The molecules of a given mass of a gas have r.m.s. velocity of 200 ms-1 at 27oC and 1.0 x 105 Nm-2 pressure. When the temperature and pressure of the gas are respectively, 1270C and 0.05 × 105 Nm-2, the r.m.s. velocity of its molecules in ms-l is:
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a
1002
b
4003
c
10023
d
1003
answer is B.
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Detailed Solution
v ∝ T ⇒ v200 = 400300 ⇒ v = 200×23m/s⇒ v = 4003m/s