JCECE Engineering JCECE Engineering Solved Paper-2003

  • question_answer
    The volume of a gas is reduced to \[1.0\,\,L\] at \[{{25}^{o}}C\] and \[1\,\,atm\] pressure. Its pressure at \[{{35}^{o}}C\] would be:

    A) \[0.96\,\,atm\]                 

    B) \[1.03\,\,atm\]

    C) \[2.04\,\,atm\]                 

    D)  \[3.08\,\,atm\]

    Correct Answer: B

    Solution :

    Key Idea:\[\frac{{{P}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}}{{{T}_{2}}}\] Given    \[{{P}_{1}}=1\,\,atm,\,\,{{P}_{2}}=?\]                 \[{{T}_{1}}={{25}^{o}}C=25+273=298\,\,K\]                 \[{{T}_{2}}={{35}^{o}}C=35+273=308\,\,K\]                 \[\frac{1}{298}=\frac{{{P}_{2}}}{308}\] or            \[{{P}_{2}}=\frac{308}{298}=1.03\,\,atm\]


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