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 $4u$ whereas Gas $B$ consists of rigid diatomic molecules, each with atomic mass of $20u$. 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 $A,m_A=4u$
and temperature of gas, $T_A=27^{\circ }C$
For diatomic gas $B$,
atomic mass of molecules of gas $B$,
$m_B=2\times 20u=40u$
Temperature of gas, $B=T_B$
Since, rms speed of the molecules of gas, $A=rms$
speed of molecules of gas, $B$
${\left(v_{rms}\right)}_A={\left(v_{rms}\right)}_B$
$\Rightarrow \sqrt{\frac{3R{T}_A}{m_A}}=\sqrt{\frac{3R{T}_B}{m_B}}$
$\Rightarrow \frac{T_A}{m_A}=\frac{T_B}{m_B}$
$\Rightarrow \frac{27}{4u}=\frac{T_B}{40u}$
$\Rightarrow T_B=270^{\circ }C$