Gases- Khan COPY

Question 1

where does water freeze/boil?

Answer 1

0 degrees C bp= 100 degrees C, where water boils from K, C= K- 273.15*

Question 2

Kelvin scale water freeze/boil

Answer 2

freezes at 273.15 K boils at 373.15 K so differs 0 points htey use, but btw water’s bp and fp is still 100 temperature units, so C and K using same size unit to measure temperature so converting btw two scales means make an adjustment btw two different scaling points - do not need degrees for Kevin scale, just call them K so all we need is upper case K** so K= 273.15 K + C

Question 3

Fahrenheit

Answer 3

water freezes at 32 F boils at 212 F span is 180 diff than C/K which is 100 for both! so say 180 F~ 100 C ratio F to C= 180/100= 9/5 Temp C= Temp F- 32 F X (unit ration 5 degrees C/ 9 degrees F) OR TEmp F= Tc X (9F/5C) + 32 F

Question 4

C and F scales…..

Answer 4

can both have negative or positive values! where differs from Kelvin scale, which can only have a positive value! turns out absolute coldest temperature is 0 K its absolute zero, cant get colder because t this point no particles have kinetic energy, coldest can get is no KE what so ever since temp= kinetic energy of molecules

Question 5

pressure =

Answer 5

Force/Area

Question 6

pressure ex of ballon

Answer 6

ex. particiles in ballon moving around container,w hen collide exert a force aupon area of collision; while one little collision doesn’t have a huge amount of pressure, but huge amount f particles and collision typically do bot fill them up a little bit fill them up a lot tighter sdies get indicating more pressure pressure = F/A F directly proportional to P, F= ma, so if inc magnitude of a can inc force because F and acceleration are akso directly proportional, acc is change in velocity, so greater change in velocity/acceleration can inc force, greater force (when particles movign) greater pressure inc temp inc pressure gas exerts! temp average measurement of kinetic energy of particles

Question 7

how do you inc amount of collisions?

Answer 7

pressure=F/A more collisions= more pressure = if add more particles = more mols of gas, mols is simply referring to the number of particles, more mols of gas means more pressure! can inc frequency of collisions by making container smaller, more space to move around and hti sides of container more dec volume inc pressure, so can change pressure of gas!

Question 8

so pressure inc when…..

Answer 8

P= F/a, F=ma so inc temp inc pressure n* inc moles inc pressure v** dec volume dec pressure! dec space of container!

Question 9

how do we measure changes of gas……Barometer

Answer 9

Former student of Gallieo, Torricelli asked same question when trying to measure changing gas in our atm, solved problem by creating barometer - device that measures pressure!

Question 10

how does a Barometer work?

Answer 10

• He took a glass tube and filled it up with mercury -quickly flipped tube over and stuck open end into open dish of mercury, so stuck tube into open dish of Hg opening down and interestingly enough most of the mercury stayed in the tube even though it was trying to flow out because as flows out exerts a pressure on mercury on dish which then causes mercury in dish to push upward against air
• when pressure of rising mercury meets pressure of atm pushign down on liquid surface, mercurcy in tube cannot flow anymore, pressure in atmosphere traps some of the mercury in the tube!
• can see at sea level =height of mercury left in column is about 760 mm**
• if measured on top of mountain would be shorter column, not as much air pushing down on liquid so more mercury from column could escape, ex in breckenridge height of colmun in this case would only be about 520 mm, so less weight of atm pushing down on surface of dish, so more can escape and amount of column is shorter!! in order to get a baseline for comparison, pressure of atm at sea level = 1 atm**

Question 11

manometers

Answer 11

generic divide measures pressure, using Hg

pressure often measured in mm of Hg** and we knw 1 atm (at sea level) = 760 mmHg

Question 12

Torr

Answer 12

because of Torrecilli

1 atm= 760 mmHg = 760 Torr

usually measure pressure of gases in 1 mm or 1 Torr, 1 atm becuase Pascals big numnber!

Question 13

Pressure in a balloon…. 2

Answer 13

• ex pressure inside ballon really a measure of force attributed to all gas particules colliding with sides of balloon walls, have particles colliding with sides of ballon
• temp measure of KE of those collisions
• inc temp inc pressures, because all particules moving faster with temperature, liek speeding car crashes into wall with more force and inc pressure!
• so P and Temperature are directly related! as temp is inc so is pressure

Question 14

Pascals

Answer 14

b/c pressure is measure of F/A , force= Newton, area= squared meter, another unit for pressure = N per m^2= Pascal

ig measure pressure of atm at sea level=

1 atm= 101, 325 Pascals**

can use to translate into other standard units!

