JEE Main & Advanced Chemistry States of Matter Question Bank Ideal gas equation and Related gas laws

  • question_answer A 10 g of a gas at atmospheric pressure is cooled from \[{{273}^{o}}C\] to \[{{0}^{o}}C\] keeping the volume constant, its pressure would become

    A)                 1/2 atm               

    B)                 1/273 atm

    C)                 2 atm   

    D)                 273 atm

    Correct Answer: A

    Solution :

               \[{{T}_{1}}={{273}^{o}}C=273+{{273}^{o}}K={{546}^{o}}K\]                    \[{{T}_{2}}={{0}^{o}}C=273+{{0}^{o}}C={{273}^{o}}K\]                    \[{{P}_{1}}=1\] ; \[{{P}_{2}}=?\]                    According to Gay-Lussac?s law                                 \[\frac{{{P}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}}{{{T}_{2}}}\] \[\therefore \] \[{{P}_{2}}=\frac{{{P}_{1}}{{T}_{2}}}{{{T}_{1}}}=\frac{1\times {{273}^{o}}K}{{{546}^{o}}K}\] atm;  \[\frac{1}{2}\] atm.


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