Chemical Thermodynamics Question 125
Question: Work done during isothermal expansion of one mole of an ideal gas from 10 atom. to 1 atm at 300K is [BHU 2004]
Options:
A) 4938.8 J
B) 4138.8 J
C) 5744.1 J
D) 6257.2 J
Show Answer
Answer:
Correct Answer: C
Solution:
Given that $ P_1=10,atm $ , $ P_2=1,atm $ , $ T=300K $ , $ n=1 $
$ R=8.314,J/K/,mol $ Now, by using $ W=2.303,nRT{\log_{10}}\frac{P_2}{P_1} $
$ =2.303\times 1\times 8.314\times 300{\log_{10}}\frac{1}{10} $
$ W=5744.1,Joule $
