A) \[9.22\times {{10}^{3}}J{{K}^{-1}}\]
B) \[8.22\times {{10}^{2}}J{{K}^{-1}}\]
C) \[2.31J{{K}^{-1}}\]
D) \[10.00\times {{10}^{3}}J{{K}^{-1}}\]
Correct Answer: A
Solution :
Pressure\[p=nRT/V\]. The work done by the gas during the isothermal expansion is \[W=\int_{{{V}_{1}}}^{{{V}_{2}}}{pdV}=nRT\int_{{{V}_{1}}}^{{{V}_{2}}}{\frac{dV}{V}}=nRT\]in\[\frac{{{V}_{2}}}{{{V}_{1}}}\] Substitute\[{{V}_{2}}=2{{V}_{1}}\]to obtain \[W=nRT\]in\[2\] \[=(4.00)(8.314)(400)in\,2\] \[=9.22\times {{10}^{3}}J{{K}^{-1}}\]
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