$A (l) ⇌ A (g), Δ_{v a p} H = 460.6 c a l m o l^{- 1}$ , boiling point = $50 K$ . What is the boiling point of 10 atm

150 k

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75 k

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100 k

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Solution

The correct option is C 100 k
$Δ H = 2.303 R [\frac{1}{T_{1}} - \frac{1}{T_{2}}] l o g \frac{P_{2}}{P_{1}}$
$460.6 = 2.303 \times 2 [\frac{1}{50} - \frac{1}{T_{2}}] l o g 10$
$∴$ $T_{2} = 100 K$

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Similar questions

A (l) ⇌ A (g), Δ_{v a p} H = 460.6 c a l m o l^{- 1}

50 K

Δ H_{v a p} = 460.6

353.23

1.8

90

354.11

K_{0}

2.53 K k g m o l^{- 1}

Δ H_{v a p} = 460.6

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