Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of

Question

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 C.

You may want to reference (Pages 478 - 487) Section 11.5 while completing this problem.

Part A

What is the vapor pressure of ethanol at 19 C?

Express your answer using two significant figures.

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 C.

You may want to reference (Pages 478 - 487) Section 11.5 while completing this problem.

Part A

What is the vapor pressure of ethanol at 19 C?

Express your answer using two significant figures.

Explanation / Answer

T1 = 78.4 oC
=(78.4 + 273)K
= 351.4 K
T2 = 19.0 oC
=(19.0 + 273)K
= 292.0 K
At normal boiling point, pressure is 1 atm
So,
P1 = 1 atm
H = 38.56 KJ/mol
= 38560 J/mol

use:
ln(P2/P1) = (H/R)*(1/T1 - 1/T2)
ln(P2/1) = (38560/8.314)*(1/351.4 - 1/292.0)
ln(P2/1) = 4638*(-5.789*10^-4)
P2 = 6.8*10^-2 atm
Answer: 0.068 atm

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