Question

Pressure of an ideal gas is obtained from kinetic gas equation. The kinetic gas equation is: $PV=\frac{1}{3}\left(m{N}_A\right)u^2=\frac{1}{3}M{u}^2$, where, $m{N}_A=M$(molar mass)
$N_A=\text{Avogadro's number}$
$u=\text{root mean square velocity}$
$\text{Translational kinetic energy of n mole}\frac{1}{2}M{u}^2=\frac{3}{2}\left(PV\right)=\frac{3}{2}\left(nRT\right)$
$\text{Average translational kinetic energy per molecule}=\frac{3}{2}\left(\frac{RT}{N_A}\right)=\frac{3}{2}\left(KT\right)$
$u_{rms}=\sqrt{\frac{3PV}{M}}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3P}{d}}$
and $u_{av}=\sqrt{\frac{8RT}{\pi M}}$
The mass of molecule A is twice the mass of molecule B. The rms speed of A is twice the rms speed of B. If two samples A and B contain same number of molecules, what will be the ratio of pressures of 2 samples in 2 separate containers of equal volume?

• A. $\frac{P_A}{P_B}=2$
• B. $\frac{P_A}{P_B}=3$
• C. $\frac{P_A}{P_B}=4$
• D. $\frac{P_A}{P_B}=8$

Answer 1

The correct option is D $\frac{P_A}{P_B}=8$

Given $m_A=2m_B$
$(u_{rms}{)}_A=2(u_{rms}{)}_B$
$\Rightarrow \text{molecular weight of A}=2\text{(molecular weight of B)}$
Also number of molecules of A = number of molecules of B
For gas A, $P_A{V}_A=\frac{1}{3}{M}_A(u_{rms}{)}_A^2$
For gas B, $P_B{V}_B=\frac{1}{3}{M}_B(u_{rms}{)}_B^2$
So, $\frac{P_A{V}_A}{P_B{V}_B}=\frac{M_A}{M_B}\times \frac{(u_{rms}{)}_A^2}{(u_{rms}{)}_B^2}$
Given, $V_A=V_B$
$\Rightarrow \frac{P_A}{P_B}=\frac{2M_B}{M_B}\times \frac{(2u_{rems}{)}_B^2}{(u_{rms}{)}_B^2}$
$\Rightarrow \frac{P_A}{P_B}=2\times 4=8$