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Q.

The enthalpy of sublimation of ice at 273 K is 50.8 kJ mol^-1. Given the vapour pressure of water $⇌$ ice at 273 K is 6 x 10^-3 atm, the sublimation pressure of ice at 263 is in Pa

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a

211.8

b

259.6

c

326.9

d

165.5

answer is C.

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Detailed Solution

$\log (\frac{6 \times 10^{- 3}}{p_{1}}) = \frac{50 .8}{2 .303 \times 8 .314 \times 10^{- 3}} [\frac{10}{273 \times 263}] = 0 .3695$
$\frac{6 \times 10^{- 3}}{p_{1}} = 2 .342; p_{1} = \frac{6 \times 10^{- 3}}{2 .342} = 2 .56 \times 10^{- 3} atm$
$= 2.56 \times 10^{- 3} \times 101.325 = 259.6 kPa$

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