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AFMC AFMC Solved Paper-2006

  • question_answer
    The volume of a gas measured at \[{{27}^{o}}C\] and 1 atm pressure is 10 L. To reduce the volume to 2 L at 1 atm. pressure, the temperature required is:

    A) 60 K                                       

    B) 75 K

    C) 150 K                                    

    D) 225 K

    Correct Answer: A

    Solution :

    Here   \[{{V}_{1}}=10L,\,\,{{V}_{2}}=2L\] \[{{P}_{1}}=1atm,\,{{P}_{2}}=1\,atm\] \[{{T}_{1}}=300K,\,{{T}_{2}}=?\] \[\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\] \[\frac{1\times 10}{300}=\frac{1\times 2}{{{T}_{2}}}\] \[{{T}_{2}}=\frac{300\times 2}{10}=60K\]


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