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What is the work done when 1 mol of water at 373 K vapourises against atmospheric pressure of 1.0 atmosphere? [Solve the question assuming ideal gas behaviour]

(a) -6200 J
(b) -306 J
(c) -3100 J
(d) -1550 J
Can you answer this question?
 
 

1 Answer

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Answer: -3100 J
 
Volume of 1 mol of water at 373 K, 1 atm ($V_1$) = $18 cm^3$ = 0.018 L
Volume of 1 mol of water in vapour state at 373 K, 1 = $\frac{RT}{P} = \frac{0.0821 cm^3 atm K^{-1} mol ^{-1} \times 373 K}{\text{1 atm}} = 30.62 $ L
Now $w = -P(V_2 - V_1)$ = 1(30.62 - 0.018) = -30.6 L-atm = -30.6 $\times$ 101.3 = -3100 J (Hydrated salt)
answered Feb 25, 2014 by mosymeow_1
edited Feb 25, 2014 by mosymeow_1
 

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