Question

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at -$20\,{}^{\circ}C$ ? (Atomic mass of $Ar = 39 \,u$ and $He = 400 \,u$)

• A. $2.52 \times 10^3\,K$
• B. $2.52 \times 10^2\,K$
• C. $4.03 \times 10^3\,K$
• D. $4.03 \times 10^2\,K$

Answer 1

Answer: (A)

Let $1$ and $2$ represent for Argon atom and Helium atom.
rms speed of Argon, $v_{rms_1 } = \sqrt{\frac{3RT_{1}}{M_{1}}}$
rms speed of Helium, $v_{rms_2} = \sqrt{\frac{3RT_{2}}{M_{2}}}$
According to question,
$V_{rms_1} = V_{rms_2}$
$\therefore \quad\sqrt{\frac{3RT_{1}}{M_{1}}} = \sqrt{\frac{3RT_{2}}{M_{2}}}, \frac{T_{1}}{M_{1}} = \frac{T_{2}}{M_{2}}$
or $\quad T_{1} = \frac{T_{2}}{M_{2}} \times M_{1} = \frac{253}{4} \times39.9 = 2.52 \times10^{3}\,K$