1

Chapter 1 - The Properties of Gases I. The perfect gas.

A. The states of gases. (definition)

1. The state variables:

volume=V amount of substance, moles = n pressure = p temperature = T

2. Knowledge of these defines the state of any pure gas. 3. These variables do not vary independently, but are connected by an

equation of state. In general, pressure p = f(T,V,n) e.g., simplest: pV = nRT (perfect or ideal gas) 4. Therefore, knowledge of any three plus the equation of state defines

the state of the gas. 5. V and n are extensive properties, proportional to size of system.

T and p are intensive properties, independent of size of system. 6. Pressure p = force per unit area that a system exerts on its

surroundings. Two major sources of pressure: i) downward gravitational force, and

ii) impacts of molecules on surroundings.

Pressure exerted by a gas is a manifestation of incessant impact of particles on their surroundings.

Above system is in mechanical equilibrium with its surroundings when the force per unit area of system particles pushing the piston out equals force per unit area pushing in.

psystem = psurroundings

Piston stops moving when this is achieved.

2

7. Units of pressure: SI unit of force = N (Newton) = kg m/s2 area = m2 (meters2) so SI unit of pressure = N/m2 1 Pascal (Pa) = 1 N/m2 8. Other more familiar and useful units of pressure: 1 atmosphere (atm) = the pressure required to support a column of

Hg 760 mm high at 0°C. 1 atm = 760. mm Hg = 760. Torr

1 atm = 101.325 kPa (kiloPascals) 1 bar = 100 kPa so 1 atm ~ 1 bar 9. Definition of standard pressure for reporting thermodynamic data: = 1 bar = 750.1 Torr 10. Measurement of pressure - The Torricelli barometer:

Hg rises until the forces balance and atmospheric p = pgrav

This height h is 760. mm at sea level on an average day.

3 Pgrav

= gdh

g = acceleration of gravity = 9.81 m/s2

d = density of Hg = 13.6 g/cm3 or 13.6 x 10

3 kg/m

3

h = 760 mm = 0.760 m

3

11. Temperature T: A measure of thermal energy per unit sample of matter, and a

manifestation of molecular motion, difficult to define in words. It is the property that tells which direction heat will flow.

12. Zeroth Law of thermo = if A is in thermal equilibrium with B, and B

with C, then C is also in thermal equilibrium with A. 13. Measurement of temperature – Thermometer. Simplest are based on thermal expansion of a liquid in a capillary (Hg

or alcohol). Celsius scale: 0 mark in system where water and ice are in equilibrium 100 mark in system where liquid water is boiling at sea level (in

equilb with steam) Above is the typical Hg or alcohol thermometer. Ambiguity: Hg and alcohol don't expand perfectly linearly with T. Better thermometer uses expansion of an ideal gas (The ideal gas

thermometer): since p ∝T at fixed V

14. Temperature scales (K = Kelvin) T/K = θ/°C + 273.15 (θ is symbol used for temperature in °Celsius)

4

B. The gas laws.

1. The ideal gas law (or perfect gas law):

PV = nRT R = universal gas constant = 8.31447 J K-1mol-1 = 0.0820574 L atm K-1mol-1 = 1.98721 cal K-1mol-1 Note: 1 J = 1 Pa-m3 Works well for most gases at room T and 1 atm. Works well for all gases at lower pressures or higher temperatures. 2. The response to pressure - Boyle's Law (1662): p ∝ 1/V (at constant n, T) or p1V1 = p2V2 (at constant n, T) or pV = constant (at constant n, T)

p-V isotherms:

lines of constant T

5

3. The response to temperature: V ∝ T (at constant n, p) - Charles' Law V1/T1 = V2/T2 V-T isobars (lines of constant Pressure) Q: How would this plot look if one plotted vs Celsius temperature? p ∝ T (at constant n, V) - Gay-Lussac's Law p1/T1 = p2/T2

p-T isochores (lines of constant volume)

6

4. Avogadro's principle: V ∝ n (at constant p, T)

molar volume = Vm = volume occupied per mole of molecules =

€

Vn

.

is independent of type of molecule. P Vm = RT Vm = RT/P Problem: Calculate Vm in liters (L) for an ideal gas at 0° C and 1.00

atm pressure. Vm = R(0°C + 273.15)/1.00atm [must use Kelvin!] What R to use? L atm K-1mol-1 Vm = 0.08206 * 273.15K/1.00 atm Vm = 22.415 L 5. Standard conditions for reporting data. STP - old definition, standard temp. and pressure: 0° C, 1 atm SATP - standard ambient temp. & pressure: 298.15K (25° C), 1 bar 6. Mass density of ideal gases.

