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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank First law of thermodynamics and Hess law

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    One mole of an ideal gas is allowed to expand reversibly and adibatically from a temperature of \[{{27}^{o}}C\]. If the work done during the process is \[3\,kJ\], then final temperature of the gas is \[({{C}_{V}}=20\,J/K)\]    [Pb. CET 2002]

    A)                 100 K    

    B)                 150 K

    C)                 195 K    

    D)                 255 K

    Correct Answer: B

    Solution :

               Given number of moles =1                    Initial temperature \[={{27}^{o}}C=300K\]                    Work done by the system \[=3\,KJ=3000K\]                    It will be \[(-)\]because work is done by the system.                    Heat capacity at constant volume \[(Cv)=20\,J/k\]                    We know that work done                    \[W=-n{{C}_{V}}\,({{T}_{2}}-{{T}_{1}})\];  \[3000=-1\times 20\,\,({{T}_{2}}-300)\]                    \[3000=-20{{T}_{2}}+6000\]                                 \[20{{T}_{2}}=3000\];  \[{{T}_{2}}=\frac{3000}{20}=150K\]


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