Hydrogen would leak out faster as r.m.s. speed of hydrogen is greater than the r.m.s. speed of CO₂,

Both gases have same value of average translational kinetic energy per mole because their temperatures are equal and \(\bar { { E } } =\frac { 3 }{ 2 } RT\)

Generally a molecule of a diatomic gas possesses 5 degrees of freedom at room temperature. 3 due to translational motion and 2 due to rotational motion.

Mean free path of a gas is directly proportional to its temperature on kelvin scale.

Yes, the number of degrees of freedom of a gas molecule may increase with rise in temperature. It is because at higher temperatures vibrational motion may also take place in gas molecules

The ratio of r.m.s velocities is given as
\(\frac { { { C }_{ 1 } } }{ { C }_{ 2 } } =\sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } =\sqrt { \frac { { \rho }_{ 2 } }{ { \rho }_{ 1 } } } \)
\(\frac { { C }_{ 1 } }{ { C }_{ 2 } } =\sqrt { \frac { 9 }{ 6 } } =\sqrt { 3 } :\sqrt { 2 } \)

Given, T = 27⁰C = 273 + 27 = 300K,
p = 1atm = 1.01 x 10⁵N/m²
d = 2 x 2 x 10^-10m = 4 x 10^-10
\(\therefore \ Mean\ free\ path,\ \lambda ={ K }_{ B }T/\sqrt { 2 } \pi { d }^{ 2 }p\)
\(\\ =\frac { 1.38\times { 10 }^{ -23 }\times 300 }{ 1.414\times 3.14(4\times { 10 }^{ -10 })^{ 2 }1.013\times { 10 }^{ 5 } } =5.75\times { 10 }^{ -8 }m\)

The law of equipartition of energy for any dynamical system in thermal equilibrium, the total energy is distributed q = equally amongst all the degrees of freedom.
The energy associated with each molecule per degree of freedom is \(\frac { 1 }{ 2 } { k }_{ B }T\), where K_B is Boltzmann's constant and T is temperature of the system.

Oxygen is a diatomic gas.
Number of moles of O₂ gas
\(=\frac { Atomic\ wt. }{ Molecular\ wt. } =\frac { 8 }{ 32 } \)
\(\\ =\frac { 1 }{ 4 } =0.25\)
\(\\ \therefore \ Energy\ associated\ with\ 1\ mole\ of\ oxygen\)
\(\\ U=\frac { 5 }{ 2 } RT\)
\(\\ \therefore \ Internal\ enreyg\ of\ 8g\ of\ oxygen=0.25\times \frac { 5 }{ 2 } \times 8.31\times 273=1417.9J\)

Number of degrees of freedom of diatomic gas at 10000K = 7.
Number of degrees of freedom of diatomic gas at 300K = 5
\(\therefore \frac { { E }_{ 1 } }{ { E }_{ 2 } } =\frac { (\frac { 7 }{ 2 } ){ k }_{ B }{ T }_{ 1 } }{ (\frac { 5 }{ 2 } ){ k }_{ B }{ T }_{ 2 } } =\frac { 7 }{ 5 } \times \frac { { T }_{ 1 } }{ { T }_{ 2 } } =\frac { 7 }{ 5 } \times \frac { 10000 }{ 300 } =\frac { 140 }{ 3 } \)

As three vessels are identical i.e., they have same volume now at constant pressure, temperature and volume the three vessels will contain equal number of molecules (by Avogadro’s law) and is equal to Avogadro's number, NA = 6.0²³ x 10²³
\(\because { V }_{ rms }=\sqrt { \frac { 3{ k }_{ B }T }{ m } } \Rightarrow { V }_{ rms }\propto \frac { 1 }{ \sqrt { m } }\)
where, m is mass of single gas molecule as neon has the smallest mass, so rms speed will be greatest in case of neon.

We have , 0.25 x 6 x 10²³ molecules, each of volume
Molecular volume = 2.5 x 10^-7 m³
Supposing, ideal gas law is valid.
Final volume =  \(\frac { { V }_{ in } }{ 100 }\) = \(\frac { { (3) }^{ 3 }\times { 10 }^{ -6 } }{ 100 }\)
\(\approx\) 2.7\(\times\)  10 ^-7 m³ 
Which is about the molecular volume. hence, intermolecular forces cannot be neglected. Therefore, the ideal gas situation does not hold.

68.25 K

390 m/s

\(\simeq 200\quad K\)