Question

The temperatureof an Ideal Gas is increased from 140K to 560 K. If at 140 K the r.m.s. velocity of the gas molecule is V, at 560 K it becomes

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The $V_{rms}$ at 140 K is V
$\therefore V=\sqrt{\frac{3R\times 140}{M}}$
At 560 K, V' = $\sqrt{\frac{3R\times 560}{M}}$
$=\sqrt{\frac{3R\times 140\times 4}{M}}=2\sqrt{\frac{3R\times 140}{M}}$
=2V