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  4. 1 cm3 of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671 cm3 . If the atmospheric pressure =1.013× 105 N/m2 and the mechanical equivalent of heat= 4.19J/cal, the energy spent in this process in overcoming intermolecular forces is

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Q. $ 1\,c{{m}^{3}} $ of water at its boiling point absorbs 540 cal of heat to become steam with a volume of $ 1671\,c{{m}^{3}} $ . If the atmospheric pressure $ =1.013\times {{10}^{5}}\,N/{{m}^{2}} $ and the mechanical equivalent of heat= 4.19J/cal, the energy spent in this process in overcoming intermolecular forces is

Rajasthan PMTRajasthan PMT 2009Thermodynamics

Solution:

Change in volume $ =1671-1=1670\,\,c{{m}^{3}} $ $ \therefore $ Work done $ (W)=p\cdot dV $ $ =(1.013\times {{10}^{5}}\times 1670)J $ $ =\frac{1670\times 1.013\times {{10}^{5}}}{4.2}cal $ $ =39.7cal $ Heat given $ (Q)=540\,\,cal $ . From first law of thermodynamics, $ \Delta U=Q-W $ $ =540-39.7 $ $ =500.3\,\,cal $ $ \approx 500\,\,cal $ .