Question

The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the root mean square velocity of the gas molecules is v, at 480 K it becomes

• A. 4v
• B. 2v
• C. v/2
• D. v/4

Answer 1

Answer: (B)

2v

Answer 2

$v_{rms} \propto \sqrt{T}$̃ $\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{120}{480}} = \frac{1}{2} \Rightarrow v_{2} = 2v_{1}$