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Q.

A sample of 0.1 g of water at l00°C and normal pressure.(1.013 x 10⁵ Nm^-2 )requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is:

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a

42.2 J

b

208.7 J

c

104.3 J

d

84.5 J

answer is B.

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Detailed Solution

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$Q = 54 cal = 54 \times 4.18 joule = 225.72 joule W = P [V_{steam} - V_{water}] [For water 0.1 gram = 0.1 cc] = 0.013 \times 10^{5} [167.1 \times 10^{- 6} - 0.1 \times 10^{- 6}] joule = 1.013 \times 167 \times 10^{- 1} = 16.917 joule By FLOT \Rightarrow ∆ U = Q - W = 225.72 - 16.917 = 208.8 joule .$

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