A certain substance has a heat of vaporization of 37.51 kJ / mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 307 K?

Question

Wolfie

Physics

22 October, 14:29

A certain substance has a heat of vaporization of 37.51 kJ / mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 307 K?

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Answers (1)

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Jamari Ayala · Answer 1

T2=336K

Explanation:

Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:

where:

In (P2/P1) = ΔvapH/R (1/T1 - 1/T2)

p1 and p2 are the vapour pressures at temperatures

T1 and T2

ΔvapH = the enthalpy of vaporization of the liquid

R = the Universal Gas Constant

p1=p1, T1=307K

p2=3.50p1; T2=?

ΔvapH=37.51kJ/mol=37510J/mol

R=8.314J. K^-1moL^-1

In (3.50P1/P1) = (37510J/mol) / (8.314J. K^-1) * (1/307 - 1/T2)

P1 and P1 cancelled out:

In (3.50) = 4511.667 (T2 - 307/307T2)

1.253=14.696 (T2 - 307/T2)

1.253 = (14.696T2) - (14.696*307) / T2

1.253T2=14.696T2 - 4511.672

Therefore,

4511.672=14.696T2 - 1.253T2

4511.672=13.443T2

So therefore, T2=4511.672/13.443=335.61

Approximately, T2=336K