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Question

ΔfG at 500 K for substance S in liquid state and gaseous state are +100.7 kcal 1 and +103 kcal 1, respectively. Vapour pressure of liquid S at 500 K is approximately equal to:
(R=2 cal K11)

A
0.1 atm
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B
10 atm
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C
100 atm
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D
1 atm
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Solution

The correct option is B 10 atm
Solution:- (B) 10atm
As we know that, at equilibrium,
ΔG=2.303RTlogKp.....(1)
S(l)S(g)
From the above reaction,
ΔG=ΔGf(product)ΔG(reactant)
ΔG=103100.7=2.3kcal/mol=2.3×103kcal/mol
Given:-
T=500K
R=2cal/molK
From eqn(1), we have
2.3×103=2.303×2×500×logKp
logKp=1
Kp=101atm
Now, from the above reaction,
Kp=1PS(l)
PS(l)=1101=10atm
Hence the vapou pressure of liquid S is approximately equal to 10atm.

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