If 1 cm3 of water is vaporized (latent heat of vaporization = 540 cal/g0C ) at P = 1 atm. If the volume of steam formed is 1671 cm3, calculate increase in internal energy.

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204 cal

167 cal

540 cal

373 cal

answer is D.

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Detailed Solution

∆Q = ∆U + ∆Wm = 1 gmLv = 540 cal/gm∆Q = 1×540 = 540540 = ∆U + P∆V540 = ∆U + 105×(1671-1)×10-6540 = ∆U + 167∆U = 540-167 = 373 cal

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