Let ρ = mass density = m/V = mass/Volume

moles n = m/M = mass/Molar_mass

pV = nRT

pV = (m/M) RT

m/V = Mp/RT

ρ= Mp/RT

Therefore ρ ∝ M

7

7. Combined plot of equation of state at fixed n:

Combined gas laws for fixed number of moles n

𝑝!𝑉!𝑇!

=𝑝!𝑉!𝑇!

Combined gas laws for variable number of moles

𝑝!𝑉!𝑛!𝑇!

=𝑝!𝑉!𝑛!𝑇!

8. Variation of atmospheric pressure with altitude h given by the

barometric formula:

€

p = poe−h /H

where H = RT /MgHere :M = average molar mass of air

g=acceleration of gravity

T is varying with altitude

8

9. Dalton's law - mixtures of gases. The pressure exerted by a mixture of perfect gases is the sum of

partial pressures exerted by the individual gases. total pressure p = nRT/V = (nA + nB + ...)RT/V p = nA RT/V + nB RT/V + ... p = pA + pB + ... pA and pB... are partial pressures, n = nA + nB + … 10. Mole fractions and partial pressures. mole fraction of component J defined as: xJ = nJ/n so xA + xB + ... = 1; therefore: pJ = xJp proof: pJ = nJ RT/V = (nJ/n) (n RT/V) = (nJ/n) P so pJ = xJp

9

II. Kinetic Theory of Gases.

Molecular (microscopic & classical) theory of gases which predicts:

a) equilibrium thermodynamic properties - Pressure b) dynamical properties - rates of processes: effusion through openings,

viscosity of gases, collision frequencies (used in collision theory of chemical reactions)

A. Model of Perfect Gas.

1. Elementary kinetic theory assumptions.

a. Gas sample is mostly empty space (molecules themselves occupy small volume).

b. No attractions or repulsions between molecules in free flight (i.e.

travel in straight lines) c. Perfectly elastic collisions between molecules (no energy

transferred into internal degrees of freedom: rot, vib, elec) d. Pressure exerted by gas = time-averaged force per unit area

exerted by many individual impacts of molecules on a surface. e. Classical variables describe gas. - position

€

x of molecule - velocity

€

v of molecule - vx, vy, vz are Cartesian components of velocity

2. Molecular Speed. molecular speed v = |

€

v| = scalar quantity, not vector v = magnitude of the velocity vector of an individual molecule

€

v2 = vx2 + vy

2 + vz2

v = vx2 + vy

2 + vz2

10

3. Root-mean-square (rms) speed of sample = c =

€

v21/2

... denotes average

(c – whole sample; v – single molecule)

€

v2 =vi

2

Ni=1

N

∑

But

€

vi2 = v

xi

2 + yyi

2 + vzi

2 (

€

vxi veloc in x-dir for molecule i)

v2 =vxi

2 + vyi

2 + vzi

2( )Ni

∑

=vxi

2∑N

+vyi

2∑N

+vzi

2∑N

= vx2 + v

y2 + v

z2

↑ mean-square speed in x-direction But since motion is random:

vx2 = v

y2 = v

z2

So: v2 = 3 v

x2

root-mean-square speed:

v2 or v21/2

has units speed or velocity

11

3. Pressure exerted by N molecules.

Container volume V = a x b x c follow 1 typical molecule and its motion in x-direction time interval between collisions with face A = Δt Δt = time required to travel distance 2a velocity x time = distance

vxΔt = 2a

Δt = 2avx

To calculate pressure P, need force per unit area 1st, find time-averaged force F exerted by 1 molecule avg force in time Δt (during which we have 1 collision)

F = force of 1 collision F = momentum change in 1 collision Δt

F = 2mv

x

Δt=mv

x2

a

P1 = pressure due to 1 molecule = ForceArea

P1=mv

x2 / a( )bc

where area of face A = bc

P1=mv

x2( )

abc=mv

x2

V where V is container volume

Now the pressure due to N molecules in volume V:

P =Nm v

x2

V

12

But we already showed that v2 = 3 vx2

So: P =Nm v2

3V key result

Recognize NV= N = number density (not mass density)

P = N m v2

3

Could also write: PV=

€

13

Nmc2

Now, avg translational (kinetic) energy Et of a single molecule of mass

m is:

€

Et = 12mv2 = 1

2m v2

So: PV = 23NE

t

But we also know that:

PV= nRT or PV= NkBT

(where n = moles and N = molecules) So:

23NE

t= Nk

BT

Et= 32kBT Equipartition theorem result for translational energy

So avg k.e. of a molecule depends only on T, not on mass of particle

13

Problem:

Calculate

€

v2 and

€

v21/2

of He atoms and Cl2 molecules at 300 K.

PV =Nm v2

3= Nk

BT

So:

€

m v2

3= kBT

€

v2 =3kBTm

c = v21/2

=3k

BT

m

!

"##

$

%&&

1/2

or = 3RTM

!

"##

$

%&&

1/2

€

v2

He=

3 ×1.38 ×10−23JK−1 × 300K

4.002 g

mol×10−3 kg

g× 1 mol

6.022×1023

= 1.869 ×106 m2

s2

cHe= v2

He

1/2

= 1,367ms

€

v2

Cl2= v2

He×

4.00270.90

=1.055×105 m2

s2

cCl2

= v2Cl2

1/2

= 325ms

14

B. Distribution of Molecular Velocities. Although every identical gas molecule in a sample may have same

average speeds and average Cartesian velocity components (averaged over period of time), at any instant of t there is a wide range of velocities possible.

Let f(vx)dvx = fraction of molecules having x-component of velocity in

range vx to vx+dvx. Also = prob that a single given molecule has x-comp of velocity in

rangevx to vx+dvx. f(vx) = probability density that a single given molecule has x-comp of

velocity in range vx to vx+dvx.

This distribution is the Maxwell-Boltzmann distribution, which can be derived from the Boltzmann equation

f vx( ) = 2πkBT

m

"

#$$

%

&''

−1/2

e−mvx2 /2kBT

Have similar probability densities in y and z directions

So: F(vx, vy, vz) = prob density that a given molecule has vx, vy, vz components

5

Also = prob that a single given molecule has x-comp of velocity in range vx to vx+dvx.

f(vx) = probability density that a single given molecule has x-comp of velocity in range vx to vx+dvx.

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

dvx

vx

f(vx)

This distribution is the Maxwell-Boltzmann distribution, which can be derived from the

Boltzmann equation

€

f vx( ) =2πkBT

m

−1/ 2

e−mvx

2 / 2kBT

Have similar probability densities in y and z directions So: F(vx, vy, vz) = prob density that a given molecule has vx, vy, vz components Max-Boltz in 3D

€

F vx,vy,vz( ) =2πkBT

m

−3 / 2

e−m vx

2 +vy2 +vz

2( ) / 2kBT

where

€

vx2 + vy

2 + vz2( ) = v2

distribution of molecular velocities F(vx, vy, vz)dvx,dvy,dvz = a probability or fraction

Normalized

€

dvx−∞

∞

∫ dvy−∞

∞

∫ dvz−∞

∞

∫ F vx,vy,vz( ) = 1

€

dvx−∞

∞

∫e−mvx

2 / 2kBT

2πmkT

m

1/2dvy

−∞

∞

∫e−mvy

2 / 2kBT

2πmkT

m

1/2dvz

−∞

∞

∫e−mvz

2 / 2kBT

2πmkT

m

1/2

=1 =1 =1

15

Max-Boltz in 3D

F vx,vy,vz( ) = 2πkBT

m

"

#$$

%

&''

−3/2

e−m vx

2+vy2+vz

2( )/2kBT

where vx2 + v

y2 + v

z2( ) = v2

distribution of molecular velocities F(vx, vy, vz)dvx,dvy,dvz = a probability or fraction, is normalized

Meaning that dvx

−∞

∞

∫ dvy

−∞

∞

∫ dvz

−∞

∞

∫ F vx,vy,vz( ) = 1

dv

x−∞

∞

∫ e−mvx2 /2kBT

2πkTm

%

&''

