If for two gases of molecular weights ${\mathrm{M}}_{\mathrm{A}} \mathrm{and} {\mathrm{M}}_{\mathrm{B}} \mathrm{at} \mathrm{temperature} {\mathrm{T}}_{\mathrm{A}} \mathrm{and} {\mathrm{T}}_{\mathrm{B}}, {\mathrm{T}}_{\mathrm{A}}{\mathrm{M}}_{\mathrm{B}} = {\mathrm{T}}_{\mathrm{B}}{\mathrm{M}}_{\mathrm{A}},$ then which property has the same magnitude for both the gases.

(i) density of a gas $\left(\mathrm{\rho }\right)=\frac{\mathrm{PM}}{\mathrm{RT}}$

Since $\frac{{\mathrm{M}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{B}}} = \frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{T}}_{\mathrm{A}}},$

$\therefore$ at the same pressure ${\rho }_A={\rho }_B$

But if pressure is different, then ${\rho }_A\ne {\rho }_B.$

(ii) Pressure of the gases would be equal if their densities are equal otherwise not.

(iii) KE per mol = $\frac{3}{2}RT$

$\therefore$ It will be different for the two gases.

(iv) $$\begin{gathered} {\mathrm{V}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}}, \mathrm{since} \frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{A}}}=\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}}; \\ {\mathrm{V}}_{\mathrm{rms}} \mathrm{of} \mathrm{A} = {\mathrm{V}}_{\mathrm{rms}} \mathrm{of} \mathrm{B} \end{gathered}$$