Chemical Thermodynamics Question 112

Question: Work done during isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 300 K is (Gas constant = 2) [AIIMS 2000]

Options:

A) 938.8 cal.

B) 1138.8 cal.

C) 1381.8 cal.

D) 1581.8 cal.

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Answer:

Correct Answer: C

Solution:

$ -,W=+,2.303,nRT, $ log $ \frac{p_1}{p_2} $

$ -,W=2.303\times 1\times 2\times 300 $ log $ \frac{10}{1} $

$ =1381.8,cal. $



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