(

)**

1/2dv

y−∞

∞

∫ e−mvy

2 /2kBT

2πkTm

%

&''

(

)**

1/2dv

z−∞

∞

∫ e−mvz2 /2kBT

2πkTm

%

&''

(

)**

1/2

=1 =1 =1

Calculation of averages using M-B distribution

vx2 = dv

x−∞

∞

∫ dvy

−∞

∞

∫ dvz

−∞

∞

∫ vx2F v

x,vy,vz( )

dvx

−∞

∞

∫vx2e−mvx

2 /2kBT

2πkTm

%

&''

(

)**

1/2dv

y−∞

∞

∫ e−mvy

2 /2kBT

2πkTm

%

&''

(

)**

1/2dv

z−∞

∞

∫ e−mvz2 /2kBT

2πkTm

%

&''

(

)**

1/2

≠1 =1 =1

integral over vx is integral of type x2

−∞

∞

∫ e−αx2

dx = π

2α3/2

where

α =m2kBT

16

So: vx2 =

π

2 m2kBT

"

#$$

%

&''

3/22πk

BT

m

"

#$$

%

&''

1/2• 1( ) • 1( )

vx2 =

kBT

m

v2 = 3 vx2 =

3kBT

m=3RTM

c2 = v2

c2 =3kBT

m

!

"##

$

%&& =3RTM

!

"##

$

%&&

C. Maxwell Distribution of Speeds.

Let F(v)dv = fraction of molecules having speed in range v to v+dv or = prob that a given molecule has speed in range v to v+dv (v= length of the velocity vector) F(v)dv = F(vx,vy,vz) 4πv2 dv

Maxwell Boltzmann distribution of speeds

F(v)dv = 2πk

BT

m

"

#$$

%

&''

−3/2

e−mv2/2kBT 4πv2dv ≈ v2e−αv

2

RMS speed v21/2

which = c

v2 = dv0

∞

∫ F(v)v2

v2 =3RTM

!

"##

$

%&&

17

v21/2

=3RTM

!

"##

$

%&&

1/2

= c

v21/2

≠ v ≠ c *

3RTM

!

"##

$

%&&

1/2

8RTπM

"

#$$

%

&''

1/2

2RTM

!

"##

$

%&&

1/2

Speed of sound ~ average speed in gas

D. Types of Average Speeds.

1. Most probable speed c*

maximum of F(v)

occurs at dF(v)/dv = 0

Solve for v

c* = 2RTM

!

"##

$

%&&

1/2

2. Mean speed v or = c

v = dv0

∞

∫ F(v)v

v =2πkTm

"

#$$

%

&''

−3/2

dv0

∞

∫ e−mv2/2kT4πv2 × v

is integral of type dx x3∫ e−αx2

v =8RTπM

"

#$$

%

&''

1/2

=2.546 RTM

"

#$$

%

&''

1/2

3. Mean relative speed = c

rel

The mean speed at which a molecule approaches another molecule.

Collision rates between molecules depends on mean relative speed c

rel= 21/2c

18

and for the case of two dissimilar molecules:

crel=8RTπµ

"

#$$

%

&''

1/2

where µ is reduced mass

Proof:

mean relative speed = crel= v

1

2+ v

2

2!

"#

$

%&

1/2

mean speed mean speed of 1 of 2

Now since vi=8k

BT

πmi

"

#$$

%

&''

1/2

We have:

crel=8k

BT

πm1

+8k

BT

πm2

"

#$$

%

&''

1/2

=8k

BT

π

"

#$$

%

&''

1/2

1m1

+1m2

"

#$$

%

&''

1/2

=1/µ12

reduced mass

crel=8k

BT

πµ12

"

#$$

%

&''

1/2

µ12=m1m2

m1+m

2

Note: if m1 = m2 crel = 28k

BT

πm1

"

#$$

%

&''

1/2

= 2c

E. Kinetic Energy Distribution. FE(E)dE = fraction of molecules with K.E. value in interval E to E+dE

Given E = 12mv2 = 1

2m v

x2 + v

y2 + v

z2( )

FE(E)dE = e−E/kBT2π E

πkBT( )3/2

dE

19

Problem: Gas phase reaction has activation energy = Ea. What fraction f of gas sample has kinetic energy greater than some required activation energy Ea for chemical reaction?

