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 $-{20}^o$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)

$2.52\times {10}^3$K

Let $1$ and $2$ represent for Argon atom and Helium atom.

rms speed of Argon, $v_{rm{s}_1}=\sqrt{\frac{3R{T}_1}{M_1}}$

rms speed of Helium, $v_{rm{s}_2}=\sqrt{\frac{3R{T}_2}{M_2}}$

According to question,
$v_{rm{s}_1}=v_{rm{s}_2}$

$\therefore \sqrt{\frac{3R{T}_1}{M_1}}=\sqrt{\frac{3R{T}_2}{M_2}},\frac{T_1}{M_1}=\frac{T_2}{M_2}$

or $T_1=\frac{T_2}{M_2}\times M_1=\frac{253}{4}\times 39.9=2.52\times {10}^3$K