A) 4.01 kJ
B) -8.02 kJ
C) 18.02 kJ
D) -14.01 kJ
Correct Answer: A
Solution :
W = - 2.303 nRT \[\log \frac{{{p}_{1}}}{{{p}_{2}}}\] Here, \[n=1\,mol,\,R=8.314\,J{{K}^{-1}}mo{{l}^{-1}}\] T = 300K, \[{{p}_{1}}=1\,bar=1\,atm=1\times 1.01\times {{10}^{5}}Pa\] \[{{p}_{2}}=4\,bar=4\,atm=4\times 1.001\times {{10}^{5}}Pa\] \[\therefore \] Work done \[=-2.303\times 1\times 8.314\times 300\,\log \frac{1}{4}\] \[=-2.303\times 1\times 8.314\times 300\] \[\times (-2\times 0.3010)\] \[=3457.9\,J=3.46\,kJ\approx 4.01\,kJ\]
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