JEE Main & Advanced Physics Thermodynamical Processes Question Bank Adiabatic Process

  • question_answer
     Two moles of an ideal monoatomic gas at \[{{27}^{o}}C\] occupies a volume of V. If the gas is expanded adiabatically to the volume \[2V,\] then the work done by the gas will be \[[\gamma =5/3,\,R=8.31J/mol\,K]\] [RPET 1999]

    A)            \[-2767.23J\]                        

    B)            \[2767.23J\]

    C)            \[2500J\]

    D)            \[-2500J\]

    Correct Answer: B

    Solution :

                       \[W=\frac{\mu R({{T}_{1}}-{{T}_{2}})}{(\gamma -1)}=\frac{\mu R{{T}_{1}}}{(\gamma -1)}\left[ 1-\frac{{{T}_{2}}}{{{T}_{1}}} \right]\]      \[=\frac{\mu R{{T}_{1}}}{(\gamma -1)}\left[ 1-{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}} \right]\]               \[=\frac{2\times 8.31\times 300}{\left( \frac{5}{3}-1 \right)}\left[ 1-{{\left( \frac{1}{2} \right)}^{\frac{5}{3}-1}} \right]\]\[=+2767.23\ J\]


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