Question

Ethanol has a heat of vaporization of 38.56kJ/mol and a normal boiling point of 78.4 ∘C. What is the vapor pressure of ethanol at 19 ∘C?

Answer 1

Answer:

0.06819 atm or 51.8559 torr

Explanation:

To solve this problem we will use the following equation:

ln(P1/P2) = H (vap) / R (1/T2- 1/T1)

Given data:

P1= 760 torr

P2= ?

T1= 78.4 °C+273 = 351.4 K

T2= 19°C + 273 = 292 k

heat of vaporization = 38.56kJ/mol

Now we will put the values in equation:

ln (760 torr/ P2) = 38.56 × 1000 j/ 8.314 j k-1 (1/ 292 - 1/ 351.4)

ln (760 torr/ P2)  = 2.6849

760 torr/ P2 = 14.656

P2 = 51.8559 torr

or

P2= 51.8559 torr × 0.001315 atm

P2 = 0.06819 atm

Answer 2

The vapor pressure of ethanol at 19 ∘C is 0.0682 atm or 51.8559 torr

What is Vapor pressure?

Vapor pressure is the pressure which occurs when liquids evaporate and

can be calculated by using the formula below:

ln(P1/P2) = H (vap) / R (1/T2- 1/T1)

We were given the following parameters :

P1= 760 torr

P2= ?

T1= 78.4 °C+273 = 351.4 K

T2= 19°C + 273 = 292 K

Heat of vaporization = 38.56kJ/mol

Substitute the the values into the equation

ln (760 torr/ P2) = 38.56 × 1000 J/ 8.314 Jk⁻¹ (1/ 292 - 1/ 351.4)

ln (760 torr/ P2)  = 2.6849

760 torr/ P2 = 14.656

P2 = 51.8559 torr

We can convert to atm

P2= 51.8559 torr × 0.001315 atm

P2 = 0.0682 atm

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