InterviewSolution
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InterviewSolution
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A volume of `10m^(3)` of a liquid is supplied with `100 kal` of heat and expands at a constant pressure of `10 atm` to a final volume of `10.2m^(3)`. Calculate the work done and change in internal energy. |
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Answer» Correct Answer - `48 Kcal. 52 Kcal`. Here, `V_(1)= 10m^(3), dQ=100 Kcal = 10^(5)cal.` `P= 10 atm = 10^(6)N//m^(2)` `V_(2)= 10.2 m^(3), U=?, dW=?` `dW=pdV=(10xx10^(5))(10.2-10)= 2xx10^(5)J` `= (2xx10^(5))/(4186)Kcal= 48 Kcal` From first law of thermodynamics, `dU=dQ-dW= 100-48= 52 Kcal` |
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