AFMC AFMC Solved Paper-2010

  • question_answer
    Ten moles of an ideal gas at constant temperature 600 K is compressed from 100 L to 10 L. The work done in the process is

    A) \[4.11\times 104\text{ }J\]                          

    B) \[-4.11\times 104\text{ }J\]

    C) \[11.4\times 104\text{ }J\]                          

    D) \[-11.4\times 104\text{ }J\]

    Correct Answer: D

    Solution :

    Work done in an isothermal process\[W=2.3026nRT{{\log }_{10}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] \[=2.3026\times 10\times 8.3\times 600\text{ lo}{{\text{g}}_{10}}\left( \frac{10}{100} \right)\]\[=-11.4\times {{10}^{4}}\text{J}\]


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