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The temperature of an ideal gas is incr...

The temperature of an ideal gas is increase from 120 K to 480 K. If at 120 K the rms velocity of the gas molecules is v, then what will be its rms velocity at T = 480K?

Text Solution

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The correct Answer is:
2V
`(v_1)/(v_2) = sqrt((3RT)/(M))/sqrt((3RT_2)/(M)) =sqrt((T_1)/(T_2)) " " v_2 = 2v`
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Knowledge Check

  • The temperature of an ideal gas is increased from 120K to 480K. If at 120K the root-mean-squre velocity of the gas molecules is v, at 480K it becomes

    A
    `4v`
    B
    `2v`
    C
    `v//2`
    D
    `v//4`

  • The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is v, at 480 K it becomes

    A
    4 v
    B
    2 v
    C
    v/2
    D
    v/4

  • The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the root mean square velocity of the gas molecules is v, then at 480 K, it will be

    A
    4v
    B
    2v
    C
    `v//2`
    D
    `v//4`

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