Calculate change in enthalpy when 2 moles of liquid water at 1 bar and 100° C is converted into steam at 2 bar and 300° C. Assume $H_{2} O$ vapours to behave ideally.
[Latent heat of vaporization of $H_{2} O$ (l) at 1 bar and 100° C is 10.8 Kcal per mole]
[R = 2 cal / mol K]

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21.6 kcal

11.8 kcal

24.8 kcal

23.6 kcal

answer is C.

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Detailed Solution

In the reaction we have,

$\begin{aligned} H_{2} O (l) ⟶ H_{2} O (g) (100^{\circ} C) ⟶ H_{2} O (g) (300^{\circ} C) \\ 1 bar 2 bar \end{aligned}$

$\begin{array}{l} q_{1} = 2 \times 10.8 kcal \\ = Δ H_{1} = 21.6 kcal \\ Δ H_{2} = n C_{p} Δ T = 2 \times \frac{4 \times 2 \times 200}{1000} = 3.2 kcal \\ Δ H = Δ H_{1} + Δ H_{2} = 24.8 kcal \end{array}$

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