Question 16

as volume goes down….

Answer 16

pressure goes up! in ballon

makes senes less space to move around, particles have more collisions with walls so pressure will inc, again written mathematically means pressure is inversely related to volume!!!

P is inverse of vol

as vol dec, pressure is inc!

Question 17

P and moles

Answer 17

Pressure is directly proportional to moles!

moles= measure of number of particles, more particles more collisions

Question 18

empirical obervations=

Answer 18

obsevations we can actually see versus theroy

Question 19

ideal gas law 1

equation

Answer 19

Pressure is directly related to moles and temperature

Pressure is inversely related to volume

P=NT/V

add constant R…… rearrange

PV= nRT

***for any ideal gas, always get same value R***

so if know, P, V and T can use ths formula to find n (number of particles)

while no gases are ideal, helps us relate before and after conditions when explain this for a gas!

Question 20

ideal gas obeys 5 conditions

Answer 20

3 rules=

1. no intermolecular forces= because if intermolecular forces will interfer with our assumption that all of our kinetic energy is completely directed to pressure!
2. NO VOLUME- molecules occupy no microscopic volume, in other words meaning molecules are point masses and have no volume
3. all collisions are perfectly elastic= gas molecules are in constant motion and randomly collide- elastic collisions- no KE is lost!
4. A gas completely fills it container and exerts pressure in all directions

PRESSURE= is caused by collisions against the container

5. The average KE of gas molecules depends on the TEMPERATURE (move faster with higher temp)

temp increases= KE inc

Question 21

ideal gas law

Rule number 1 explanation

Answer 21

1. no intermolecular forces=

=because if intermolecular forces will interfer with our assumption that all of our kinetic energy is completely directed to pressure!

Gas molecules are far away enough so there are no intermolecular forces between them= no attraction between molecules!!

Question 22

Ideal gas law

Rule 2

Answer 22

2. molecules occupy no microscopic volume, in other words meaning molecules are point masses and have no volume, make this condition becuase in our equation moles and volume are directly related, n directly related to V; if particles as added containers V would go down not up, particules extremely tiny so assuming they are point mases makes sense= NO VOLUME

the volume of gas molecules is so small that it is ignored when compared to the total volume of the container***

The total volume is mostly space- thats why gases have such low densities and are highly compressible!!!

Question 23

ideal gas law 3

Rule explained

Answer 23

3. Perfectly elastic collisions= Gas molecules are in constant motion and randomly collide

ELASTIC collisions= NO KINETIC ENERGY IS LOST

Question 24

real gases 1

Answer 24

non ideal

CONDITIONS that cause nonideal behavior

1. HIGH PRESSURE! molecules are pushed together, creating more IMF’s
2. LOW TEMPERATURE- molecules slow down, creating more IMFs**

remember for ideal gases=

• 1. there are no intermolecular forces
• 2. the space between molecules is ENORMOUS compared to the volume of the molecules themselves

Question 25

STP

Answer 25

STP =273 K and 1 atm

Question 26

for ideal gas constant,

1 mol=

Answer 26

22.4 L

use 1 m³/ 1000 L to convert! = 0.224 m³

Question 27

R=

Answer 27

= 0.821 (atm X L) / mol X K

OR

R= 8.314 (N X M/ mol X K) *remember N X M= J* so can insert a J per mol X K**

for any ideal gases PX V/ nT= R which may look like either of these just have diff values for diff units!

