The molecules of a given mass of a gas have r.m.s. velocity of 200 m s-1 at 27°C and 1.0×105 N m-2 pressure. When thetemperature and pressure of the gas are respectively, 127°Cand 0.05×105 N m-2 the r.m.s. velocity of its molecules in ms-1

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10023ms-1

1003ms-1

1002ms-1

4003ms-1

answer is D.

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Detailed Solution

As, vrms=3kBTm∴v27v127=27+273127+273=300400=32 or v127=23×v27=23×200 m s-1=4003 m s-1

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