + v dp the cross terms have opposite signs, hence the minus in (∂s/∂p)T = −(∂v/∂T)p.

Uses. Combine T ds = du + p dv with the third relation at constant T: (∂u/∂v)T = T(∂s/∂v)T − p = T(∂p/∂T)v − p. In the same way (∂h/∂p)T = v − T(∂v/∂T)p. Also (∂c_p/∂p)T = −T(∂²v/∂T²)p. For an ideal gas the first gives T·R/v − p = 0 (u depends on T only), the second gives v − R T/p = 0 (h depends on T only), and the third gives 0. So only a real-gas equation of state makes them non-zero.

Worked example. One mole of a van der Waals gas, (p + a/v²)(v − b) = RT, expands isothermally. Then p = RT/(v − b) − a/v², so (∂p/∂T)v = R/(v − b) and (∂s/∂v)T = R/(v − b). Integrating at constant T, s₂ − s₁ = R ln[(v₂ − b)/(v₁ − b)]. The constant a drops out of the entropy change; the molecular volume b alone matters. For v₁ = 1.0×10⁻³ and v₂ = 5.0×10⁻³ m³/mol with b = 4.0×10⁻⁵ m³/mol, Δs = 8.314 ln(4.96/0.96) = 13.65 J/mol·K, slightly more than the ideal-gas value R ln 5 = 13.38.

3. Expansion coefficient, compressibilities and c_p − c_v

Three measurable coefficients describe how volume responds to T and p:
β = (1/v)(∂v/∂T)p (volume expansion coefficient, K⁻¹)
κ_T = −(1/v)(∂v/∂p)T (isothermal compressibility, Pa⁻¹)
κ_s = −(1/v)(∂v/∂p)s (adiabatic compressibility, Pa⁻¹).
For an ideal gas β = 1/T and κ_T = 1/p; κ_s = 1/(γp), since p vᵞ = constant along an isentrope. Stable substances have κ_T > 0 and κ_s > 0: squeezing never increases the volume.

Writing s(T,v) and using the third Maxwell relation, the general result is
c_p − c_v = T v β² / κ_T
and the ratio of the two compressibilities equals the ratio of the specific heats: c_p/c_v = κ_T/κ_s. Because T, v, κ_T > 0 and β² ≥ 0, c_p ≥ c_v always. The two are equal when β = 0 (for liquid water near its density maximum, close to 4 °C at 1 atm) and as T → 0. For an ideal gas, T v (1/T)²/(1/p) = pv/T = R, recovering c_p − c_v = R.

Worked example. A liquid at 300 K has v = 1.0×10⁻³ m³/kg, β = 2.0×10⁻⁴ K⁻¹ and κ_T = 5.0×10⁻¹⁰ Pa⁻¹. Then c_p − c_v = 300 × 1.0×10⁻³ × (2.0×10⁻⁴)² / 5.0×10⁻¹⁰ = 24 J/kg·K. Compared with c_p of about 4000 J/kg·K for a typical liquid the gap is under 1%, which is why liquids and solids are taken to have c_p ≈ c_v. For a gas c_p − c_v = R is a large fraction of c_v.

4. The Joule-Thomson coefficient and the inversion curve

Steady adiabatic throttling through a valve or porous plug, with negligible changes of kinetic and potential energy and no work, keeps enthalpy constant: h₁ = h₂. The Joule-Thomson coefficient is the temperature change per unit pressure drop at constant enthalpy, μ_JT = (∂T/∂p)h. Writing h(T,p) and using (∂h/∂p)T = v − T(∂v/∂T)p from section 2:
μ_JT = [T(∂v/∂T)p − v] / c_p = (v/c_p)(Tβ − 1).

