Butane (C4 H10) has a heat of vaporization of 22.44 kJ / mol and a normal boiling point of -0.4^

Butane (C4 H10) has a heat of vaporization of 22.44 kJ / mol...

Key Concepts

Phase Equilibrium

Phase equilibrium refers to the condition in which a substance exists simultaneously in more than one phase (such as liquid and vapor) with no net change in the amount of each phase over time. This concept is central in understanding how substances behave when confined in a sealed environment at a given temperature and pressure, leading to a dynamic balance between the rate of evaporation and condensation.

Vapor Pressure

Vapor pressure is the pressure exerted by the vapor when it is in equilibrium with its liquid (or solid) phase at a given temperature. It reflects the tendency of molecules to escape from the liquid phase into the gas phase and plays a crucial role in determining the amount of substance that exists as vapor versus liquid in a closed system.

Boiling Point

The boiling point is the temperature at which the vapor pressure of a liquid equals the external pressure surrounding the liquid. At this temperature, the liquid can change to vapor rapidly. Understanding the boiling point helps in determining whether a substance will predominantly be in the vapor or liquid phase under given conditions.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert one mole of a liquid into vapor at constant temperature and pressure. This thermodynamic property is essential in understanding how energy input affects the phase change process, and it is integral to calculations involving the temperature-dependent equilibrium between liquid and vapor states.

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation provides a quantitative relationship between the vapor pressure of a substance and its temperature, using the heat of vaporization. This relation is useful for predicting how the vapor pressure (and consequently the phase distribution) changes with temperature, which is vital when analyzing systems undergoing heat-induced phase transitions.

Transcript

00:01 So before we start solving this question, let's review what is being given to us.

00:04 It is always good to write those down.

00:07 So we know the heat of vaporization of butane is 22 .44 kilojoules per mole.

00:13 The normal boiling point of butane is negative 0 .4 degrees celsius.

00:18 The butane is in a 250 milliliter flask, and there are 0 .55 grams of it.

00:24 And the two equations you need to know are the clausius -clyperian equation, which is used to calculate vapor pressure given the heated vaporization and the ideal gas law, which is the equation used to calculate ideal gas behavior under certain constant conditions.

00:45 So let's begin with the first part of the question where we need to know the amount of liquid of liquid butane at negative 22 degrees celsius.

01:01 Now the very first thing you want to do is convert.

01:04 This temperature into kelvin's because when you're dealing with these two equations, they calculate using kelvin's.

01:11 So in order to convert celsius into calvins, all you have to do is just add 273 to it.

01:19 And that gives us 251 kelvin.

01:25 And the equation you'll need to use first is the clausius -clyperon equation, which is as follows.

01:39 It is the natural log of pressure 2 over pressure 1, which is equal to the negative heat of vaporization over a constant r, parentheses 1 over temperature 2 minus temperature 1.

02:03 So notice that there are two pressures and two temperatures.

02:08 And we already know pressure 1.

02:10 Pressure 1 is going to be one atmosphere.

02:15 This is going to be one atmosphere because we are given the normal boiling point.

02:21 And normal boiling point is always when the temperature, the pressure is at one atmosphere.

02:30 While we're up here, we need to convert all of these values as well.

02:35 So the heat of vaporization is in kilojoules, but we need it in joules.

02:39 So all they have to do is just multiply it by 1 ,000.

02:44 And we get 22 ,440 joules per mole.

02:53 The boiling point should also be in kelvin, so we just add 273, and we get 272 .6 kelvin.

03:03 The flask volume should be in liters, so we just divide by 1 ,000, and that gives us 0 .250 liters.

03:26 And butane should be in moles.

03:29 But the moles, well first we need to know the molar mass.

03:35 And the molar mass is 58 .12 grams per mole.

03:51 We'll calculate the moles of butane in a second, but i'll just write the molar mass here.

03:57 This is just found by adding the molar masses of all the individual elements, the atoms, that make up a butane molecule.

04:04 So back down here, let's start filling in these gaps.

04:10 So we need to find pressure 2.

04:12 Pressure 2 is going to be the pressure at this temperature at negative 22.

04:20 So down here, pressure 2, that's what we're looking for.

04:27 Pressure 1 is going to be 1 atmosphere equals negative.

04:36 We know the heat of vaporization, so that's 22 ,000.

04:41 440 joules per mole.

04:51 R is going to be 8 .314.

05:00 Actually, i'll write that in red to make it a little clearer.

05:03 This is a value that we get indirectly.

05:06 So this is 8 .314.

05:10 And we know this because this is the value of r when we have jewels and moles and kelvin in our calculations.

05:24 It would be a different number if we calculated with different values, or using a different equation rather.

05:31 And then for the parentheses, temperature 2 is going to be 251 kelvin because that is the temperature that we're looking the pressure for.

05:49 And temperature 1 is going to be 272 .6.

06:00 This is 272 .6 because because that is the initial boiling point.

06:10 This is the initial temperature where butane is going to start to boil.

06:17 So then we solve this part.

06:26 We can also cross out like values, like units.

06:30 So we can cross out jewels, and cross out kelvin's, and we can cross out moles.

06:37 So we just get atmospheres as our targeted value, our targeted unit.

06:44 So solving this whole thing gives us 0 .852, and then you multiply this by 1, which gives us the same value, and then you solve for the natural log.

07:02 So then p2 should be 0 .427 atmospheres.

07:14 So this is the vapor pressure.

07:23 Okay? so now we use the ideal gas law, which is pv equals nrt.

07:44 And so we want to know the amount of moles that are present as a gas.

07:53 So we rearrange the equation to solve for n...

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