The normal boiling of water is $373 K$ . Vapour pressure of water at temperature $T$ is 19 mm $H g$ . If enthalpy of evaporation is $40.67 k J / m o l$ , then temperature $T$ would be:
(Use $l o g 2 = 0.3, R = 8.3 {J K}^{- 1}$ )

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Solution

By Clausius- Clayperon Equation
$l n \frac{P_{2}}{P_{1}} = \frac{Δ H}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
Here, $T_{2} = 373 K, P_{2} = 760 m m H g, T_{1} = T$ & $P_{1} = 19$
$∴ l n (\frac{760}{19}) = \frac{40670}{8.3} [\frac{1}{T} - \frac{1}{373}]$
$\Rightarrow T = 291.3 K$

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(Use :

l o g 2 = 0.3, R = 8.3 J K^{- 1} m o l^{- 1})

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Normal boiling point of water is 373 K (at 760mm). Vapour pressure of water at 298 K is 23 mm. If the enthalpy of evaporation is 40.656 kJ/mole, the boiling point of water at 23 mm pressure will be

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The normal boiling of water is 373K. Vapour pressure of water at temperature T is 19 mm Hg. If enthalpy of evaporation is 40.67kJ/mol, then temperature T would be: (Use log2=0.3,R=8.3JK−1)

By Clausius- Clayperon EquationlnP2P1=ΔHR[1T1−1T2]Here, T2=373K,P2=760mmHg,T1=T & P1=19∴ln(76019)=406708.3[1T−1373]⇒T=291.3K

The normal boiling of water is $373 K$ . Vapour pressure of water at temperature $T$ is 19 mm $H g$ . If enthalpy of evaporation is $40.67 k J / m o l$ , then temperature $T$ would be:
(Use $l o g 2 = 0.3, R = 8.3 {J K}^{- 1}$ )

By Clausius- Clayperon Equation
$l n \frac{P_{2}}{P_{1}} = \frac{Δ H}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
Here, $T_{2} = 373 K, P_{2} = 760 m m H g, T_{1} = T$ & $P_{1} = 19$
$∴ l n (\frac{760}{19}) = \frac{40670}{8.3} [\frac{1}{T} - \frac{1}{373}]$
$\Rightarrow T = 291.3 K$