Question

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of helium gas atom at –20°C ? (Atomic mass of Ar = 39 u and He = 400 u)

• A. 2.52 × 103 K
• B. 2.52 × 102 K
• C. 4.03 × 103 K
• D. 4.03 × 102 K

Answer 1

Answer: (A)

2.52 × 103 K

Answer 2

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 questions,$V_{rms_{1}} = V_{rms_{2}}$

$\therefore\sqrt{\frac{3RT_{1}}{M_{1}}} = \sqrt{\frac{3RT_{2}}{M_{2}}}$

$\frac{T_{1}}{M_{1}} = \frac{T_{2}}{M_{2}}$ or $T_{1} = \frac{T_{2}}{M_{2}} \times M_{1} = \frac{253}{4} \times 39.9$

$= 2.52 \times 10^{3}K$