I plan to use slighty different formulations of the Clausius-Clapeyron Equation in the first several questions. Look to see how they are different. Why do I do this? As you encounter different presentations, you will probably see whatever form of the equation the instructor (or the textbook writer) learned. There is a decent chance that it will be different from the form you learned. I would like for you to see that they are the same equation. As Ludwig Wittgenstein said:

Understanding means seeing that the same thing said different ways is the same thing.

Some brief notes on the units:

1) The natural log term on the left-hand side is unitless. It does not matter what units of pressure you use; the only restriction is that P_{1}and P_{2}must be expressed using the same pressure unit.2) The unit on the temperature term will be K¯

^{1}. The unit on R is J mol¯^{1}K¯^{1}.3) The K¯

^{1}in the temperature term will cancel with the K¯^{1}associated with R.4) This means that the unit on ΔH must be J/mol. This then makes the right-hand side unitless.

Below, I plan to just solve a few problems. If you want more on this equation please see here. Or, fire up Teh Great Googlizer.

**Problem #1:** Determine ΔH_{vap} for a compound that has a measured vapor pressure of 24.3 torr at 273 K and 135 torr at 325 K.

**Solution:**

1) Let us use the Clausius-Clapeyron Equation:

with the following values:

P _{1}= 24.3 torrT _{1}= 273 KP _{2}= 135 torrT _{2}= 325 K

2) Set up equation with values:

ln (135/24.3) = (x / 8.31447) (1/273 minus 1/325)1.7148 = (x / 8.31447) (0.00058608)

1.7148 = 0.000070489x

x = 24327 J/mol = 24.3 kJ/mol

**Problem #2:** A certain liquid has a vapor pressure of 6.91 mmHg at 0 °C. If this liquid has a normal boiling point of 105 °C, what is the liquid's heat of vaporization in kJ/mol?

**Solution:**

1) Let us use the Clausius-Clapeyron Equation:

ln (P_{1}/ P_{2}) = (ΔH / R) (1/T_{2}- 1/T_{1})with the following values:

P _{1}= 6.91 mmHgT _{1}= 0 °C = 273.15 KP _{2}= 760.0 mmHgT _{2}= 68.73 °C = 378.15 K

2) Set up equation with values:

ln (6.91/ 760) = (x / 8.31447) (1/378.15 minus 1/273.15)-4.70035 = (x / 8.31447) (-0.0010165)

4.70035 = 0.00012226x (notice I got rid of the negative signs)

x = 38445 J/mol = 38.4 kJ/mol

**Problem #3:** Carbon tetrachloride has a vapor pressure of 213 torr at 40.0 °C and 836 torr at 80.0 °C. What s the enthalpy of vaporization in kJ/mol?

**Solution:**

1) Let us rearrange the Clausius-Clapeyron Equation:

ln (P_{1}/P_{2}) = (-ΔH_{vap}/R) x (1/T_{1}- 1/T_{2})ΔH

_{vap}= [-R x ln (P_{1}/P_{2})] / (1/T_{1}- 1/T_{2})

2) Insert values and solve:

ΔH_{vap}= [(-8.314 J/mole K) x ln (213 torr / 836 torr)] / (1/313.15 K - 1/353.15 K)ΔH

_{vap}= 31.4 kJ/mole

**Problem #4:** The molar enthalpy of vaporization of hexane (C_{6}H_{14}) is 28.9 kJ/mol, and its normal boiling point is 68.73 °C. What is the vapor pressure of hexane at 25.00 °C?

**Solution:**

1) Let us use the Clausius-Clapeyron Equation:

ln (P_{1}/ P_{2}) = - (ΔH / R) (1/T_{1}- 1/T_{2})with the following values:

P _{1}= xT _{1}= 25.00 °C = 298.15 KP _{2}= 760.0 mmHgT _{2}= 68.73 °C = 341.88 K

2) Set up equation with values:

ln (x / 760) = - (28900 / 8.31447) (1/298.15 minus 1/341.88)ln (x / 760) = - 1.4912

x /760 = 0.2251

x = 171 torr

Comment: note that no pressure is given with the normal boiling point. This is because, by definition, the vapor pressure of a substance at its normal boiling point is 760 mmHg.

Not indicating the above in this type of question is common. You have been warned!