Condition	Sign of μ_JT	Effect of throttling (pressure falls)
Tβ > 1	μ_JT > 0	gas cools
Tβ = 1 (inversion point)	μ_JT = 0	temperature unchanged
Tβ < 1	μ_JT < 0	gas warms

For an ideal gas Tβ = 1 exactly, so μ_JT = 0 at every T and p: throttling an ideal gas does not change its temperature, in agreement with h = h(T). The effect is a real-gas effect, and the line on a T-p diagram where μ_JT = 0 is the inversion curve; inside it gas cools, outside it warms. Refrigeration and gas liquefaction use the cooling region, so a gas must first be brought below its maximum inversion temperature.

van der Waals gas. At low pressure v ≈ RT/p + b − a/(RT), so T(∂v/∂T)p − v = 2a/(RT) − b and
μ_JT ≈ (2a/(RT) − b)/c_p,
which is positive below the zero-pressure inversion temperature T_inv = 2a/(Rb) and negative above it. Since the critical temperature is T_c = 8a/(27Rb), T_inv = (27/4)T_c = 6.75 T_c. This is twice the Boyle temperature a/(Rb) = (27/8)T_c. The model overestimates real inversion temperatures, but it gives the right form: attraction (a) cools, finite molecular size (b) warms.

⚠️ Throttling is not isentropic

Throttling is constant enthalpy, not constant entropy: it is highly irreversible and s rises. Reversible adiabatic expansion in a turbine always cools the gas (μ_s = (∂T/∂p)s = Tvβ/c_p > 0 for β > 0), while throttling can cool, leave unchanged or warm it. Also note that μ_JT = 0 for an ideal gas does not make the gas unusual: it is the sign of Tβ − 1 that decides.

5. The Clapeyron and Clausius-Clapeyron equations

Along a saturation curve g_f = g_g at every point, so dg_f = dg_g, i.e. −s_f dT + v_f dp = −s_g dT + v_g dp. Rearranging, with s_fg = h_fg/T:
dp_sat/dT = s_fg/v_fg = h_fg / (T v_fg) (the Clapeyron equation).
It is exact for any phase change (vaporisation, sublimation, melting) with h and v the differences between the two phases, and it needs only measurable quantities. For melting use h_sf and v_sf = v_liquid − v_solid.

Clausius-Clapeyron equation. Far below the critical point, for liquid-vapour or solid-vapour change, make three approximations: v_g ≫ v_f so v_fg ≈ v_g, the vapour is an ideal gas so v_g = RT/p, and h_fg is constant over the range. Then dp/p = (h_fg/R) dT/T², and
ln(p₂/p₁) = (h_fg/R)(1/T₁ − 1/T₂),
with R the gas constant of the vapour (R_u/M) and T in kelvin. A plot of ln p against 1/T is therefore a straight line of slope −h_fg/R.

Worked example. Water boils at 100 °C (373.15 K) at 101.325 kPa with h_fg = 2257 kJ/kg and R = 0.4615 kJ/kg·K for steam. At 90 °C (363.15 K): ln(p₂/p₁) = (2257/0.4615)(1/373.15 − 1/363.15) = 4890.6 × (−7.38×10⁻⁵) = −0.361, so p₂ = 101.325 × e^(−0.361) = 70.6 kPa. For melting, water is exceptional: v_sf = 1.000×10⁻³ − 1.0905×10⁻³ = −9.05×10⁻⁵ m³/kg < 0, so with h_sf = 333.7 kJ/kg at 273.15 K, dp/dT = 333.7×10³/(273.15 × (−9.05×10⁻⁵)) = −13.5 MPa/K. The melting curve slopes backwards, and about 13.5 MPa lowers the melting point by 1 K.

⚠️ Where Clausius-Clapeyron fails

Using the Clausius-Clapeyron form for melting is wrong, because v_fg ≈ v_g does not apply to a solid-liquid change. It also degrades near the critical point, where v_g is not ≫ v_f, the vapour is far from ideal and h_fg falls to zero. Use temperatures in kelvin and R of the vapour (not R_u unless h_fg is per mole).

