Suppose the vapor pressure of a substance is measured at two...

Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius-Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, $P_{1}$ and $P_{2}$, and the absolute temperatures at which they were measured, $T_{1}$ and $T_{2}$ : $$ \ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ (b) Gasoline is a mixture of hydrocarbons, a component of which is octane $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)$. Octane has a vapor pressure of $1.86 \mathrm{kPa}$ at $25^{\circ} \mathrm{C}$ and a vapor pressure of $19.3 \mathrm{kPa}$ at $75^{\circ} \mathrm{C}$. Use these data and the equation in part (a) to calculate the heat of vaporization of octane. $(\mathbf{c})$ By using the equation in part (a) and the data given in part (b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.81 . (d) Calculate the vapor pressure of octane at $-30^{\circ} \mathrm{C}$.

Suppose the vapor pressure of a substance is measured at two different temperatures.
(a) By using the Clausius-Clapeyron equation (Equation 11.1) derive the following relationship between the vapor pressures, $P_{1}$ and $P_{2}$, and the absolute temperatures at which they were measured, $T_{1}$ and $T_{2}$ :
$$
\ln \frac{P_{1}}{P_{2}}=-\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)
$$
(b) Gasoline is a mixture of hydrocarbons, a component of which is octane $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)$. Octane has a vapor pressure of $1.86 \mathrm{kPa}$ at $25^{\circ} \mathrm{C}$ and a vapor pressure of $19.3 \mathrm{kPa}$ at $75^{\circ} \mathrm{C}$. Use these data and the equation in part (a) to calculate the heat of vaporization of octane. $(\mathbf{c})$ By using the equation in part (a) and the data given in part
(b), calculate the normal boiling point of octane. Compare your answer to the one you obtained from Exercise 11.81 .
(d) Calculate the vapor pressure of octane at $-30^{\circ} \mathrm{C}$.

Key Concepts

Clausius-Clapeyron Equation

This equation relates the change in vapor pressure with temperature to the heat of vaporization. It provides a quantitative means to understand how vapor pressure varies as a function of temperature under equilibrium conditions, allowing one to derive relationships between pressures at different temperatures, and to calculate thermodynamic properties such as the enthalpy of vaporization.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form at a given temperature. It is an essential concept in understanding phase transitions, as it dictates the conditions under which a liquid will evaporate or boil.

Heat of Vaporization

Heat of vaporization is the energy required to convert a unit amount of a substance from the liquid phase to the vapor phase at constant temperature and pressure. It appears directly in the Clausius-Clapeyron equation and is a key parameter for calculating changes in vapor pressure with temperature.

Boiling Point

The boiling point is defined as the temperature at which a liquid's vapor pressure equals the external or ambient pressure. Using the Clausius-Clapeyron equation, one can determine the boiling point by finding the temperature where the vapor pressure reaches a specified value, such as atmospheric pressure.

Transcript

00:01 This is sort of a tough problem for me because i'm not real good with the whole natural logarithm things.

00:08 Whoops see.

00:09 We're given the following information.

00:13 Suppose the vapor pressure of a substance is measured at two different temperatures.

00:18 By using the closcius -clperian equation, we're going to derive the following relationship.

00:24 And here's the relationship we're going to be deriving.

00:50 Okay.

00:53 So the closcius -cliparin equation is as follows.

01:22 Now since i have two different pressures that we're given here, since we have p1, p2, and t1, t2, we will have two equations.

01:30 We're going to call them equation one and equation two.

01:33 And we're going to subtract equation two from equation one.

01:37 Let's write them down.

01:40 First, p1.

01:58 And again, we're going to subtract from that equation two.

02:17 Now we can cross out this part, and we are going to get the following.

02:56 So this part of the equation turns into this.

03:03 This is just one of those things i even had to look up.

03:07 And i'm going to pull out the parts of this that i'm about to highlight.

03:14 These are the same for each and the negatives.

03:18 So i'm going to pull those out as follows.

03:24 Negative delta h.

03:31 And that leaves be 1 over t1 minus 1 over t2.

03:39 And if you'll check that is the equation that we're deriving.

03:48 For part 2, or i guess b, gasoline is a mixture of hydrocarbons, a component at which is octane.

03:58 We're given sort of the structural formula in sort of a modified structural formula, but i'm just going to tell you that it's c8, h18, so we don't need it for anything else.

04:19 We're told that it has a vapor pressure of 1 .86 kpas.

04:27 At 25 degrees celsius, which is 298 .15k.

04:41 And it has a vapor pressure of 19 .3 kilopascals at 75 degrees celsius, which is 343.

05:00 So 343, 348.

05:15 .15 kelvin.

05:18 We're going to use these data in the equation from part a to calculate the heat of vaporization of octane.

05:25 So let me go ahead and put these values into our equation.

05:30 First i'll write the equation again.

05:38 And we're going to be solving for the heat of vaporization.

05:42 That's our unknown.

05:51 And let me put all the values of dory equation.

06:23 That's my gas constant.