Question

The temperature of an ideal gas is increased from 140 K to 560 K. If at 140 K the root mean square velocity of the gas molecule is V, at 560 K it becomes

• A. 5V
• B. 2V
• C. V/2
• D. V/4

Answer 1

Answer: (B)

2V

Answer 2

The V_rms at 140K is V

∴ V = $\sqrt{\frac{\text{3R} \times \text{140}}{M}}$

At 540 K = V′ = $\sqrt{\frac{\text{3R} \times \text{560}}{M}}$ = $\sqrt{\frac{3R \times 140 \times 4}{M}}$= 2 $\sqrt{\frac{3R \times 140}{M}}$ = 2V