ThermodynamicsEasy

Question

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

Options

A.

104.3 J

B.

208.7 J

C.

42.2 J

D.

84.5 J

Solution

∆Q = 54 cal = 54 × 4.18 joule = 225.72 joule
∆W = P[Vsteam – Vwater] [For water 0.1 gram=0.1 cc]
= 1.013 × 105[167.1 × 10–6 – 0.1 × 10–6] joule
= 1.013 × 167 × 10–1 = 16.917 joule
By FLOT
⇒∆U = ∆Q – ∆W = 225.72 – 16.917
∆U = 208.8 joule

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