How Do You Calculate Heat of Vaporization Using the Clausius Clapeyron Equation?

[mygymnst]

Homework Statement

slope: -8.000 x 10^3 K
The graph is of ln vapor pressure of a gas vs. invert kelvin temperature
Using the Clausius Clapeyron equation, find Hvap (heat of vaporization)

Homework Equations

Clausius Clapeyron equation: ln P= (-delta Hvap/R)*(1/T) + C

The Attempt at a Solution

I tried using the given slope in the equation as T. Then I tried using y=mb. I multiplied the slope by 100K thinking I could find lnP.

 

[chemisttree]

In this example you have the equation in the form of a line already.

y=mx + b

where y is lnP, m is -deltaHvap/R, x is 1/T and b is C. You are asked to determine the value of -deltaHvap given that the slope (m) is -8000 K.

Can you find a way to tease out the -deltaHvap from the expression for the slope (-8000 K = -deltaHvap/R)?

 

[Blueshift5]

I also tried converting the slope to lnP by taking the natural log of the slope. However, I am not sure how to use the given information to find Hvap.

The Clausius Clapeyron equation is a fundamental relationship in thermodynamics that relates the vapor pressure of a substance to its temperature and heat of vaporization. In this case, the given slope of -8.000 x 10^3 K represents the change in temperature (T) over the change in ln vapor pressure (ln P). To solve for Hvap, we can rearrange the equation to isolate for Hvap:

Hvap = -R*(slope)*(1/T) + C

Where R is the gas constant and C is the y-intercept of the graph.

To find the heat of vaporization, we need to plug in the values for slope, temperature, and gas constant into the equation. The given temperature is in Kelvin, so we do not need to convert it. However, we need to convert the slope from K to Kelvin^-1 by dividing it by 1000, since the slope represents the change in temperature over 1000 units of ln vapor pressure.

Hvap = -(8.000 x 10^3 K)/1000 * (1/T) + C

Next, we need to determine the value of ln P at a specific temperature. We can do this by using the given information of the graph, which shows ln P on the y-axis and 1/T on the x-axis. We can choose a specific point on the graph, such as (1/T = 0.001, ln P = -5), and plug these values into the equation to solve for C.

C = ln P + (delta Hvap/R)*(1/T)
C = -5 + (8.000 x 10^3 K)/1000 * (0.001)
C = -5 + 8 = 3

Now, we can plug in the values for slope, temperature, gas constant, and C into the equation to solve for Hvap.

Hvap = -8.000 x 10^3 K/Kelvin * (1/0.001 Kelvin^-1) + 3

Hvap = -8.000 x 10^3 J/mol + 3

Hvap = -7.997 x 10^3 J/mol

Therefore, the heat of vaporization for