JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    One mole of an ideal gas at an initial temperature of T K does 6R joules of word adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be

    A) \[(T-4)K\]          

    B)        \[(T+2.4)K\]

    C) \[(T-2.4)K\]        

    D)        \[(T+4)K\]

    Correct Answer: A

    Solution :

    [a] \[{{T}_{1}}=T,W=6R\text{ joules},\gamma =\frac{5}{3}\] \[W=\frac{{{P}_{1}}{{V}_{1}}-{{P}_{2}}{{V}_{2}}}{\gamma -1}=\frac{nR{{T}_{1}}-nR{{T}_{2}}}{\gamma -1}\] \[=\frac{nR\left( {{T}_{1}}-{{T}_{2}} \right)}{\gamma -1}\] \[n=1,{{T}_{1}}=T\Rightarrow \frac{R\left( T-{{T}_{2}} \right)}{5/3-1}=6R\] \[\Rightarrow {{T}_{2}}=\left( T-4 \right)K\]


You need to login to perform this action.
You will be redirected in 3 sec spinner