6. Reducing a partial derivative: a method

Exam questions ask for a derivative that no table lists, such as (∂s/∂p)v or (∂T/∂v)u. The reduction is mechanical.
(i) If s appears in the numerator or denominator at fixed T, use a Maxwell relation to remove it.
(ii) If a derivative is taken at constant s, use the chain rule with T ds = c_p dT − T(∂v/∂T)p dp or T ds = c_v dT + T(∂p/∂T)v dv.
(iii) If u or h appears, use (∂u/∂v)T or (∂h/∂p)T from section 2.
(iv) Finally express everything through p, v, T, c_p, c_v, β and κ_T. The cyclic rule (∂x/∂y)z (∂y/∂z)x (∂z/∂x)y = −1 and the reciprocal rule help to exchange fixed variables.

Worked example. Find μ_s = (∂T/∂p)s. From T ds = c_p dT − T(∂v/∂T)p dp with ds = 0: c_p dT = T(∂v/∂T)p dp, so (∂T/∂p)s = T(∂v/∂T)p/c_p = Tvβ/c_p. For an ideal gas this is v/c_p = RT/(p c_p), which on integration gives T ∝ p^(R/c_p) = p^((γ−1)/γ), the familiar isentropic relation. Another check: if a gas obeys p(v − b) = RT, then (∂u/∂v)T = T R/(v − b) − p = 0, yet μ_JT = (T R/p − v)/c_p = −b/c_p < 0, so such a gas warms on throttling at every temperature.

Key takeaways

T ds = du + p dv = dh − v dp hold between any two equilibrium states; da = −s dT − p dv and dg = −s dT + v dp, so (∂g/∂T)p = −s, (∂g/∂p)T = v, and g_f = g_g at saturation.
Maxwell relations swap entropy for measurable derivatives: (∂s/∂v)T = (∂p/∂T)v and (∂s/∂p)T = −(∂v/∂T)p. They give (∂u/∂v)T = T(∂p/∂T)v − p and (∂h/∂p)T = v − T(∂v/∂T)p, both zero for an ideal gas.
β = (1/v)(∂v/∂T)p, κ_T = −(1/v)(∂v/∂p)T, c_p − c_v = Tvβ²/κ_T ≥ 0 and c_p/c_v = κ_T/κ_s. For an ideal gas β = 1/T, κ_T = 1/p and c_p − c_v = R.
μ_JT = [T(∂v/∂T)p − v]/c_p is zero for an ideal gas and changes sign at the inversion curve (Tβ = 1); a van der Waals gas at low pressure inverts at 2a/(Rb) = 6.75 T_c. Throttling is isenthalpic, not isentropic.
Clapeyron: dp/dT = h_fg/(T v_fg), exact for any phase change. Clausius-Clapeyron, ln(p₂/p₁) = (h_fg/R)(1/T₁ − 1/T₂), needs v_g ≫ v_f, ideal vapour and constant h_fg, with T in kelvin.
Water melts with a negative slope dp/dT because ice is less dense than liquid (v_sf < 0); Clausius-Clapeyron is never used for melting.

