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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    1 gm of water at a pressure of \[1.01\times {{10}^{5}}\] Pa is converted into steam without any change of temperature. The volume of 1 g of steam is 1671 cc and the latent heat of evaporation is 540 cal. The change in internal energy due to evaporation of 1 gm of water is

    A) \[\approx 167\text{ }cal\]

    B) \[\approx 500\text{ }cal\]

    C) 540 cal

    D) 58 cal

    Correct Answer: B

    Solution :

    [b] \[dW=P\Delta V=1.01\times {{10}^{5}}\left[ 1671-1 \right]\times 10-6\text{ }Joule\] \[=\frac{1.01\times 167}{4.2}cal.=40\text{cal}\text{. nearly}\] \[\Delta Q=mL=1\times 540.\] \[\Delta Q=\Delta W+\Delta U\] \[\text{or }\Delta U=540-40=500\text{ cal}\text{. }\]


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