Question

The molecules of a given mass of gas has r.m.s speed $400\text{m/s}$ at ${27}^{\circ }\text{C}$ and ${10}^5{\text{N/m}}^2$ pressure . When the absolute temperature is doubled and the pressure is halved, the r.m.s. speed of the molecules of the same gas is

• A. $400\text{m/s}$
• B. $200\sqrt{2}\text{m/s}$
• C. $400\sqrt{2}\text{m/s}$
• D. $1200\text{m/s}$

Answer 1

The correct option is C $400\sqrt{2}\text{m/s}$

According to ideal gas equation
$PV=nRT$
$\Rightarrow \frac{PV}{T}=$ constant
Let,
$P_1,V_1$ and $T_1$ as initial pressure, volume and temperature of gas respectively.
$P_2,V_2$ and $T_2$ as final pressure, volume and temperatue of gas respectively.
$\therefore \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$......(1)

We know that,
Pressure $P=\frac{1}{3}\frac{M}{V}{v}_{\text{rms}}^2$
$PV=\frac{1}{3}M{v}_{rms}^2$.......(2)
Where $M=$ Molar mass, $V=$ Volume
Also,
$P_1{V}_1=\frac{M}{3}{v}_{\text{rms(1)}}^2$
$P_2{V}_2=\frac{M}{3}{v}_{\text{rms(2)}}^2$

From eqn 1.
$\Rightarrow \frac{v_{\text{rms(1)}}^2}{T_1}=\frac{v_{\text{rms(2)}}^2}{T_2}$
$\Rightarrow v_{\text{rms(2)}}^2=\frac{T_2}{T_1}\times v_{\text{rms(1)}}^2$
$v_{\text{rms}\left(2\right)}=\sqrt{\frac{2T_1}{T_1}}\times v_{rms\left(1\right)}$
[$\because T_2=2T_1$]
$v_{\text{rms}\left(2\right)}=v_{rms\left(1\right)}\sqrt{2}\text{m/s}=400\sqrt{2}\text{m/s}$