Chemical Thermodynamics Question 380
Question: One mole of an ideal gas is expanded isothermally and reversibly to half of its initial pressure. $ \Delta S $ for the process in $ J{{K}^{-1}}mo{{l}^{-1}} $ is $ [ln2=0.693,and,R=8.314,J/(mol,K)] $
Options:
A) 6.76
B) 5.76
C) 10.76
D) 8.03
Show Answer
Answer:
Correct Answer: B
Solution:
- For isothermal process $ ( \Delta T=0 ) $
$ \Delta S=R\ell n\frac{P_1}{P_2}=8.314\ell n,2, $
$ =8.314\times 0.693=5.76 $
