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Suppose there are N molecules each of ma...

Suppose there are `N` molecules each of mass `m`, of an ideal gas in a container. The `x` component of velocity of a molecule is denoted by `u_(x)`. The gas is enclosed using a horizontal piston of area `A` as shown.

The pressure of the gas is

A

`(m(u_(x)^(2))N)/(AL)`

B

`(m(u_(x)^(2))N)/(3AL)`

C

`(3m(u_(x)^(2))N)/(AL)`

D

`(3m(u_(x)^(2)))/(AL)`

Text Solution

Verified by Experts

The correct Answer is:
A
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Knowledge Check

  • Suppose there are N molecules each of mass m, of an ideal gas in a container. The x component of velocity of a molecule is denoted by u_(x) . The gas is enclosed using a horizontal piston of area A as shown. The pressure of the gas can also be written in terms of momentum transferred per collision (Deltap) and collision frequency (f) on the wall of area A as

    A
    `Deltapxxf`
    B
    `(Deltap)/(Axxf)`
    C
    `(Deltapxxf)/(A)`
    D
    none of these

  • Suppose there are N molecules each of mass m, of an ideal gas in a container. The x component of velocity of a molecule is denoted by u_(x) . The gas is enclosed using a horizontal piston of area A as shown. If the temperature of gas is doubled keeping volume constant, we know from the gas law that the pressure will be doubled. On microscopic level this increase in pressure is because

    A
    momentum change per collision is doubled while the frequency of collision remains constant
    B
    momentum change per collision remain constant while the frequency of collision is doubled
    C
    momentum change per collision and the frequency of collision both are increased
    D
    none of these two physical quantities are changed. It is due to some other reason.

  • Suppose there are N molecules each of mass m, of an ideal gas in a container. The x component of velocity of a molecule is denoted by u_(x) . The gas is enclosed using a horizontal piston of area A as shown. If the piston is moved so as to reduce the volume of gas by half, keeping the temperature of gas constant , we know from the gas law that the pressure will be doubled. On microscopic level the increase in pressure on the piston is because

    A
    momentum change per collision is doubled while the frequency of collision remains constant
    B
    momentum change per collision remain constant while the frequency of collision is doubled
    C
    momentum change per collision and the frequency of collision both are increased
    D
    none of these two physical quantities are changed. It is due to some other reason.