Average translational kinetic energy of an ideal gas molecule is $$\frac{3}{2}\,kT$$  which depends on temperature only. Therefore, if temperature is same, translational kinetic energy of $${O_2}$$ and $${N_2}$$ both will be equal.

From ideal gas equation $$PV = \mu RT$$
⇒ Slope of $$P.T$$  curve $$\frac{P}{T} = \frac{{\mu R}}{V}$$
$$ \Rightarrow {\text{Slope}} \propto \frac{1}{V}$$
It means line of smaller slope represent greater volume of gas.
For the given problem figure
Point 1 and 2 are on the same line so they ill represent same volume i.e. $${V_1} = {V_2}$$
Similarly point 3 and 4 are on the same line so they will represent same volume i.e. $${V_3} = {V_4}$$
Also slope of line 1-2 is less than 3-4.
Hence $$\left( {{V_1} = {V_2}} \right) > \left( {{V_3} = {V_4}} \right)$$

$$\eqalign{ & {\rho _A} = 1.5\,{\rho _{B}}\,\,\,\,{\rho _B} \cr & {\rho _A} = 2{\rho _B}\,\,\,\,\,\,\,{p_B} \cr} $$
According to ideal gas equation, we have
Pressure, $$p = \frac{{\rho RT}}{M},$$   where $$M$$ is molecular weight of ideal gas.
Such that, $$\frac{p}{\rho } = \frac{{RT}}{M}$$
$$ \Rightarrow M = \frac{{\rho RT}}{P}$$
where, $$R$$ and $$T$$ are constants.
So, $$M \propto \frac{\rho }{p} \Rightarrow \frac{{{M_A}}}{{{M_B}}} = \frac{{{\rho _A}}}{{{\rho _B}}} \times \frac{{{p_B}}}{{{p_A}}}$$
$$ = 1.5 \times \frac{1}{2} = 0.75 = \frac{3}{4}$$

At high temperature for diatomic gas $$f = 7$$
$${C_p} = \frac{{fR}}{2} + R = \frac{{9R}}{2}$$

Since pressure and volume are not changing, so temperature remains same.

Given, $${V_1} = V,{V_2} = 2V$$
$${T_1} = {27^ \circ } + 273 = 300\,K\,{\text{so,}}\,{T_2} = ?$$
From charle’s law
$$\eqalign{ & \frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}}\,\,{\text{or,}}\,\,\frac{V}{{300}} = \frac{{2V}}{{{T_2}}}\,\,\left( {\because {\text{Pressure is constant}}} \right) \cr & \therefore {T_2} = 600\,K = 600 - 273 = {327^ \circ }C \cr} $$

$$\frac{{K.E.}}{{vol}} = \frac{3}{2}P.$$
Here $$P$$ is constant

All three vessels are at same temperature. According to Maxwell's distribution of speed, average speed of molecules of a gas $$v \propto \sqrt T .$$
∴ The velocity of oxygen molecules will be same in $$A$$ as well as $$C$$ where $$M$$ is the mass of an oxygen molecule.

On raising the temperature, the average velocity of the gas molecules increases. As a result of which more molecules collide with the walls or number of collisions per unit time will increase.

Here $${v_1} = 200\,m/s;$$
temperature $${T_1} = {27^ \circ }C = 27 + 273 = 300k$$
temperature $${T_2} = {127^ \circ }C = 127 + 273 = 400k,$$
$$V = ?$$
$$R.M.S.$$  Velocity, $$V \propto \sqrt T $$
$$ \Rightarrow \frac{v}{{200}} = \sqrt {\frac{{400}}{{300}}} \Rightarrow v = \frac{{400}}{{\sqrt 3 }}m/s$$