**Problem #5:** What is the vapor pressure of benzene at 25.5 °C? The normal boiling point of benzene is 80.1 °C and its molar heat of vaporization is 30.8 kJ/mol. Answer in units of atm

**Solution:**

1) Let us use the Clausius-Clapeyron Equation:

ln (x / 1.00) = (30800 / 8.31447) (1/353.25 minus 1/298.65)

Comment: I used the form of the equation shown in this image:

I assigned the unknown value to be associated with P_{2}. That puts x in the numerator and a 1.00 in the denominator, making my calculation a bit easier.

2) Let us calculate:

ln (x / 1.00) = (3704.385) (-0.000517545)ln x = -1.9172

x = 0.147 atm

Comment: I used 273.15 to convert Celsius to Kelvin. It's normal to use 273. I just decded to walk on the wild side for a moment.

**Problem #6:** The normal boiling point of Argon is 83.8 K and its latent heat of vaporization is 1.21 kJ/mol. Calculate its boiling point at 1.5 atmosphere.

**Solution:**

1) Let us use the Clausius-Clapeyron Equation:

ln (P_{1}/ P_{2}) = - (ΔH / R) (1/T_{1}- 1/T_{2})with the following values:

P _{1}= 1.0 atmT _{1}= 83.8 KP _{2}= 1.5 atmT _{2}= x

2) Set up equation with values:

ln (1.0/ 1.5) = - (1210 / 8.31447) (1/83.8 minus 1/x)0.405465 = 145.53 (0.011933 minus 1/x)

0.405465 = 0.17366 minus 145.53 / x

0.231805 = -145.53 / x

**Problem #7:** Chloroform, CHCl_{3}has a vapor pressure of 197 mmHg at 23.0 °C, and 448 mmHg at 45.0 °C. Estimate its heat of vaporization and normal boiling point.

**Solution:**

1) Let us use the Clausius-Clapeyron Equation:

ln (P_{1}/ P_{2}) = - (ΔH / R) (1/T_{1}- 1/T_{2})with the following values:

P _{1}= 197 mmHgT _{1}= 296 KP _{2}= 448 mmHgT _{2}= 318 K

2) Set up equation to solve for the enthalpy of vaporization:

ln (197 / 448) = - (x / 8.31447) (1/296 minus 1/318)x = 29227.66 J = 29.2 kJ

3) Let us use the Clausius-Clapeyron Equation:

ln (P_{1}/ P_{2}) = - (ΔH / R) (1/T_{1}- 1/T_{2})with the following values:

P _{1}= 760 mmHgT _{1}= xP _{2}= 448 mmHgT _{2}= 318 K

4) Set up equation to solve for the normal boiling point:

ln (760 / 448) = - (29227.66 / 8.31447) (1/x minus 1/318)0.5285252 = - 3515.2764 (1/x minus 1/318)

0.5285252 = 11.05433 minus (3515.2764 / x)

10.5258048 = 3515.2764 / x

x = 334 K

You might be interested in a collection (from the literature) of enthalpy of vaporization values for chloroform. The author of the above problem (not the ChemTeam!) obviously used the fourth of the four listed values. Interestingly, Wikipedia uses the 31.4 value.

**Problem #8:** A 5.00 L flask contains 3.00 g of mercury. The system is at room temperature of 25.0 °C. By how many degrees should we increase the temperature of the flask to triple the mercury vapor pressure. The enthalpy of vaporization for mercury is 59.11 kJ/mol?

**Solution:**

1) Let us use the Clausius-Clapeyron Equation:

with the following values:

P _{1}= 1T _{1}= 298 KP _{2}= 3T _{2}= x

Comment: I don't care what the actual vapor presssure value is at either temperature. I just care that it triples in value from P_{1} to P_{2}. Wy can I do this? Because I will be using a ratio of P_{2} to P_{1}. I only care that that ratio is 3.

3) Set up equation with values:

ln (3/1) = (59110 / 8.31447) (1/298 minus 1/x)1.0968 = 7109.2926 (1/298 minus 1/x)

1.0968 = 23.8567 minus 7109.2926/x)

x = 312.4 K = 39.4 °C

Clicking this link will take you to a NIST paper that has a table of calculated mercury vapor pressures. See page 20.

Also, notice the kinda, sorta vapor pressure assumption. I assumed that only some of the 3.00 g of Hg evaporated at 25 °C. That's a fair assumption, I would think.