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NEET Sample Paper NEET Sample Test Paper-23

  • question_answer
    Temperature of 1 mole of an ideal gas is increased from 300 K to 310 K under isochoric process. Heat supplied to the gas in this process is Q = 25 R where R = universal gas constant. What amount of work has to be done by the gas if temperature of the gas decreases from 310 K to 300 K adiabatically

    A)  10 R                

    B)  50 R

    C)  25 R               

    D) \[\left( \frac{25}{2} \right)R\]

    Correct Answer: C

    Solution :

    \[\Delta Q=n{{C}_{v}}\Delta T\] \[\therefore \] \[25R=(1){{C}_{v}}(310-300)\]or \[{{C}_{v}}=\frac{5}{2}R\] i.e., the gas is diatomic or \[\gamma =1.4\] Now, work done in adiabatic process \[W=\frac{nR({{T}_{i}}+{{T}_{f}})}{\gamma -1}\] \[=\frac{1.R(310-300)}{1.4-1}=25R\] Hence, the correction option is [c].


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