10 grams of ethanol arc placed in a 10 L sealed container at 292.15 K. At this t

Question

10 grams of ethanol arc placed in a 10 L sealed container at 292.15 K. At this temperature, the equilibrium vapor pressure of ethanol is 10 torr. If we assume that the enthalpy of vaporization of ethanol is 43 kJ/mol and independent of temperature over the range of discussion, calculate the vapor pressure of ethanol inside the container as a function of temperature from 292 K up to the temperature at which all of the ethanol is in the vapor phase. Plot the pressure inside the container as a function of temperature from 292 K to 450 K.

Explanation / Answer

a)
T1= 292 K
P1 = 10 torr

T2=T (general)
P2 = P(general)

delta H=43 KJ/mol=43000 J/mol

use:
ln(P2/P1) = (- delta H/R)*(1/T2 - 1/T1)
ln(P/10) = (-43000/8.314)* (1/T - 1/292)
ln P - 2.303 = - 5172/T + 17.7
ln P = - 5172/T + 20.

It can be written as:

P = e^(- 5172/T + 20)

P= e^(- 5172/T)

b)

ln P = - 5172/T + 20.

When all are in vapour phase,

P =760 torr

ln (760) = - 5172/T + 20.

6.6333 = - 5172/T + 20.

T= 387 K

You need to plot the graph upto 387 K

Please plot is.

You can use excel.

Thanks

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