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NEET Sample Paper NEET Sample Test Paper-80

  • question_answer
    Minimum amount of work done is required to compress 5.00 mol of an ideal gas isothermally from 200 L to 40 L is

    A)  +20.1 kJ                     

    B)  - 20.1 kJ

    C)  -20.1 kJ                       

    D)  +20.1 kJ

    Correct Answer: A

    Solution :

    Work done in isothermal compression can be calculated as follows  \[W=-nRT\,\,In\,\,\frac{{{V}_{2}}}{{{V}_{1}}}\] \[=\,\,-5\times 8.314\times 300\times 2.303 log \frac{40}{200}\] \[=\,\,\,-1500 \times \,\,8.314\,\,\times \,\,2.303\,\,\times \,\, \left( -\,log 5 \right)\] \[=\,\,\,+ 20.1 kJ\]


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