Practice questions (16)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which pair correctly defines the Helmholtz function a and the Gibbs function g of a simple compressible substance?
1. a = u + Ts and g = h + Ts
2. a = u − Ts and g = h − Ts
3. a = h − Ts and g = u − Ts
4. a = u − pv and g = h − pv
Show answer
Answer: B — a = u − Ts and g = h − Ts
The Helmholtz function is a = u − Ts and the Gibbs function is g = h − Ts; these are the Legendre transforms that make T, v and T, p the natural variables. The swapped version pairs the wrong energy with each function, and subtracting pv from u or h gives an enthalpy-like quantity, not a free energy.
Which of the following Maxwell relations, derived from dg = −s dT + v dp, is correct?
1. (∂s/∂p)T = +(∂v/∂T)p
2. (∂s/∂v)T = −(∂p/∂T)v
3. (∂T/∂p)s = −(∂v/∂s)p
4. (∂s/∂p)T = −(∂v/∂T)p
Show answer
Answer: D — (∂s/∂p)T = −(∂v/∂T)p
From dg = −s dT + v dp, equality of the cross derivatives gives −(∂s/∂p)T = (∂v/∂T)p, so (∂s/∂p)T = −(∂v/∂T)p. The version with a plus sign is the tempting slip; the relation (∂s/∂v)T = +(∂p/∂T)v comes from da, not dg, and has no minus sign; and dh gives (∂T/∂p)s = +(∂v/∂s)p.
The Joule-Thomson coefficient μ_JT = (∂T/∂p)h of a pure substance is equal to
1. [v − T(∂v/∂T)p]/c_p
2. [T(∂v/∂T)p − v]/c_v
3. [T(∂v/∂T)p − v]/c_p
4. T(∂v/∂T)p/c_p
Show answer
Answer: C — [T(∂v/∂T)p − v]/c_p
With h(T,p), dh = c_p dT + (∂h/∂p)T dp = 0 gives (∂T/∂p)h = −(∂h/∂p)T/c_p = −[v − T(∂v/∂T)p]/c_p = [T(∂v/∂T)p − v]/c_p. The version with the sign reversed is the negative of the answer, and the one with c_v uses the wrong specific heat. The form T(∂v/∂T)p/c_p, with no −v, is (∂T/∂p)s, the isentropic coefficient, not the Joule-Thomson one.
Which statement about c_p − c_v = T v β²/κ_T is correct for a stable pure substance?
1. c_p − c_v is never negative, and it vanishes whenever β = 0, as for liquid water at its density maximum near 4 °C
2. c_p − c_v can be negative for liquids that contract on heating, because β is then negative
3. c_p − c_v equals R for every substance, since the relation is general and R is a universal constant
4. c_p − c_v is zero only for an ideal gas, because only an ideal gas has β = 1/T
Show answer
Answer: A — c_p − c_v is never negative, and it vanishes whenever β = 0, as for liquid water at its density maximum near 4 °C
β enters squared, and T, v, κ_T are positive, so c_p − c_v ≥ 0 even when β < 0; it is zero when β = 0, as for water at its density maximum. A negative β does not make the difference negative. The value R holds only for an ideal gas, where it is not zero but R; and the claim that it vanishes only for an ideal gas confuses the ideal-gas value β = 1/T with the condition β = 0.
The Clausius-Clapeyron equation ln(p₂/p₁) = (h_fg/R)(1/T₁ − 1/T₂) for a liquid-vapour equilibrium rests on which set of assumptions?
1. v_f ≪ v_g, the vapour behaves as an ideal gas, and h_fg is constant over the temperature range
2. v_f = v_g at the critical point, the liquid is incompressible, and s_fg is constant over the range
3. the process is isenthalpic, the vapour is incompressible, and the pressure is constant
4. the vapour is ideal, the liquid and vapour have equal specific volumes, and T is constant
Show answer
Answer: A — v_f ≪ v_g, the vapour behaves as an ideal gas, and h_fg is constant over the temperature range
Starting from the exact Clapeyron equation dp/dT = h_fg/(T v_fg), the three approximations v_fg ≈ v_g, v_g = RT/p and constant h_fg give the integrated form. Equal specific volumes would make v_fg zero and dp/dT infinite, and isenthalpic or constant-T conditions are not part of the derivation.
