Thermodynamics

JEE Main level, single correct, medium

38 of 190

Question 38 of 190

1g1g of a liquid is converted to vapour at 3×105Pa3\times {10}^{5}Pa pressure. If 1010% of the heat supplied is used for increasing the volume by 1600cm31600{cm}^{3} during this phase change, then the increase in internal energy in the process will be :

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Answer: (A) 4320J4320J

As pressure against which expansion is happening is constant, work done can be written as =PΔV=P\Delta V
=(3×105)×(1600×10−6)=\left(3\times {10}^{5}\right)\times \left(1600\times {10}^{-6}\right)
=480J=480J
As only 1010% of heat is used in work done.
Hence ΔQ=W0.1=4800J\Delta Q=\dfrac{W}{0.1}=4800J
The rest goes in increase in internal energy, which is 9090% of heat.
Change in internal energy =0.9×4800=4320J=0.9\times 4800=4320J

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