A) \[1.9518\times {{10}^{4}}J\]
B) \[2.6472\times {{10}^{7}}J\]
C) \[1.3527\times {{10}^{5}}J\]
D) \[2.2226\times {{10}^{6}}j\]
Correct Answer: A
Solution :
Key Idea: Work done in reversible process under isothermal conditions is calculated by using following formula \[W=-2.303\text{ }nRT\text{ }\log \frac{{{P}_{1}}}{{{P}_{2}}}\] Given, \[n=2,\text{ }T=27{}^\circ C=27+273=300\text{ }K\] \[{{p}_{1}}=1.01\times {{10}^{5}}N{{m}^{-2}},\text{ }{{P}_{2}}=5.05\times {{10}^{6}}N{{m}^{-2}}\] Substituting the values \[=-2.303\times 2\times 8.314\times 300\log \frac{1.01\times {{10}^{5}}}{5.05\times {{10}^{6}}}\] \[=-11488.28\log \frac{1.01\times {{10}^{5}}}{5.05\times {{10}^{6}}}\] \[=1.9518\times {{10}^{4}}J\]
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