For ordinary ice melting to liquid water at 0 °C, the density of ice is lower than that of water. The melting curve on the p-T diagram therefore has a slope dp/dT that is
1. positive, because h_sf is positive and the Clapeyron equation then always gives a positive slope
2. zero, because melting occurs at constant temperature and pressure
3. positive, because the Clausius-Clapeyron equation applies and v_g is large
4. negative, because v_sf = v_liquid − v_solid is negative while h_sf and T are positive
Show answer
Answer: D — negative, because v_sf = v_liquid − v_solid is negative while h_sf and T are positive
Clapeyron's dp/dT = h_sf/(T v_sf) has positive h_sf and T, so the sign is that of v_sf. Ice is less dense, so v_liquid < v_solid and v_sf < 0 and the slope is negative: raising pressure lowers the melting point. The claim of a positive slope ignores v_sf; Clausius-Clapeyron is not applicable to melting because no vapour is involved.
A van der Waals gas at vanishing pressure has maximum Joule-Thomson inversion temperature T_inv = 2a/(Rb). In terms of its critical temperature T_c = 8a/(27Rb), the ratio T_inv/T_c is
1. 3.375
2. 6.75
3. 0.296
4. 2.000
Show answer
Answer: B — 6.75
T_inv/T_c = [2a/(Rb)]/[8a/(27Rb)] = 54/8 = 6.75. The value 3.375 is the Boyle-temperature ratio (a/Rb)/T_c = 27/8, half the inversion value; 0.296 is its reciprocal-type slip 8/27; and 2 is just the numerator coefficient.
A hypothetical gas obeys p(v − b) = RT, with b > 0 and constant c_p. Which statement about its behaviour on throttling is correct?
1. μ_JT = −b/c_p, so the gas warms on throttling at every temperature
2. μ_JT = 0, because u of this gas is independent of volume, as for an ideal gas
3. μ_JT = +b/c_p, so the gas cools on throttling at every temperature
4. μ_JT = 2b/c_p at low temperature and −b/c_p at high temperature
Show answer
Answer: A — μ_JT = −b/c_p, so the gas warms on throttling at every temperature
v = RT/p + b, so T(∂v/∂T)p − v = RT/p − (RT/p + b) = −b and μ_JT = −b/c_p < 0: the gas warms. Although (∂u/∂v)T = T R/(v − b) − p = 0 here (so u depends on T alone), h = u + pv = u + RT + bp still depends on p, so μ_JT is not zero; that is the tempting error. There is no sign change because T cancels in the numerator.
Which of the following statements are correct for a simple compressible substance?
1. (∂g/∂T)p = −s and (∂g/∂p)T = v
2. Saturated liquid and saturated vapour at the same T and p have equal specific Gibbs functions
3. A spontaneous change at constant T and v has da ≤ 0
4. A spontaneous change at constant T and p has da ≤ 0
Show answer
Answer: A — (∂g/∂T)p = −s and (∂g/∂p)T = v; B — Saturated liquid and saturated vapour at the same T and p have equal specific Gibbs functions; C — A spontaneous change at constant T and v has da ≤ 0
From dg = −s dT + v dp the two partial derivatives follow. At saturation the phases are in equilibrium at fixed T and p, so g_f = g_g. At constant T and v the criterion is the fall of the Helmholtz function. The criterion at constant T and p is the fall of g, not a; da ≤ 0 is not the condition there.
Which of the following statements about throttling and the Joule-Thomson effect are correct?
1. An ideal gas has μ_JT = 0 at all temperatures and pressures
2. A real gas throttled at a state where μ_JT > 0 is cooled as its pressure falls
3. Steady adiabatic throttling is an isentropic process
4. The inversion curve is the locus where T(∂v/∂T)p = v
Show answer
Answer: A — An ideal gas has μ_JT = 0 at all temperatures and pressures; B — A real gas throttled at a state where μ_JT > 0 is cooled as its pressure falls; D — The inversion curve is the locus where T(∂v/∂T)p = v
