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The molecules of a given mass of a gas have r.m.s. velocity of 200 ms–1 at 27°C and 1.0 × 105 Nm–2 pressure. When the temperature and pressure of the gas are respectively, 127°C and 0.05 × 105 Nm–2, the r.m.s. velocity of its molecules in ms–1 is :
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- 100√2
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400 √3 -
100√2 3 -
100 3
Correct Option: B
Here v1 = 200 m/s;
temperature T1 = 27°C = 27 + 273 = 300 k
temperature T2 = 127° C = 127 + 273 = 400 k, V = ?
R.M.S. Velocity, V ∝ √T
| ⇒ | = √ | ||
| 200 | 300 |
| ⇒ v = | m/s ⇒ v = | m/s | ||
| √3 | √3 |
