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Q.

The boiling point of a substance X at 1 atm pressure is 500 K. The enthalpy of vapourisation at the boiling point of X(l) is 80 kl/mol.The molar specific heat of X (l) = 5 × 10-3 kJ/mol and Cp of X(g) = 5 × 10-4 kJ / mol. What is the molar latent heat of vaporization of X(1) at 800 K and at 1 atm?

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a

Lesser than 80 kJ

b

Greater than 80 kJ

c

Equal to 80 kJ

d

Since 800 K is the higher temperature than the boiling point of X(l); therefore, enthalpy of vaportzation is not applicable.

answer is A.

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Detailed Solution

ΔH1=5×10−3×(500−800)×10−3kJΔH2=80kJ;ΔH3=5×10−4×(800−500)×10−3kJΔH1=−0.0015kJ;ΔH2=80kJΔH3=0.00015kJ;ΔH=ΔH1+ΔH2+ΔH3

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The boiling point of a substance X at 1 atm pressure is 500 K. The enthalpy of vapourisation at the boiling point of X(l) is 80 kl/mol.The molar specific heat of X (l) = 5 × 10-3 kJ/mol and Cp of X(g) = 5 × 10-4 kJ / mol. What is the molar latent heat of vaporization of X(1) at 800 K and at 1 atm?