Question
Asked Dec 4, 2019
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If the normal boiling point of a liquid is 82 °C, what is the vapor pressure (in atm) at 50 °C? Enthalpy of vaporization = 37.2 kJ/mol K.    

A. 0.61 atm
B. 0.73 atm
C. 0.29 atm
D. 0.11 atm
E. 1.0 atm
 
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Expert Answer

Step 1

If the normal boiling point of a liquid is 82 °C, the vapor pressure (in atm) at 50 °C is to be calculated.

Enthalpy of vaporization = 37.2 kJ/mol K.    

Step 2

Clausius-Clapeyron equation: It is used to relate the vapor pressures of a substance at two different temperatures if the enthalpy of vaporization is known.

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Clausius-Clapeyron Equation: P, In P, vap Т, т, R Where P, is the vapor pressure at temperature T,. P, is the vapor pressure at temperature T,. AHa - Enthalpy of vaporization. vap R- Universal gas constant

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Step 3

Given conditions are:

ΔHvap =37.2 kJ/mol.K = 37200 J/mol.K

T1 = 82°C (boiling point) = 82°C + 273 = 355 K

T2 = 50°C = 50°C +273 = 323 ...

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ΔΗ. P, In P, vap Т, т, R 1 atm In (37200 J/mol.K) (8.314 J/mol.K)( 323 K P2 355 K 1 atm = (4474.38) 355- 323 In =1.25 P, 323x 355 Taking anti-log of both the sides: 1 atm = e25 = 3.4 P, ´1 atm :. P2 = 0.29 atm 3.4

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