Question 28

Boyle’s Law

Answer 28

Robert Boyle irish scientist

volume is inversely proportional to pressure

INVERSE RELATIONSHIP

If temperature is constant= PiVi= PfVf

i= initial; f= final

P~ 1/V

if graph V vs 1/P straight line, y= mx+b, y intercept is V and X value is inverse of pressure, PV= m (product is constant value)

Question 29

Charles’s Law

Answer 29

100 yrs after Robert Boyle, Jacques Charles french physicist, experiments with gas first person to fill up balloon with gas and fly solo

= Volume is direcrtly proportional to temp

If pressure is constant, closed system with constant mols*

= Vi/Ti = Vf/Tf

**when use ideal gas law need it to be kelvin so not negative values to temp, so convert Celsius degrees to K (C + 273.15)**

As heat system of gas vol inc directly with temperature, or proportionally to inc in temperature

Remember all gases have diff boiling points, turn into liquid at different points**

Also more proof 0 K is absolute zero, can’t have negative volume for gas all gases have to take up some volume!**

So lowest temp we can achieve for all of these gases is 0 K or -273.15 C

Question 30

Avogadro’s Law

Answer 30

-Spent time experimenting with tiny particles so in honor of him number of particles in one mol was named after him

1 mol= 6.02 X 10²³

-Postulated equal volumes of gas at same T and P is equal amount of particles: For instance, if filled up 4 ballons to exactly 1 L at 25 C with diff gases, one Ar (argon), one N2, one H2, one Ch4 (methane) saying at 1 L each of the 4 balloons has same number of particles! so at one liter, each of these balloons would contain 0.041 moles, using Avogadro’s numbers = 2.5 X 10²² particles, 1 L v similar to at stp 1 mol = 6.02 X 10^{23 *****} at 22.4 L

V1/n1=V2/n2

as put in more air particles, put in more air into balloon ballon gets large,r 1 mol of air particles baloon would be abotu 22.4 L, if blew in another 3rd of a mol inc volume to 29.88 L, if blow even more air into this system would inc volume even more…. so another 3rd of a mole of air volume no 27.35 L adn 1.67 mol, volume expandedd proportonally to air put into system

Question 31

Avogadro’s Law 2

Answer 31

volume is directly proportional to the amount (mol) of gas

V1/n1 = V2/n2

ratio! see inc in V is directly proportional to inc in moles, for that reason percent change in V = V2-V1/V1 (1= initial) volume for % change* divided by initial volume gives us % change we know this is equal to % change for moles

V2-V1/V1 = n2-n1/n1

**Doubling the moles of gas will double its volume*** etc….

Question 32

Presure in ideal gas law….

Answer 32

Pressure is directly proportional to temperature

If volume is constant: P1/T1 = P2/T2

Question 33

van der waals equation

Answer 33

P + a (n/v)² X (V-nb) = nRT

ideal gas law modified for intermolecular forces and actual volume

Question 34

Partial Pressure 1 ex find total pressure

Answer 34

ex. 2.1 kg og gas
30. 48% O2, 2.86% H2, 66.67% N2= percentages of total mass
1. if 66.67 % N2 X 2100= 1400 g of N2

molar mass of N2 is 14 X2; 1 mole of N2 = 28 g/ mole

1400 g/ 28 g per mol = 50 moles of N2

2. 30.48 % O2 of 2100 g = 640 grams, 1 mol of O2= (16 X 2) = 32 g/ mole

so 640 g/ 32 g per mole = 20 moles of Oxygen

3. 2.86 % of H2 X 2100 g = 60 g, 1 mole of H2 = 2 g/ mol

60 g/ 2 g/ mol= 30 moles of H2

so even though H super small fraction of total gas we have inside of container, actually have more particles, molecules of H then we do of O* particles are what matter not hte mass when talk about partial pressrue!

20 mol + 30 + 50 = 100 moles of gas** so to find out total pressure

= 100 moles. so PV= nRT =

P 4m³= 100 moles X 8.3145 (m³X Pa/ moles/K) X 273 K

= 25 X 8.3145 X 273= 56746 Pa, actually Pa very small amout of pressure since 101,325 = 1 atm

so if 56,746/ 101, 325 = 56.746 kPa

Question 35

Partial Pressure 1 ex cnt

Answer 35

If total pressure= 56.746 kPa, what is the partial pressure due to N2 molecules its 50 moles?

