The Maxwell Distribution Laws

What is Root mean square speed?

It is defined as the square root of the mean of squares of the speed of different molecules.

ie. vms=v12+v22+v32+v42+…N=v¯2

As the Pressure due to an ideal gas is given as

P=13ρvrms2⇒vrms=3Pρ=3PV Mass of gas =3RTM=3kTm

Where,

R = Universal gas constant

M = molar mass

P = pressure due to gas

$\rho$ = density

• vrmsαT I.e With the rise in temperature, rms speed of gas molecules increases.
• vrmsα1M I.e With the increase in molecular weight, rms speed of the gas molecule decreases.
• The rms speed of gas molecules does not depend on the pressure of the gas (if the temperature remains constant)

What is the Most probable speed?

This is defined as the speed which is possessed by maximum the fraction of the total number of molecules of the gas.

I.e;vmps=2Pρ=2RTM=2kTm

What is Average speed?

It is the arithmetic mean of the speeds of molecules in a gas at a given temperature.

vavg=v1+v2+v3+v4+…N

And according to the kinetic theory of gases

vavg=8Pπρ=8πRTM=8πkTm

The relation between RMS speed, average speed, and most probable speed

Vrms>Vavg >Vmps

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What is Maxwell’s Law?

The vrms (Root mean square velocity) gives us a general idea of molecular speeds in a gas at a given temperature. So, it doesn't mean that the speed of each molecule is vrms.

Many of the molecules have speeds less than vrms and many have speeds greater than vrms. So, Maxwell derived an equation that describes the distribution of molecules at different speeds as -

dN=4πN(m2πkT)3/2v2e−mv22kTdv

where, dN= Number of molecules with speeds between v and v+dv

So, from this formula, you have to remember a few key points

1. dNdv∝N
2. dNdv∝v2

Conclusions From This Graph

1. This graph is between a number of molecules at a particular speed and the speed of these molecules.

2. You can observe that the dNdv is maximum at the most probable speed.

3. This graph also represent that vrms>vav>vmp.

4. This curve is the asymmetric curve.

5. From this curve we can calculate the number of molecules corresponding to that velocity range by calculating the area bonded by this curve with the speed axis.

Effect of temperature on velocity distribution

With the rising of temperature, the curve starts shifting right side and becomes broader as shown as -

Solved Examples Based on The Maxwell Distribution Laws

Example 1: The value closest to the thermal velocity of a Helium atom at room temperature (300 K)ms−1 is : [kB=1.4×10−23 J/K;mHe=7× 10−27 kg]

1) 13000

2) 1300

3) 130000

4) 130

Solution:

v=3kTm=3∗1.4∗10−23∗3007∗10−27v=1.8∗106=1.3∗103 m/s

Hence, the answer is option (2).

Example 2: At room temperature, a diatomic gas is found to have an r.m.s. speed of 1930 ms^-1. The gas is :

1) H2
2) Cl2
3) O2
4) F2

Solution:

Root mean square velocity

Vrms=3RTM=3Pρ
At room temperature T = 300 K

For diatomic gas,

Vrms=3RTM=⇒M=3RTV2M=3×8.3×300(1930)2=2 g/mole

The diatomic gas is H2

Hence, the answer is the option (1).

Example 3: An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300 K. The mean time between two successive collisions is 6∗10−8 s. If the pressure is doubled and the temperature is increased to 500 K, the mean time (in sec) between two successive collisions will be close to:

1) 5×10−8 s
2) 2×10−8 s
3) 3×10−8 s
4) 4×10−8 s

Solution:

Root mean square velocity

Vrms=3RTM=3Pρ
wherein

R = Universal gas constant

M = molar mass

P = pressure due to gas

$\rho$ = density

Vrms∝TVrms∝ mean freepath timebetweensuccessivecollision
and mean free path =
Y=kT2πσ2P

Vrms∝YbVrms∝Tp×t…….1 but Vrms∝T……….2

From 1 and 2,

T∝TP×tt∝Tpt2t1=(T2T1)×(P1P2)=500300×P12P1=56t2=56t1t2≈4×10−8 s

Hence, the answer is option (4).

Example 4: A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27∘C. The amount of heat (in kJ ) transferred to the gas, so that rms velocity of molecules is doubled is about: [ Take R = 8.3J/K mole ]

1) 10

2) 6

3) 14

4) 0.9

Solution:

Root mean square velocity

Vrms=3RTM=3Pρ

As gas is a closed vessel

Q=nCVΔTV∝T

RMS velocity is doubled

⇒>T2=4T1∴Q=nCV(4T1−T1)=nCV⋅3T1=1528×52R×3×300=10000 J=10 kJ

Hence, the answer is the option (1).

Example 5: A mixture of 2 moles of helium gas (atomic mass =4u ) and 1 mole of argon gas (atomic mass =40u ) is kept at 300 K in a container. The ratio of their ds [Vrms( helium )Vrms( argon )] is close to:

1) 3.16

2) 0.32

3) 0.45

4) 2.24

Solution:

Root mean square velocity -

Vrms=3RTMVrms(He)Vrms(Ar)=MArMHe=404=10

= 3.16

Hence, the answer is the option (1).

Summary

The Maxwell Distribution Laws provide a general description of the spread of gas-specific particle velocities and their effects on the pressure of a gas. This paper reports the existence of various molecular speeds with a greater number of particles that move at a mean speed while others have slow or very high velocities. This realization is important in understanding gas properties under various conditions such as pressure and heat.