Problems on Kinetic Energy
Q1

What is the average translational kinetic energy of molecules in an ideal gas at 40°c ?

Sol:

Average translational KE = KT

K = Boltzmann's constant = 1.38 × 10–23 J/k
T = absolute temperature = 40 + 273
= 313 k
∴ = (1.38 × 10– 23)(313)
= 6.48 × 10–21 J
Q2

Calculate the temperature at which rms velocity of gas molecules is double its value at 27° c, pressure of the gas remaining constant.?

Sol:

Let T2 be the required temperature
Initial temperature T1 = 27°c = 27 + 273 = 300 k
Let initial rms velocity Vrms1 = V
Given that rms velocity at T2 is twice Vrms2 = 2V

We know that Vrms =
here = constant
=
T2 = 4T1
= 1200 k
= 927°c
Q3

Calculate the total number of degrees of freedom possessed by the molecules in one cm3 of H2 gas at NTP

Sol:

We know that at NTP condition 22.4 lts = 2.4 × 103 cm3 of any gas contains 6.023 × 1023 molecules
∴ Number of molecules in 1 cm3 of H2 gas = = 0.26875 × 1020
Number of degrees of freedom of a H2 gas molecule is 5

∴ Total number of degrees of freedom of 0.26875 × 1020 molecules = 0.26875 × 1020 × 5
= 1.34375 × 1020
Q4

Find the mean free path of air molecules at STP. The diameter of O2 and N2 molecules is about 3 × 10–10 m

Sol:

Mean free path, λ =
here n = number of molecules per unit volume
we know that 1 mole of an ideal gas at STP occupies a volume 22.4 × 10–3 m3

∴ n =
= 2.69 × 1025 molecules/m3
∴ Mean free path λ =
λ = 9 × 10–8 m
Q5

How many degrees of freedom are associated with 2g of He at NTP? Calculate the amount of heat energy required to raise the temperature of this amount from 27°c to 127°c.

Sol:

Molecular weight of He = 4
∴Number of molecules in 4g of He = 6.02 × 1023 molecules
Number of molecules in 2g of He = × 6.02 × 1023 = 3.01 × 1023 molecules
As He is a monoatomic gas, degrees of freedom associated with each of its molecule is 3.

Now total degrees of freedom of 2g of He is
f = (Total number of molecules) × (Degrees of freedom per molecule)
f = 3.01 × 1023 × 3
= 9.03 × 1023
We know that energy associated with one degree of freedom per molecule
U = KT
Energy associated with 2g of He is
U = (Total degrees of freedom) × KT
U = f × KT
U = 9.03 × 1023 × 1.38 × 10 -23 × T
Energy at 27° c = 27 + 273
= 300 k
U1 = 9.03 × 1023 × 1.38 × 10 -23 × 300
= 1869.2 J
Energy at 127° c = 127 + 273
= 400k
U2 = 9.03 × 1023 × 1.38 × 1023 × 400
= 2492.3 J
Heat energy required to raise the temperature from 27° c to 127° c is
U2 – U1 = 2492.3 – 1869.2
= 623.1 J
Q6

The pressure of sulphur dioxide (SO2) is 2.12 × 104 pa. There are 420 moles of this gas in a volume of 50 m3. Find the translational rms speed of the SO2 molecules

Sol:

Translational rms speed is related to the pressure and density of the gas by the equation

Vrms =
ρ = density of the gas
=
Given volume of the gas V = 50 m3
Molar mass of SO2 = 64 g/mol
But there are 420 moles in 50 m3 gas
∴ Total mass m = 64 × 420
Pressure of the gas is P = 2.12 × 104 pa
Substituting all these values, we get
Vrms =
=
=
= 10.8 m/s