JCECE Engineering JCECE Engineering Solved Paper-2006

  • question_answer
    If the volume of gas is changed from \[{{V}_{1}}\] to \[{{V}_{2}}\] isothermally, then work done is:

    A) \[RT\ln \frac{{{V}_{1}}}{{{V}_{2}}}\]                        

    B) \[RT\ln \frac{{{V}_{2}}}{{{V}_{1}}}\]

    C) \[R({{T}_{2}}-{{T}_{1}})\ln \frac{{{V}_{2}}}{{{V}_{1}}}\]  

    D)  \[R({{V}_{2}}-{{V}_{1}})\ln \frac{{{T}_{2}}}{{{T}_{1}}}\]

    Correct Answer: B

    Solution :

    Key Idea: Ideal gas equation is                 \[PV=RT\] Work done in isothermal process is                 \[W=\int_{{{V}_{1}}}^{{{V}_{2}}}{p\,\,dV}\] From ideal gas law\[PV=RT\] \[\therefore \]  \[W=\int_{{{V}_{1}}}^{{{V}_{2}}}{\left( \frac{RT}{V} \right)}dV=RT\int_{{{V}_{1}}}^{{{V}_{2}}}{\frac{dV}{V}}\] \[\Rightarrow \]               \[W=RT\ln \left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\]


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