μ_JT ∝ T(∂v/∂T)p − v, which vanishes identically for an ideal gas, is positive where the gas cools, and is zero on the inversion curve, T(∂v/∂T)p = v. Throttling keeps enthalpy constant and generates entropy, so it is not isentropic.
Which of the following relations are valid for a pure simple compressible substance?
1. (∂u/∂v)T = T(∂p/∂T)v − p
2. (∂c_p/∂p)T = −T(∂²v/∂T²)p
3. c_p/c_v = κ_T/κ_s
4. (∂s/∂p)T = +(∂v/∂T)p
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Answer: A — (∂u/∂v)T = T(∂p/∂T)v − p; B — (∂c_p/∂p)T = −T(∂²v/∂T²)p; C — c_p/c_v = κ_T/κ_s
The u-relation follows from T ds = du + p dv and (∂s/∂v)T = (∂p/∂T)v. Differentiating (∂s/∂p)T = −(∂v/∂T)p with respect to T and using c_p = T(∂s/∂T)p gives the c_p relation. The ratio c_p/c_v = κ_T/κ_s is the standard compressibility-ratio result. The (∂s/∂p)T relation with a plus sign is wrong; the correct Maxwell relation carries a minus.
A liquid at 300 K has specific volume 1.0×10⁻³ m³/kg, volume expansion coefficient β = 2.0×10⁻⁴ K⁻¹ and isothermal compressibility κ_T = 5.0×10⁻¹⁰ Pa⁻¹. Using c_p − c_v = Tvβ²/κ_T, the value of c_p − c_v is ____ J/kg·K (nearest integer).
Numerical answer — type the value.
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Answer: 24
c_p − c_v = 300 × 1.0×10⁻³ × (2.0×10⁻⁴)² / 5.0×10⁻¹⁰ = 300 × 1.0×10⁻³ × 4.0×10⁻⁸ / 5.0×10⁻¹⁰ = 1.2×10⁻⁸/5.0×10⁻¹⁰ = 24 J/kg·K. Omitting the square on β would give 0.12.
Water boils at 373.15 K at 101.325 kPa. Take h_fg = 2257 kJ/kg (constant), steam as an ideal gas with R = 0.4615 kJ/kg·K, and v_f negligible. Using the Clausius-Clapeyron equation, the saturation pressure at 363.15 K is ____ kPa (one decimal place).
Numerical answer — type the value.
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Answer: 70.6
ln(p₂/p₁) = (2257/0.4615)(1/373.15 − 1/363.15) = 4890.6 × (−7.380×10⁻⁵) = −0.3609, so p₂ = 101.325 × e^(−0.3609) = 70.63 kPa, i.e. 70.6. Using Celsius temperatures would give a meaningless answer.
At 273.15 K, melting ice has latent heat h_sf = 333.7 kJ/kg, specific volume of liquid water 1.000×10⁻³ m³/kg and of ice 1.0905×10⁻³ m³/kg. By the Clapeyron equation, the magnitude of the pressure rise needed to lower the melting point by 1 K (treating dp/dT as constant) is ____ MPa (one decimal place).
Numerical answer — type the value.
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Answer: 13.5
dp/dT = h_sf/(T v_sf) = 333.7×10³ / [273.15 × (1.000×10⁻³ − 1.0905×10⁻³)] = 333.7×10³ / (−2.472×10⁻²) = −1.350×10⁷ Pa/K. The magnitude is 13.5 MPa per kelvin; the negative sign shows pressure lowers the melting point.
A gas obeys the van der Waals equation with a = 0.137 Pa·m⁶/mol² and b = 3.87×10⁻⁵ m³/mol. At 300 K and vanishing pressure its molar c_p is 30.0 J/mol·K. Using μ_JT ≈ (2a/(RT) − b)/c_p with R = 8.314 J/mol·K, the Joule-Thomson coefficient is ____ K/MPa (two decimal places).
Numerical answer — type the value.
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Answer: 2.37
2a/(RT) = 2 × 0.137/(8.314 × 300) = 1.0986×10⁻⁴ m³/mol. Subtract b: 1.0986×10⁻⁴ − 3.87×10⁻⁵ = 7.116×10⁻⁵ m³/mol. Divide by c_p: 7.116×10⁻⁵/30.0 = 2.372×10⁻⁶ K/Pa = 2.37 K/MPa. Positive, so the gas cools on throttling at 300 K; T_inv = 2a/(Rb) = 852 K is above 300 K.
One mole of a van der Waals gas with b = 4.0×10⁻⁵ m³/mol expands isothermally from 1.0×10⁻³ m³ to 5.0×10⁻³ m³. Using the Maxwell relation (∂s/∂v)T = (∂p/∂T)v and R = 8.314 J/mol·K, the entropy change of the gas is ____ J/K (two decimal places).
Numerical answer — type the value.
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Answer: 13.65
p = RT/(v − b) − a/v², so (∂p/∂T)v = R/(v − b). Then Δs = R ln[(v₂ − b)/(v₁ − b)] = 8.314 ln(4.96×10⁻³/9.6×10⁻⁴) = 8.314 × 1.6422 = 13.65 J/K. The constant a plays no role. The ideal-gas value R ln 5 = 13.38 is the tempting but different result.