Question

Ethanol has a heat of vaporization of 38.56 kJ/mol and a vapor pressure of 760 torr at 78.4 oC. What is the vapor pressure of ethanol at 38.8 oC?

Answer 1

The vapor pressure of ethanol is higher

It will decrease

It will increase by a factor of 2

It will be half as large

Step-by-step explanation:

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Answer 2

To develop this problem it is necessary to apply the concepts developed by Clausius - Claperyron.

This duet found the relationship between temperature and pressure expressed as,

`ln P = constant - (\Delta H)/(RT)`

For the two states that we have then we could define the pressure and temperature in each of them as

`ln((P_2)/(P_1)) = (-\Delta H)/(R)((1)/(T_2)-(1)/(T_1))`

Where,

`P_(1,2)`= Pressure at state 1 and 2

`T_(1,2)`= Temperature at state 1 and 2

`\Delta H`= Enthalpy of Vaporization of a substance

R = Gas constant (8.134J/mol.K)

Our values are given by,

`P_1 = 1atm \\\Delta H = 38.56*10^(-3) J/mol \\R = 8.134J/mol.K\\T_1 = 78.4\°C = 351.55K\\T_2 =38.8\°C = 311.95K`

Therefore replacing we have that,

`ln((P_2)/(P_1)) = (-\Delta H)/(R)((1)/(T_2)-(1)/(T_1))`

`ln((P_2)/(1atm)) = (-38.56*10^3)/(8.314)((1)/(311.95)-(1)/(351.55))`

`ln(P_2) - Ln(1atm) = (-38.56*10^3)/(8.314)((1)/(311.95)-(1)/(351.55))`

`P_2 = e^{(-38.56*10^3)/(8.314)((1)/(311.95)-(1)/(351.55))}`

`P_2 = 0.187355atm`

Therefore the pressure of the Ethanol at 38.8°C is 0.187355atm