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).

The lowest temperature for R-12 is -90°C, so it must be extended to -140°C using the Clapeyron equation.
at T1= -90°C = 183.2 K, P1 = 2.8 kPa
R = 8.3145/120.914 = 0.068 76 kJ/kg K
ln P/P1 = (hfg/R)(T-T1)/(T*T1)
= (189.748/0.068 76)[(133.2 – 183.2)/(133.2 × 183.2)] = -5.6543
P = 2.8 exp(-5.6543) = 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.

Water, triple point T = 0.01°C, P = 0.6113 kPa, vf = 0.001 m³/kg,
hf = 0.01 kJ/kg, vi= 0.001 0908 m³/kg, hi = -333.4 kJ/kg
dPif/dT = (hf – hi)/[(vf – vi)T] = 333.4/(-0.0000908 × 273.16) = -13442 kPa/K
∆P = (dPif/dT)*∆T = -13442(-3 – 0.01) = 40460 kPa
P = P(tp) + ∆P = 40461 kPa.