Question

The temperature of an ideal gas is increased from 200 K to 800 K. If r.m.s. speed of gas at 200K is v0. Then, r.m.s. speed of the gas at 800 K will be

• A. \(\frac{v_o}{4}\)
• B. \(2v_0\)
• C. \(v_0\)
• D. \(4v_0\)

Answer 1

The Correct Option is B

Using \(v_{\text{rms}} = \sqrt{\frac{3RT}{m}}\)
At \(200 \, K\), the r.m.s. speed is \(v_0\): \[ v_0 = \sqrt{\frac{3R \times 200}{m}} \quad \cdots (1) \] At \(800 \, K\), the r.m.s. speed is \(v'\): \[ v' = \sqrt{\frac{3R \times 800}{m}} \quad \cdots (2) \] Dividing equation (2) by equation (1): \[ \frac{v'}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2 \] Therefore, \(v' = 2v_0\).