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Q.

Molar enthalpy change for vapourisation of 1 mole of water at 1 bar and $100 ° c$ is 41KJ mol^-1. Calculate the internal energy change, when 1 mole of water is vaporised at
1bar pressure and $100 ° c$ .

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a

44.904 KJ mol^-1

b

37,904 J mol^-1

c

41 KJ mol^-1

d

37.904 J mol^-1

answer is D.

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Detailed Solution

$H_{2} O_{(l)} \to H_{2} O_{(g)} ∆ H = ∆ U + ∆ n_{g} RT ∆ U = ∆ H - ∆ n_{g} RT Given, ∆ H = 41 \times 10^{3} J {mol}^{- 1}, T = 100 ° + 273 = 373 K ∆ U = 41 \times 10^{3} - 1 (8.3) 373 = 41 \times 10^{3} - 3.096 \times 10^{3} = 37.904 \times 10^{3} J {mol}^{- 1}$

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