The vapor pressure of benzene is 100.0 mmHg at 26.1^∘ C and 400.0 mmHg at 60.6^∘ C. What is the

The vapor pressure of benzene is 100.0 mmHg at 26.1^∘ C and...

The vapor pressure of benzene is $100.0 \mathrm{mmHg}$ at $26.1^{\circ} \mathrm{C}$ and $400.0 \mathrm{mmHg}$ at $60.6^{\circ} \mathrm{C}$. What is the boiling point of benzene at $760.0 \mathrm{mmHg}$ ?

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid (or solid) phase at a given temperature. It is an important property that describes the tendency of molecules to escape from the liquid and become a gas, and it increases with temperature.

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is a key thermodynamic relation that describes the exponential dependence of vapor pressure on temperature. It relates the natural logarithm of the vapor pressure to the inverse of the absolute temperature, allowing the estimation of temperature changes or enthalpy of vaporization from vapor pressure measurements.

Boiling Point

The boiling point of a substance is the specific temperature at which its vapor pressure equals the external or ambient pressure. At this temperature, bubbles of vapor can form within the liquid, leading to boiling. Determining the boiling point at different pressures is important for understanding phase transitions under varying conditions.

Transcript

00:01 So in this problem, we have to know the boiling point is when the vapor pressure equals the atmospheric pressure.

00:08 So therefore, we just want to know what is the temperature of the benzene when the pressure of the benzene is 760 millimeters of mercury.

00:21 That's going to give us the boiling point.

00:25 So we can use atmospheres or millimeters of mercury.

00:28 It doesn't really matter with the claseus -clapperyon equation.

00:31 So i'm just using millimeters of mercury, but you can also do that conversion.

00:36 It's just that the pressures need to have the same units and we'll end up with the pressure in the units that we used in the equation.

00:43 So first of all, we are given two temperatures and two pressures.

00:46 We can solve for delta h vaporization using the classiest claypron equation.

00:51 Then we can find the temperature at which the pressure is equal to 760 millimeters of mercury.

00:56 So let's plug in the values into the equation.

00:59 So we're going to use this value as a value of r and has units of joules per mole with mous and kelvin.

01:04 So add 273 to the temperatures given because we need them in calvin.

01:14 Then plug in p1 and p2.

01:16 So p1 is 100 millimeters of mercury.

01:21 And p2 is equal to 400 meters of mercury.

01:26 So it doesn't really matter which one you pick for p1 or p2, as long as it matches up with the temperatures.

01:32 So p2 matches with t2 and p1 matches with t1.

01:36 So then we're solving for delta hv8, so we don't know that.

01:39 So then divide by r and then we plug in t2 and t1.

01:45 So 1 over t2.

01:46 So t2 is corresponding to p2.

01:48 So that is 400 millimeters of mercury.

01:50 So that would be the 60 .6 degrees celsius.

01:52 So it's a higher temperature.

01:54 And then we're plugging in t1 in the next slide to 1 divided by t1.

01:58 And that's the lower temperature.

02:01 So then we will just need to solve this.

02:06 So i've simplified it a little.

02:08 I've gotten rid of the units...

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