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Q.

Latent heat of vapourisation of a liquid at 500k and 1atm pressure is 10k.Cal/mole. What is t he change in internal energy when 3 moles of the liquid is vapourised at the same temperature

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a

7kCal

b

27kCal

c

33kCal

d

25kCal

answer is A.

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Detailed Solution

3H2O(l)→3H2O(g)

∆n(g)=np-nr

=3-0

∆n(g)=3

1 mole → 10k. cal

∆H for 3 moles→30k.cal

T=500k

∆H=∆E+∆nRT

∆H=∆E-∆nRT

=30-3×2×10-3 × 500

∆E=27K.cal

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