Thermodynamics Multiple Choice Questions on “Clausius-Clapeyron Equation”.
1. During phase transitions like vaporization, melting and sublimation
a) pressure and temperature remains constant
b) volume and entropy changes
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: This is what happens during a phase transition.
2. Which of the following requirement is satisfied by a phase change of the first order?
a) there are changes of volume and entropy
b) the first-order derivative of the Gibbs function changes discontinuously
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These requirements must be satisfied for a phase change to be of first order.
3. The Clausius-Clapeyron equation is given by
a) dp/dT = l / T(vf+vi)
b) dp/dT = l / T(vf-vi)
c) dT/dp = l / T(vf+vi)
d) dT/dp = l / T(vf-vi)
Answer: b
Clarification: Here vf is the final specific volume and vi is the initial specific volume and l is the latent heat.
4. Water ____ on melting and has the fusion curve with a ____ slope.
a) contracts, negative
b) contracts, positive
c) expands, negative
d) expands, positive
Answer: a
Clarification: Unlike other substances which expands on melting, water contracts on melting and hence the slope of the fusion curve is negative.
5. The vapour pressure curve is of the form ln(p) = A + B/T + C*lnT + DT.
a) true
b) false
Answer: a
Clarification: This is the form of vapour pressure curve where A,B,C, and D are constants.
6. According to Trouton’s rule, the ratio of latent heat of vaporization to the boiling point at 1.013 bar is
a) 77 kJ/kgmol K
b) 88 kJ/kgmol K
c) 99 kJ/kgmol K
d) 100 kJ/kgmol K
Answer: b
Clarification: This is the statement of Trouton’s rule.
7. The vapour pressure p in kPa at temperature T can be given by the relation
a) p = 101.325 exp (88/R)(1+T/Tb)
b) p = 101.325 exp (88/R)(1+Tb/T)
c) p = 101.325 exp (88/R)(1-T/Tb)
d) p = 101.325 exp (88/R)(1-Tb/T)
Answer: d
Clarification: Here Tb is the boiling point at 1.013 bar and this relation comes from the latent heat of vaporization and Trouton’s rule.
8. At the triple point, l(sublimation) = l(vaporization) – l(fusion).
a) true
b) false
Answer: b
Clarification: At the triple point, l(sublimation) = l(vaporization) + l(fusion), where l is the latent heat.
9. The slope of sublimation curve is ____ the slope of the vaporization curve at triple point.
a) equal to
b) less than
c) greater than
d) none of the mentioned
Answer: c
Clarification: This is because at triple point, l(sublimation) > l(vaporization).
10. Latent heat of sublimation is given by
a) l(sublimation) = -2.303*(R)*(d(1/T)/d(log p ))
b) l(sublimation) = +2.303*(R)*(d(log p )/d(1/T))
c) l(sublimation) = +2.303*(R)*(d(1/T)/d(log p ))
d) l(sublimation) = -2.303*(R)*(d(log p )/d(1/T))
Answer: d
Clarification: This is the expression for finding the latent heat of sublimation.
11. An application requires R-12 at −140°C. The triple-point temperature is −157°C. Find the pressure of the saturated vapour at the required condition.
a) 0.0058 kPa
b) 0.0098 kPa
c) 0.0068 kPa
d) 0.0088 kPa
Answer: b
Clarification: 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.
12. Ice (solid water) at −3°C and 100 kPa, is compressed isothermally until it becomes liquid. Find the required pressure.
a) 20461 kPa
b) 30461 kPa
c) 40461 kPa
d) 50461 kPa
Answer: c
Clarification: Water, triple point T = 0.01°C, P = 0.6113 kPa, vf = 0.001 m^3/kg,
hf = 0.01 kJ/kg, vi= 0.001 0908 m^3/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.
13. Estimate the freezing temperature of liquid water at a pressure of 30 MPa.
a) -2.2°C
b) 0°C
c) -0.2°C
d) -1.2°C
Answer: a
Clarification: At the triple point,
vif = vf – vi = 0.001000 – 0.0010908 = -0.0000908 m^3/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.