JEE Main & Advanced Sample Paper JEE Main - Mock Test - 23

  • question_answer
    A polyatomic gas with six degrees of freedom does 25 J of work when it is expanded at constant pressure. The heat given to the gas is

    A) \[100\text{ }J\]                

    B)        \[150\text{ }J\]                

    C) \[200\text{ }J\]                

    D)       \[250\text{ }J\]

    Correct Answer: A

    Solution :

    [a] \[dW=25=P({{V}_{2}}-{{V}_{1}})=nR({{T}_{2}}-{{T}_{1}})\] \[dQ=dU+dW-n{{C}_{v}}dT+25\] \[=n.\frac{R}{\gamma -1}({{T}_{2}}-{{T}_{1}})+25\] \[=\frac{25}{\frac{4}{3}-1}+25=100J\left\{ \gamma =\frac{f+2}{f}=\frac{6+2}{6}=\frac{4}{3} \right\}\]

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