Question

The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the rms velocity of the gas molecules is vrms, then at 480 K, it becomes

• A. 4 vrms
• B. 2 vrms
• C. $\frac{\nu_{\text{rms}}}{2}$
• D. $\frac{\nu_{\text{rms}}}{4}$

Answer 1

Answer: (A)

4 vrms

Answer 2

$\frac{V_{rms_{2}}}{V_{rms_{1}}} = \sqrt{\frac{T_{2}}{T_{1}}} = \sqrt{\frac{480}{120}}$

$V_{rms} = 2(V_{rms1})$