fraction f = dE2π E e

−E/kBT

πkBT( )3/2

Ea

∞

∫

F. Collision rates with a wall or opening in a wall. Use: calculation of pressure, escape rate (effusion)

calculate zw = # of collisions

unit Area x unit time= "collision frequency"

consider small time interval = dt patch of area = A

# of collisions due to molecules with z-veloc will be number of molecules within distance = vzdt of patch (only those will strike)

That will = number in cylinder of volume Avzdt Which will = NAvzdt where N is number density of molecules (# per unit vol) # of collisions due to

molecules with vz velocity = NAvzdt f v

z( ) f vx( ) f vy( )dvxdvydvz

Total # = NAdt vz

0

∞

∫ f vz( )dvz dvx∫ f v

x( ) dvy∫ f vy( )

=1 =1

zw=

total # collAdt

zw = N dvz∫ vzf v

z( )

= N dvz∫2πk

BT

m

#

$%%

&

'((

−1/2

vze−mvz

2 /2kT

zw = N RT2πM

20

or using c = v =8RTπM

"

#$$

%

&''

1/2

zw = N c4

G. Collisions between molecules. Need to consider this in order to determine:

-mean free path -diffusion coefficient -heat conductivity -viscosity 1. Collision cross section σ and effective target

σ = area the molecule sweeps out as it travels, σ = πd2 , where d = hard sphere diameter, no long-range attractive forces

2. Collision frequency z of one molecule

z = σcrelN

=σcrelp

kBT

Collision cross-sections

21

3. Mean free path λ = avg distance traveled by a molecule between collisions

λ =crel

z

=kBT

σP

note λ ∝1P

Problem

A requirement for effusive flow through an opening is that radius of opening must be small compared to λ. O2 has effective spherical diameter d =3.61Å. At 25°C at what pressure will the mean free path be equal to a hole of radius = 1 mm? require λ = 1 mm = 10-3m

λ =kBT

σP P =

kBT

σλ σ = πd2

P =1.38 ×10−23J°K ×298K

π 3.61×10−10m( )2

10−3m= 10.0 pascals

x 1 atm

101325Pa= 9.91×10−5atm

4. Graham’s Law of Effusion

already have zw = collisions per unit time per unit area with wall or opening A = area of opening

Effusion rate R = zwA = AkBT

2πm

"

#$$

%

&''

1/2

N

# density

22

Compare 2 gases

R1

R2

=m2

m1

=M2

M1

molec weights

R ∝

1

M

Requirement for effusion: λ > diameter of opening Then. molecules pass independently through the opening rather than

creating a hydrodynamic flow. 5. Collision frequency between molecules.

Let z1(2) = # of collisions per unit time of one molecule of type 1 with all molecules of type 2 # of collisions in = vol of this cylinder ∗ N2 time dt ↑ number density of type 2 molecules

= creldtπ r

1+ r2( )2 N2

let d12 = r1 + r2 = collision diameter for molec 1 & 2 types

z1(2) = c

relπd122 N2

z1(2) = 8k

BT

πµ12

"

#$$

%

&''

1/2

πd122 N2

Now, total rate of collisions per unit volume between molecules of type 1 and those of type 2 is:

Z12 = z1(2) N 1 = 8k

BT

πµ12

"

#$$

%

&''

12

πd122 N1 N2

23

Collision theory of gas phase reaction rates uses this collision frequency. Rate = Z12 x (Probability of successful rxn upon collision)

≈ e−Ea/kBTf Ea = activation energy f = geometric factor

Rate = 8k

BT

πµ

"

#$$

%

&''

1/2

πd122 fe−Ea/kBT N1 N2

~rate constant k gives concentrations = k c1c2 (2nd order rate law) If type 1 and type 2 molecules are identical:

Z12 → Z11 = v πd2

2 N1

2 where d is diam of 1 v is 8k

BT

πm1

"

#$$

%

&''

1/2

24

III. Real gases.

A. Molecular Interactions.

1. In general - gases deviate from ideal behavior due to interactions between molecules.

Repulsive forces - exist over very short range, a molecular diameter. Called excluded volume forces. Causes deviation from ideality only when the density is so high that the physical volume of the molecules begins to become comparable to V.

Attractive forces - exist over several molecular diameters.