Partial pressure of N2= 50 moles X 56.746 kPa = 28, 373 Pa (or .28 atm, 28.4 kPa)

Partial pressure H2= 30 moles so 30% of molecules s(if somethign small mass average KE actually moving faster than N2 or O2) partial pressure due ot H= 30% of .56 atm= .168 atm

so total pressure= partial pressures of each gas

P=P_N2 + P_O2 + P_H2

_{.56 atm = .28 + .112 + .168}

Question 36

Boltzmann’s atomic theory

Kinetic molecular explanation of temperature

Answer 36

named after Ludwig Boltzmann, father of modern atomic theory, that world made out of atoms and molecules, obvious to us now but 120 years ago not obvious and ppl really disagreed with them

atomic theory= container of anything full of air, feels like air is continuous, cube of gold feels like continous but now kow its made out of atoms and molecules, like container of steam hand feels hot energy being transfered but not obvious what kind of energies, new energy in disguised?

gas molecules running around in here, actually particles and what you are feeling is these particles striking yoru hand, so small and so many of them can’t tel there are particles looks completely continuous, all this heat energy is that you are feeling is KE if steam its just KE of H2O molecules flying around at some rapid speed, faster they go greater impact with your hand, transfer more energy, faster they go hotter it feels

So for him, if said temp large, if hot outside its redundant, if high temperature means high KE; the average KE of gas molecules outside is larger

This is why it hurts transferring KE to molecules in hand when molecules absorb too much energy and energy starts to move around your hand can damage skin molecules

Question 37

Boltzmann’s constant

Answer 37

K_B= nR= NK_B

K_B= n/N X R = 1/ N_A R

8.31 J/mol X K divided / 6.022 X 10^23 (molescules/mole) using Avogadro’s number

= 1..38 X 10^-23 J/k

can write a more microscopicly oriented version of ideal gas law, number of molecules focus instead of number of moles*** one of the most important constants in all of thermal physics* boltzmann’s own grave stone has this number** S= k log W

entropy of a system S= k (Boltzman’s constant) log (or ln natural logarithm) W (means number of microstates)

so if had macroscopic system and wanted to know microscopicaly all the numbers can use this** particles are doing somethign different on macroscopic level though its identical!

Question 38

to find n in ideal gas law=

Answer 38

n= N (total number of molecules in gas)/ N_A (a constant A’s number)

N_A= Avogadro’s number= 6.022 x 10²³

that many molecules per mole, so 6.022 x 10^23 molecules

Question 39

if want to add internal energy for gases

Answer 39

1. HAVE TO HEAT IT UP, heat will flow into gas causing these particles to ove faster and faster so put in hot plate or something to add heat
2. is to do work on the gas= can take piston and push it down hard enough to squash gas together, impact with piston as moving down causes it to mvoe faster and faster adding internal energy to teh gas** if doign work add energy to gas, but if let gas push up on piston, gas doing work so energy leaving system so then have to subtract W done by gas if outside work is added then +W work done on gas**

work done by gas= subtract, -W

so to write down formula to change internal energy delta U really just changing KE= add Q heat + work (work done on gas) first law of thermodynamics law of conservation fo energy only two ways to add internal energy to gas**

Question 40

gas with work

Answer 40

work done can figure out by P times Delta V but only if Pressure reamins constant! if pressure is constant this gives you an exact way to find work done, how possible for gas to explain and remain at same pressur have to heat it up while gas expands allows pressure to remain constant as heat expands*

Question 41

heat capacity 1

Answer 41

defined to be= certain amount of heat gets added to your gas, how much does temperature inc? thats what heat capacity tells you

heat capacity defined as=

C= Q (amount of heat added to gas) / change in T (amount of change in temperature of that gas)

also hear about molar heat capacity

C= Q/ n X delta T

n= number of moles*

Question 42

Heat capacity at constant volume

Answer 42

C= Q/change in T

ex. if add heat but piston doesnt move, gas stucked no work can be done beucase piston can’t move, gas can’t do work and allow energy to leave, external forces cannot do forces on gas becuse cannot move, so Q only thing adding energy into system

Constant volume = Cv= delta U -W / delta T, but work has to be zero

if delta U= Q + W (from before

constant volume heat capacity = delta U/ delta T= 3/2 PV/ delta T

3/2PV = Q in this case!

so heat capacity for any monatomic ideal gas constant volume = 3/2 NK delta T/ delta T, then delta Ts cancel and constant heat capacity for any monotaumtic gas = 3/2 NK (boltzmann’s constant) N equals total number of molecules or can even write it as 3/2 n (little n number of moles) X R (gas constant)

