# Answered Essay: Clausius Clapeyron Equations: ln P = Delta H_vap/R (1/T) + c or ln P_z/P_s = – Delta

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Clausius Clapeyron Equations: ln P = Delta H_vap/R (1/T) + c or ln P_z/P_s = – Delta H_vap/R (1/T_2 – 1/T_1) a) Based on the Claussius-Clapeyron equation, a plot of _____ vs. _____ will give a linear line with the slope of the line equal to _____ b) A plot of data for a substance gives a slope of -1876 K^-1. What is the enthalpy of vaporization of the substance. c) The same substance form part b has a vapor pressure of 50.0 mmHg at 25.0 degree C. What is its normal boiling point?

a)

Clearly, there is a slope:

-HRvap/R = slope

if we assume “X-axis” to be 1/T

and get the y-axis to be ln(P)

then, there will be a striahgt line

so

a plot of ln(P) vs 1/T should give a STRAIGHT LINE

b)

If

slope = -Hvap /R

and R = 8.314 J/molK ( idea gas constant )

then

HVap = -slope*R = -(-1876)*8.314 = 15597.064 J/mol

Hvap = 15.597 kJ/mol

c)

if

P = 50 mm Hg, T = 25°C = 298 KK

then get Pnomral; i.e. 1 atm 760 mm Hg; T2 = ?

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Change negative signs

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

ln(760/50) = 15597.064 /8.314*(1/298-1/T2)

Get T2 ( in K)

ln(760/50)*8.314/15597.064 – 1/298 = -1/T2

T2 = -(-0.001905)^-1

T2 = 524.934 K

T2 = 524.934-273 C

T2 = 251.9 °C, which is normal BP

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