# Homework Help: Calculating the boiling temperature

1. Jun 12, 2013

### fluidistic

1. The problem statement, all variables and given/known data
I'm stuck on the following problem: A particular liquid boils at 127°C at a pressure of 800 mmHg. It has a heat of vaporization of 1000 cal/mole. At what temperature will it boil if the pressure is raised to 810 mmHg?

2. Relevant equations
Clapeyron's equation: $\frac{dP}{dT}=\frac{l}{T\Delta v}$. (1)
$l=T\Delta s$. (2)

3. The attempt at a solution
Since the problem is given right after introducing the Clapeyron's equation in the book, I guess I must use it. However nothing is said about the difference in volumes of the gaseous and liquid phase of the substance.
Some thoughts: Since the pressure difference is small, I guess I can consider the heat of vaporization $l$ as a constant for the problem.
So I could calculate the change in entropy of the substance when it boils at 127°C and pressure of 800 mmHg. But I don't see how this would help me to calculate the temperature I'm asked to find. I'd need to know the change in entropy (which I know is lower than the one at pressure of 800 mmHg) of the substance if it boils when the pressure is 810 mmHg.

Second thoughts. If I consider that $\frac{dP}{dT}$ is a constant between T=127°C and the temperature I'm looking for, then $T\Delta v$ is also constant (applying Clapeyron's equation as well as assuming $l$ constant). Now, $\Delta v \approx v^\text{gas}$ because 1 mole of gas occupies a somewhat much greater volume than 1 mole of liquid. So $Tv^\text{gas}=\text{constant}$. If I assume that the gas is ideal, then $v=\frac{RT}{P}$. Thus $\frac{T^2R}{P}=\text{constant} \Rightarrow \frac{T^2}{P}=\text{constant}$.
A simple plugging and chugging and isolating $T^2$ leads me to $T\approx 129.5°C$.
It looks like I've solved the problem when writting it up here. I wonder if my assumptions are correct (that the heat of vaporization as well as dP/dT are constants in the region where I'm interested) and if I've reached the correct result.
Thank you for your time and effort.

Edit: I doubt my answer is right since I've never used the value of $l=1000 cal/mol$. I considered it constant, that all I did about it.

Last edited: Jun 12, 2013
2. Jun 12, 2013

### Staff: Mentor

You did a very nice job of analyzing this. You were very perceptive to substitute the ideal gas law and neglect the volume of the liquid. This is exactly what Clausius did when he derived the Clausius-Clapeyron equation:
$$\frac{dP}{dT}=\frac{lP}{RT^2}$$
Rearranging:
$$\frac{dlnP}{d(1/T)}=-\frac{l}{R}$$

Since the change in lnP and the change in 1/T are going to be very small, you can assume that, over that small an interval, l is nearly constant. So,

$$\frac{ΔlnP}{Δ(1/T)}=-\frac{l}{R}$$

3. Jun 12, 2013

### rude man

Perhaps a bit more straightforwardly:

Δp/ΔT = lp/RT2 or
ΔT = RT2Δp/lp.
Right? Those differentials give me headaches ...

Last edited: Jun 12, 2013
4. Jun 12, 2013

### Staff: Mentor

Yes. But one of the nice features of the Clausius-Clapeyron equation is that the heat of vaporization is a relatively weak function of temperature, so that the integrated version (assuming constant l) typically applies quite accurately over fairly large changes in temperature and pressure (i.e., several tens of degrees):

$$\ln(P_2/P_1)=\frac{l(T_2-T_1)}{RT_2T_1}$$

5. Jun 13, 2013

### rude man

I think the math's the same:

Δp/ΔT = lp/RT2 or more accurately
dp/dT = lp/RT2
dp/p = l/RT2 dT
integrating from p1 to p2 on the LHS and T1 to T2 on the RHS gives

ln(p2/p1) = (-l/R)(1/T2 - 1/T1).

EDIT: never mind Chet, I approximated dp/dT as Δp/ΔT which is not as accurate as dp/dT integrated as per your result. Not a big difference in the answer but still I should have integrated.

Interesting what you pointed out about Clapeyron first publishing his relation assuming ideal gases ... Fermi managed two independent means of deriving the more general dp/dT = l/TΔv. Of course, that was in the 1930's ...

Last edited: Jun 13, 2013