Question 43

heat capacity for constant pressure

Answer 43

C p = Q/delta T = delta U- W/ delta T

this time W is not zero, W= p X delta V

so work done by gas, if monatomic ideal gas= 3/2 nR delta T + Pdelta V/ delta T

-W becuase work done on gas** since energy is leaving system,

so - P delta V, so two negative signs = positive P delta V

= 3/2 nRdelta T + n R detal T/ delta T

= Constant pressure = 5/2 nR

if want molar divide everything around by little n, which is 5/2 R so almost the same as heat capacity at constant volume is 3/2 nR and heat capacity at constant volume is 5/2 nR

so Cp- Cv = nR

if want molar, just Cp-Cv = R becuase everything divided by n

Question 44

molar heat capacity at constant volume

Answer 44

Cv= 3/2 nR/n canceling out n

so molar heat capacity at constant volume= 3/2 R

Question 45

Kinetic molecular theory of gases

Answer 45

• result of microscopic properties of gas molecules
• like position, x, speed, v; if know speeds and distributions of speeds in box of gas can figure out macroscopic properties, if we know microscopic properites how can we predict macroscopic properites, if knew temp of gas can you say what average speeds of are molecules in this gas….?
• remember one assumption these molecules do not intearct, if they do interact its an elastic collision, momentum and KE conserved* if one molecule strikes wall of container and has a collision there should also be elastic and no KE lost!
• Force on wall by this particle= mvx^2/ L, if just took x component of particles speed get force contribution to pressure on this wall, so thats the force on that wall by one particle
• if want to know force from all particles to get pressure, just add up contributions of all particles, assume same mass m and same L so only difference in contribution some have diff component of velocity in x direction so just add them all up
• = m (vx1^2 + vx2^2 + vx3^2 + …… vxn^2) / L, n= n particules in there
• Force/ N = m/L, tis whole thign divided by N = m/M (vx^2 with bar on top tells you average value of Vx^2) …. look in photo

Question 46

Kinetic molecular theory of gases 2

Answer 46

PV = Nm (vx^2) with bar over it for Vx^2

Vtotal= Vx^2 + Vy^2 + Vz^2 = pythagory theorm, also works if average them all

then… look on image

this is KE so direct relatoonship btw KE average of gas molecule and what macroscopic pressure and volume is** so 3/2PV= NKEaverage

so then put in PV= NKbT

KEaverage= 3/2KbT if know temp can figure out average KE of one of these gas molecules as long as it is an ideal gas!

and NKe average= total energy U, if monatomic gas all it has is KE only energy it can have so

3/2PV= Utotal (total energy of gas) gancy word for total KE boltzman told us nope just KE energy in there; for monatomic only KE in there, so Utotal another word for total KE**

3/2NKbT= total internal energy

or 3/2nRT= total internal energy

monatomic = molecules that make up gas only composed of one gas type lik He, Ne or any neutral gases*

Question 47

Gas phase quiz question 4

Gases found in the environment are most likely to exhibit properties similar to that of ideal gases under conditions of:

Answer 47

Think back to the assumptions of the ideal gas laws.

Hint #22 / 3

In order to obey the ideal gas laws, we would need to minimize the effects of intermolecular interactions and molecular size of these gases. Thus, the ideal conditions will be under high temperatures (lots of kinetic energy in the molecules to move around and not be experiencing forces from each other) and low pressure (less inhibited to move around).

Gases found in the environment are most likely to exhibit properties similar to that of ideal gases under conditions of high temperatures and low pressures.

Question 48

Gas phase quiz Khan Q5

Assume an experiment vessel which is well fitted with a piston. The airtight piston has negligible mass and can move up and down freely. There is 1 mol of an ideal gas contained within the vessel under the piston at pressure of 100 kPa. The piston rests at 10 cm from the base of the vessel. The pressure in the vessel increases to 200 kPa as force is applied to the piston. What would be the new resting position of the piston from the base of the vessel? Assume the temperature is held constant at 300 K throughout the experiment.

Answer 48

Try using Boyle’s Law for this question.

P1V1=P2V2 then rearranged, V2= P1V1/P2. It’s important to realize that to find the volume occupied by the gas in this experiment vessel, we would use A (area of the piston) multiplied by h (the height of the piston). We then would have Ah2= P1X Ah1/P2. The surface areas actually cancel out, and you will have h2= P1h1/ P2.

Substituting in the values, you will get the new resting position as 5 cm.

Question 49

Gas phase quiz Q6

Which properties reflected in real gases does the van der Waals equation attempt to account for by modifying the ideal gas law?

