Clausius Clapeyron Equation Practice Problems

Getting Started with the Clausius Clapeyron Equation

The Clausius Clapeyron Equation Practice Problems come up constantly in physical chemistry courses, and they also show up in engineering work when you need to estimate vapor pressures at temperatures you haven't measured. The equation itself is straightforward enough that most students breeze through the first problem, then hit a wall when the numbers stop cooperating. dP/dT = H_vap / (T × V) That's the fundamental form. You integrate it assuming H_vap is constant over your temperature range, and you get the two-point form most people actually use:

ln(P2/P1) = -(H_vap/R) × (1/T2 - 1/T1) I remember working through a batch of these problems for a phase-equilibria class back in my undergrad. The textbook example used water with nice round numbers. Then the homework switched to benzene and gave you vapor pressure data at 30°C and 80°C, asking for the enthalpy of vaporization. That's where things got messy, because the answer depended on whether you treated H_vap as temperature-independent or not. It is not, obviously, but the equation assumes it is.

Clausius Clapeyron Equation Practice Problems

Let me walk through a realistic problem. You're given two data points for liquid ethanol: P1 = 40.0 kPa at T1 = 349 K P2 = 101.3 kPa at T2 = 351 K

You need to find H_vap. Rearrange the two-point form: H_vap = -R × ln(P2/P1) / (1/T2 - 1/T1) Plugging in: ln(101.3/40.0) = ln(2.5325) = 0.9296

1/351 - 1/349 = -0.000001624 K^-1 H_vap = -8.314 × 0.9296 / (-0.000001624) = 47,600 J/mol or about 47.6 kJ/mol The literature value for ethanol is around 38.6 kJ/mol at its normal boiling point. The discrepancy is noticeable because your two temperatures are so close together that rounding errors in the reciprocal-temperature difference blow up the result. This happens all the time in practice problems where the test maker picks convenient numbers without realizing they create numerical instability.

A better approach when you have closely spaced data is to rearrange differently. Instead of computing 1/T2 - 1/T1 directly, compute it as (T1 - T2)/(T1 × T2). With T1 = 349 and T2 = 351, that gives -2/(349 × 351) = -2/122499 = -0.00001633. Now your calculation is more stable and you still get roughly the same answer, but at least you know it's not a floating-point artifact. Here's another common setup: you know H_vap and one vapor pressure point, and you need to find the pressure at a different temperature. Say you're working with diethyl ether, H_vap = 26.0 kJ/mol, and you know the vapor pressure is 442 mmHg at 20°C. What's the pressure at 40°C? Convert everything first. Temperatures to Kelvin: 293 K and 313 K. Pressure units just need to match on both sides of the ratio, so mmHg is fine. R = 8.314 J/(mol·K) and H_vap = 26000 J/mol.

ln(P2/442) = -(26000/8.314) × (1/313 - 1/293) 1/313 - 1/293 = 0.003195 - 0.003413 = -0.000218 ln(P2/442) = -3127.3 × (-0.000218) = 0.6815

P2/442 = e^0.6815 = 1.977 P2 = 874 mmHg That's above atmospheric pressure, which means ether would be boiling at 40°C. The actual vapor pressure from tables is closer to 795 mmHg, so your answer is off by about 10%. The error comes from treating H_vap as constant when it actually decreases with temperature and drops significantly over a 20-degree span for a volatile liquid like ether.

One thing most textbooks gloss over: the Clausius Clapeyron equation assumes the vapor behaves as an ideal gas and that the molar volume of the liquid is negligible compared to the molar volume of the gas. Both assumptions break down at high pressures or near the critical point. I've seen engineers use this equation at 50 atm and then wonder why their predictions were wildly off. The integrated form simply isn't valid there. Another subtlety that catches people up: the sign convention. The equation is often written with a negative sign in front, which trips students who don't track what happens to 1/T2 - 1/T1 when T2 > T1. If T2 is higher, 1/T2 is lower, so (1/T2 - 1/T1) is negative. The negative sign in front cancels that, giving a positive ln(P2/P1), which means P2 > P1. It works. But if you mess up the subtraction order, you get a negative pressure ratio, which is impossible and usually means you've set up the problem backward. I always write out the full expression with numbers before evaluating anything just to check the signs. For the most common exam format, expect one of three question types: find H_vap from two P-T pairs, find a new pressure given H_vap and one reference point, or find the boiling point at a different external pressure. The third type is the one students hate because it requires solving for temperature in the exponent, which means using the exponential function rather than logarithms. It's algebra, not a different equation, but people forget that under time pressure.

If you're looking for practice problems beyond what your textbook provides, most physical chemistry lab manuals include a dataset where you plot ln(P) versus 1/T and extract H_vap from the slope. That graphical method is worth doing even if it's not on the exam, because it shows you at a glance when your data deviates from linearity, which is your signal that H_vap isn't constant and the simple equation is breaking down. One more practical note: when you're given pressure in atmospheres or bar, make sure you're consistent. The ratio P2/P1 is unitless, so any pressure unit works as long as both points use the same one. Temperature must always be in Kelvin. Using Celsius in that equation will give you nonsense, and I've corrected enough students on that to know it's a common mistake even among people who otherwise understand the derivation.