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Test: Clausius-Clapeyron Equation - Chemistry MCQ


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10 Questions MCQ Test Physical Chemistry - Test: Clausius-Clapeyron Equation

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Test: Clausius-Clapeyron Equation - Question 1

During phase transitions like vaporization, melting and sublimation

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 1

This is what happens during a phase transition.

Test: Clausius-Clapeyron Equation - Question 2

The Clausius-Clapeyron equation is given by

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 2

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

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Test: Clausius-Clapeyron Equation - Question 3

The vapour pressure p in kPa at temperature T can be given by the relation

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 3

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

Test: Clausius-Clapeyron Equation - Question 4

The slope of sublimation curve is ____ the slope of the vaporization curve at triple point.

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 4

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

Test: Clausius-Clapeyron Equation - Question 5

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.

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 5

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.

Test: Clausius-Clapeyron Equation - Question 6

Estimate the freezing temperature of liquid water at a pressure of 30 MPa.

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 6

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.

Test: Clausius-Clapeyron Equation - Question 7

Which of the following requirement is satisfied by a phase change of the first order?

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 7

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

Test: Clausius-Clapeyron Equation - Question 8

Water ____ on melting and has the fusion curve with a ____ slope.

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 8

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

Test: Clausius-Clapeyron Equation - Question 9

According to Trouton’s rule, the ratio of latent heat of vaporization to the boiling point at 1.013 bar is

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 9

This is the statement of Trouton’s rule.

Test: Clausius-Clapeyron Equation - Question 10

Ice (solid water) at −3°C and 100 kPa, is compressed isothermally until it becomes liquid. Find the required pressure.

Detailed Solution for Test: Clausius-Clapeyron Equation - Question 10

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.

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