The hydrogen atom has one proton and one electron. The mass of electron is negligible compared to the mass of proton.
Mass of one proton = 1.67\(\times\)10⁻²⁷kg.
One hydrogen molecule = 2 hydrogen
atoms = 2\(\times\)1.67\(\times\)10⁻²⁷kg.
The average speed
\(\overset { - }{ v } =\sqrt { \frac { 8KT }{ \pi m } } =1.60\sqrt { \frac { KT }{ m } } =\)
\(=1.60\sqrt { \frac { \left( { 1.38\times 10 }^{ -23 } \right) \times \left( 300 \right) }{ 2\left( 1.67\times { 10 }^{ -27 } \right) } } =1.78\times { 10 }^{ 3 }{ ms }^{ -1 }\)
(Boltzmann Constant k = 1.38\(\times\)10⁻²³ J K^-1)
The rms speed \({ v }_{ rms }=\sqrt { \frac { 3KT }{ m } } =1.73\sqrt { \frac { kT }{ m } } \)
\(=1.73\sqrt { \frac { \left( { 1.38\times 10 }^{ -23 } \right) \times \left( 300 \right) }{ 2\left( 1.67\times { 10 }^{ -27 } \right) } } =1.9\times { 10 }^{ 3 }{ ms }^{ -1 }\)
Most probable speed \({ v }_{ mp }=\sqrt { \frac { 2KT }{ m } } =1.41\sqrt { \frac { kT }{ m } } \)
\(=1.41\sqrt { \frac { \left( { 1.38\times 10 }^{ -23 } \right) \times \left( 300 \right) }{ 2\left( 1.67\times { 10 }^{ -27 } \right) } } =1.57\times { 10 }^{ 3 }{ ms }^{ -1 }\)
Note that v_rms > \(\overset { - }{ V } \) > v_mp

The average speed
\(\overset { - }{ v } =\frac { 2+3+4+5+5+5+6+6+7+9 }{ 10 } =5.2{ ms }^{ -1 }\)
To find the rms speed, first calculate the mean square speed \(\overset { - }{ { v }^{ 2 } } \)
\(\overset { - }{ { v }^{ 2 } } =\frac { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 }+6^{ 2 }+{ 6 }^{ 2 }+{ 7 }^{ 2 }+{ 9 }^{ 2 } }{ 10 } \)
= 30.6ms²s^-2
The rms speed
\({ v }_{ rms }\sqrt { \overset { - }{ { v }^{ 2 } } } =\sqrt { 30.6 } =5.53{ ms }^{ -1 }\)
The most probable speed is 5 m s^-1 because three of the particles have that speed.

The internal energy of ideal gas = \(\frac{3}{2}\)NKT
The number of air molecules is given in terms of number of moles so, rewrite the expression as follows 
U = \(\frac{3}{2}\)\(\mu\)RT
Since Nk = μR. Here μ is number of moles.
Gas constant R = 8.31\(\frac{J}{molK}\)
Temperature T = 273 + 27 =300K
U =\(\frac{3}{2}\)\(\times\)0.5\(\times\)8.31\(\times\)300 = 1869.75J
This is approximately equivalent to the kinetic energy of a man of 57 kg running with a speed of 8 m s^-1.

The specific heat of one mole of a monoatomic gas C_V = \(\frac{3}{2}\)R
For \(\mu\)₁ mole , C_V = \(\frac{3}{2}\)\(\mu\)₁R  C_p = \(\frac{5}{2}\)\(\mu\)₁ R
The specific heat of one mole of a diatomic gas
C_v = \(\frac{5}{2}\)R
For μ₂ mole, C_V = \(\frac{5}{2}\)μ₂ R C_P = \(\frac{7}{2}\)μ₂ R
The specific heat of the mixture at constant volume C_V = \(\frac{3}{2}\)\(\mu\)₁R +\(\frac{5}{2}\)\(\mu\)₂ R
The specific heat of the mixture at constant pressure C_P = \(\frac{5}{2}\)\(\mu\)₁ R + = \(\frac{7}{2}\)\(\mu\)₂ R
The adiabatic exponent \(\gamma =\frac { { C }_{ p } }{ { C }_{ V } } =\frac { 5{ \mu }_{ 1 }+{ 7\mu }_{ 2 } }{ 3{ \mu }_{ 1 }+{ 5\mu }_{ 2 } } \)

From (9.26) \(\lambda =\frac { 1 }{ \sqrt { 2 } \pi { nd }^{ 2 } } \)
We have to find the number density n By using ideal gas law
\(n=\frac { N }{ V } =\frac { P }{ KT } =\frac { { 101.3\times 10 }^{ 3 } }{ 1.381\times { 10 }^{ -23 }\times 300 } \)
= 2.449\(\times\)10²⁵ molecues/m³
\(\lambda =\frac { 1 }{ \sqrt { 2 } \times \pi \times 2.449\times { 10 }^{ 25 }\times \left( 1.2\times { 1 }0^{ -10 } \right) ^{ 2 } } \)
\(=\frac { 1 }{ 15.65\times 10^{ 5 } } \)
λ = 0.63\(\times\)10⁻⁶m