Test: Clausius-Clapeyron Equation - Chemistry MCQ

This is what happens during a phase transition.

Here vf is the final specific volume and vi is the initial specific volume and l is the latent heat.

Here Tb is the boiling point at 1.013 bar and this relation comes from the latent heat of vaporization and Trouton’s rule.

This is because at triple point, l(sublimation) > l(vaporization).

Step 1: Choose a reference saturation point and convert to Kelvin.
We use tabulated data at −90 °C because saturation tables for R-12 list values here.
T₁ = −90 °C + 273.15 = 183.15 K, P₁ (sat. vapour pressure at T₁) = 2.8 kPa
T₂ = −140 °C + 273.15 = 133.15 K

Step 2: Compute the specific gas constant for R-12.
Universal gas constant Rᵤ = 8.3145 kJ/kmol·K
Molar mass of R-12 M = 120.914 kg/kmol
R = Rᵤ / M = 8.3145 / 120.914 = 0.06876 kJ/kg·K

Step 3: Obtain latent heat of vaporisation at T₁.
From thermodynamic tables (Çengel & Boles), h_fg at T₁ = 189.75 kJ/kg

Step 4: Apply the integrated Clausius–Clapeyron equation.
ln(P₂ / P₁) = (h_fg / R) · (T₂ − T₁) / (T₁ · T₂)
Let A = h_fg / R = 189.75 / 0.06876 ≈ 2759.7 K
ΔT = T₂ − T₁ = 133.15 − 183.15 = −50 K
T₁·T₂ = 183.15 × 133.15 = 24380 K²
Exponent = A·ΔT / (T₁·T₂) = 2759.7 × (−50) / 24380 ≈ −5.655
Therefore P₂ = P₁ × exp(−5.655) = 2.8 × exp(−5.655) ≈ 2.8 × 0.003496 = 0.0098 kPa

Answer: 0.0098 kPa

At the triple point,
vif = vf – vi = 0.001000 – 0.0010908 = -0.0000908 m³/kg
hif = hf – hi = 0.01 – (-333.40) = 333.41 kJ/kg
dPif/dT = 333.41/[(273.16)(-0.0000908)] = -13 442 kPa/K
at P = 30 MPa, T = 0.01 + (30 000-0.6)/(-13 442) = = -2.2°C.

These requirements must be satisfied for a phase change to be of first order.

Unlike other substances which expands on melting, water contracts on melting and hence the slope of the fusion curve is negative.

This is the statement of Trouton’s rule.