Question

Two gases $A$ and $B$ are contained in two separate, but otherwise identical containers. Gas $A$ consists of monatomic molecules, each with atomic mass of $4\, u$ whereas Gas $B$ consists of rigid diatomic molecules, each with atomic mass of $20\, u$. If gas $A$ is kept at $27^{\circ} C$, at what temperature should gas $B$ be kept so that both have the same rms speed?

• A. $27^{\circ} C$
• B. $54^{\circ} C$
• C. $270^{\circ} C$
• D. $62^{\circ} C$

Answer 1

Answer: (C)

For monatomic gas $A$,
atomic mass of molecules of gas $A, m_{A}=4 u$
and temperature of gas, $T_{A}=27^{\circ} C$
For diatomic gas $B$,
atomic mass of molecules of gas $B$,
$m_{B}=2 \times 20 u=40 u$
Temperature of gas, $B=T_{B}$
Since, rms speed of the molecules of gas, $A= rms$
speed of molecules of gas, $B$
$\left(v_{ rms }\right)_{A} =\left(v_{ rms }\right)_{B} $
$ \Rightarrow \sqrt{\frac{3 R T_{A}}{m_{A}}} =\sqrt{\frac{3 R T_{B}}{m_{B}}} $
$\Rightarrow \frac{T_{A}}{m_{A}} =\frac{T_{B}}{m_{B}} $
$\Rightarrow \frac{27}{4 u}=\frac{T_{B}}{40 u} $
$ \Rightarrow T_{B} =270^{\circ} C$