NEET Sample Paper NEET Sample Test Paper-19

  • question_answer One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of \[27{}^\circ C\]. The work done is 3 kJ. The final temperature of the gas is equal to\[[{{C}_{v}}=20k{{J}^{-1}}\,]:\]

    A)  450 K              

    B)  150 K

    C)  600 K                         

    D)  300 K

    Correct Answer: B

    Solution :

    Reversible work done during adiabatic expansion of ideal gas under the condition is \[W={{\,}_{n}}{{C}_{v}}({{T}_{1}}-{{T}_{2}})\] Work done by the gas\[=-3\,kJ=-3000\,J\] \[{{C}_{v}}=20\,J/mol\,K\] \[n=1\] \[{{T}_{1}}=273+27=300\,K,{{T}_{2}}=?\] 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 \[W=-2.303\,nRT\log \frac{{{V}_{2}}}{{{V}_{1}}}\] Hence, the correct option is [b].


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