A) 75 K
B) 150 K
C) 225 K
D) 300 K
Correct Answer: B
Solution :
Work done during \[\frac{\text{reversible}}{\text{adiabatic}}\] expansion of ideal gas under conditions is \[\text{W}\,\text{=}{{\,}_{n}}{{\text{C}}_{V}}({{T}_{2}}-{{T}_{1}})\] Work done by the gas\[~=-\text{ }3\text{ }kJ=-\text{ }3000\text{ }J\] \[{{C}_{v}}=20\,J/mol\,K\] \[n=1\] \[{{T}_{1}}=273+27=300\,K,{{T}_{2}}=?\] Putting the values \[-3000=1\times 20({{T}_{2}}-300)\] \[-150={{T}_{2}}-300\] \[300-150={{T}_{2}}\] \[\Rightarrow \] \[{{T}_{2}}=150\,K\] NOTE: Work done during isothermal irreversible expansion of ideal gas \[W=-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})\] Work done during reversible isothermal expansion of an ideal gas \[W=-2.303\,nRT\,\log \frac{{{V}_{2}}}{{{V}_{1}}}\]
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