2. The compression factor Z = a dimensionless quantity expressing

deviation from ideality.

€

Z ≡pVnRT

= 1 for gas obeying ideal gas law.

Since Vn

= Vm

Z =pVm

RTAlso,

Z =Vm

Vmo

where Vmo is ideal gas molar volume = RT/p

25

26

3. Virial expansions: Since ideality pertains at low p, this suggests that we mathematically

expand Z in the pressure: Z = 1 + B'p + C'p2 + ...

B', C' are called the 2nd and 3rd virial coefficients. Virial coefficients are not just curve-fitting parameters, but can be

calculated from molecular properties by statistical mechanics. Alternately, the virial expansion can be expressed in powers of 1/Vm

€

Z = (1+BVm

+C

Vm2 + ...)

4. Condensation:

Decreasing the volume of a confined gas (increasing its p and

density) at constant T below a certain T called the critical temperature Tc causes condensation.

p-V isotherms of CO2 -------------

27

5. Critical constants: Every gas has a critical point with a critical pressure Pc critical volume Vc critical temperature Tc Tc = temperature above which a gas can not be made to undergo a

condensation phase transition no matter how much it is compressed. Always one phase.

Synoptic Table 1C.2 Critical constants of gases

B. The van der Waals equation - an intuitively appealing approximate equation of state for a real gas. 1. Constructing the equation. a) First account for repulsive forces. Replace V in pV=nRT by the

effective available volume the molecules have to move in:

Modified equation becomes P(V-nb) = nRT Where b = excluded volume vdW parameter Excluded volume effect is to increase the P at fixed n, T.

2r

Excl volume

28

Derive relation between b and the physical volume of a particle for the hard sphere gas case:

Answer: b = 4Vparticle*Navo For example: if molecule is 1.0 Å in radius b = 4* (4/3)π (1 Å)3 * 6.02 x 1023=1.009 x 10-5 m3/mol ~ 1.0 x 10-2 L/mol b) Second account for attractive forces. Should be proportional to the

frequency of collisions in the sample: collision frequency ∝ (n/V)2 = (molar conc)2 (P + a n2/V2)(V-nb) = nRT Where a = vdW attractive forces parameter. Attractive term effect is to decrease the observed pressure p,

everything else remaining fixed. Every gas is characterized by parameters a and b.

Synoptic Table 1C.3 van der Waals coefficients

So what is the effective “hard sphere” radius of an Argon atom?

b = 4VArgon*N

avo

3.2 ×10−2 Lmol

= 4 43πR3

$

%&&

'

())Navo

Solve for R. (Top Hat question)

29

2. van der Waals isotherms:

3. Relation of a and b to critical constants: Since vdW equation predicts the critical point, derive relationship

between critical point and parameters a and b.

At critical point:

dpdV

= 0 d2pd2V

= 0

Use n=1 and differentiate vdW equation:V

c= 3b

Pc=

a27b2

Tc=

8a27bR

30

Table 1C.4 Selected equations of state

4. Relation of a and b to virial coefficients: By expanding p in a power series, we identify B = b - a/RT (second virial coefficient)

C. The principle of corresponding states.

Define reduced variables (dimensionless variable):

Pr = P/Pc Vr = Vm/Vc Tr = T/Tc

Observation that real gases behave virtually identically if expressed in

terms of reduced (dimensionless) variables. I.e., at same reduced V & T they exert approximately the same reduced

pressure. Works very well for spherical molecules. Why does this work so well? To see this rewrite vdW equation in terms of

reduced variables. Start with (P + an2/V2)(V-nb) = nRT

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set n = 1 (P + a/Vm

2)( Vm -b) = RT Make substitutions: (PrPc + a/Vm

2)(VrVc - b) = RTrTc Now sub in Vc = 3b, Pc = a/(27b2), Tc = 8a/(27Rb) Now reorganize, noting that a's + b's disappear, leaving:

€

pr =8Tr

3Vr −1−

3Vr

2

Since no a or b remain in the equation, the equation of state in

reduced variables is identical for every type of gas (as long as vdW equation is valid).

Example: H2 at 1 atm and 298 K behaves identically to NH3 at 8.7 atm

and 3638 K. How did we determine this? Look up the critical constants of the gases in question and use the principle of corresponding states.

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Notes: