Chemical Thermodynamics Question 49

Question: One mole of water at $ 100^{o}C $ is converted into steam at $ 100^{o}C $ at a constant pressure of 1 atm. The change in entropy is [heat of vaporisation of water at $ 100^{o}C=540,cal/gm $ ] [Pb. PMT 2004]

Options:

A) 8.74

B) 18.76

C) 24.06

D) 26.06

Show Answer

Answer:

Correct Answer: D

Solution:

The entropy change $ =\frac{\text{heat of vaporisation}}{temperature} $ Here, heat of vaporisation $ =540,cal/gm $

$ =540\times 18,cal,mo{{l}^{-1}} $ Temperature of water $ =100+273=373K $

$ \therefore $ entropy change $ =\frac{540\times 18}{373}=26.06,cal,mo{{l}^{-1}}{{K}^{-1}} $



NCERT Chapter Video Solution

Dual Pane