I. Volume

II. Pressure

III. Temperature

Answer 49

Answer= volume and pressure

Real gases experience intermolecular forces between gas molecules (van der Waals forces).

Real gases molecules possess mass.

Given that real gases experience intermolecular forces and possess mass, the van der Waals equation attempts to modify the ideal gas law by introducing the constants a and b, which adjusts for the effects of volume and pressure in real gases. (Real gases have lower pressure and higher volume compared to ideal gases). There is no term in the van der Waals equation that adjusts for temperature.

Question 50

Gas phase quiz Khan Q7

The behavior of which of these real gases will be reflected most closely with the ideal gas law?

Options: Ar, He, CH4, CO2

Answer 50

Consider the factors that distinguish real gases from ideal gases (i.e. what the van der Waals equation attempts to take into account)

Large gas molecules generally possess more volume and intermolecular forces.

Carbon dioxide and methane are large molecules, which immediately means more mass in each molecule, causing deviations from ideal gas behavior. Between helium and argon (both monomolecular noble gases), helium is the smaller gas molecule of the two, which similarly will have less of a mass effect, demonstrating closer adherence to the ideal gas law.

Question 51

Gas phase quiz Khan Q8

Assuming ideal gas properties, which of the following occupies the most volume at 273 K and 1 atm of pressure?

options: one mole of O2, one mole of n2, one mole of H2, they all occupy the same volume

Answer 51

273 K and 1 atm of pressure is known as STP, or standard temperature and pressure.

At STP, all gases occupy 22.4L volume regardless of the identity of the gas.

One mole of nitrogen gas, oxygen gas and hydrogen gas all occupy the same volume at STP assuming ideal gas properties.

Question 52

Gas phase quiz Khan Q9

Atmospheric air is comprised, roughly, of 80% nitrogen and 20% oxygen. A 100 L sample of atmospheric air is kept at 300 K and 100 kPa. How many moles of oxygen molecules are found in this gas sample? (Use R = 10 (LkPa)/(molK))

Answer 52

The law of partial pressures states that the volume of each gas in the mixture is proportional to the partial pressures of each gas in the mixture.

Given atmospheric air is 20% oxygen, the partial pressure of oxygen in this sample is 20 kPa. Now use the ideal gas equation to find n.

n=PV/RT, or n=(100)(20)/(10)(300) = 2/3 moles.

Question 53

Gas phase quiz Khan Q10

Shown below are four mercury barometers of the same height (all four barometer tubes measure one meter from the tube opening to rounded top). Which barometer shows the greatest external pressure?

Answer 53

Consider the difference in shapes between the different barometers.

Given that the force acting on the fluid in the barometer is F=mg. We can derive it thus: m=(density)(volume), so F=(density)(volume)g. Volume=(area)(height), so F=(density)(area)(height)g. We know that F/area = pressure, so we can say that the change in pressure = (the change in height)(density)g.

From this derivation, we see that pressure measurements using a barometer is only dependent on height of the fluid and the density of the fluid. The cross-sectional area of the tube does not affect the pressure measurement.

The first barometer shows the greatest external pressure as it has the tallest column of mercury out of the four barometers.

Question 54

Partial Pressure

Answer 54

For a gas that is a component in a mixture of gases, partial pressure is defined as the pressure that this gas would exert if it took up the same volume by itself. Dalton’s law states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of its components. Thus, the partial pressure is essentially a measure of the relative amount of a gas in a mixture, and it can be related to the mole fraction of a gas (Xgas) through the equation Pgas = XgasPtotal, where Pgas is the partial pressure. In turn, the mole fraction of a certain gas is defined as the number of moles of that gas present divided by the total number of moles of gas in the mixture (Xgas = ngas/ntotal).

A related concept known as vapor pressure describes what happens when a vapor, or gas, is in thermodynamic equilibrium with the liquid phase (or, theoretically, the solid phase as well). Vapor pressure is defined as the pressure exerted by the molecules of that substance that are in gas form in a closed system at a given temperature. If a solute is present in a solution, Raoult’s law states that the presence of solute decreases the vapor pressure: P = XAPA°, where P is the vapor pressure of the solution, XA is the mole fraction of the solvent, and PA° is the vapor pressure of the pure solvent. This is conceptually quite similar to Dalton’s law, as discussed above.