The molecules of a given mass of a gas have r.m.s. velocity of 200 m s-1 at 27 and 1.0 105 N m-2 pressure. When the temperature and pressure of the gas are respectively, 127 and 0.05 105 N m-2, the r.m.s. velocity of its molecules in m s-1 is
- 100 23
- 1003
- 100 2
- 400 3
Answer
(4) 400 3
Solution
vrms = 3RTM ⇒ vrms T (independent of pressure)
T1 = 300 K, T2 = 400 K
⇒ v2v1 = 400300 = 2 3
∴ v2 = 200 2 3 = 400 3 m s-1
Answer: (4)