The rms velocity of hydrogen is sqrt(7) ...

The rms velocity of hydrogen is `sqrt(7)` times the rms velocity of nitrogen. If `T` is the temperature of the gas, then

`T(H_(2))=T(N_(2))`

`T(H_(2))=sqrt(7)T(N_(2))`

`T(N_(2))=2T(H_(2))`

`T(N_(2))=sqrt(7)T(H_(2))`

Text Solution

Verified by Experts

The correct Answer is:

`u=sqrt((3RT)/(M)),(u(H_(2)))/(u(N_(2)))=sqrt((T(H_(2)))/(M(N_(2)))xx(M(N_(2)))/(T(N_(2))))`,
`sqrt(7)=sqrt((T(H_(2)))/(T(N_(2)))xx(28)/(2)),7=(T(H_(2)))/(T(N_(2)))xx14`
or `T(N_(2))=2T (H_(2))`
i.e., `T(N_(2)) gt T(H_(2))" or "T(H_(2)) lt T(N_(2))`.

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The rms velocity of hydrogen is sqrt7 times the rms velocity of nitrogen. If T is the temperature of the gas :

`T (H_(2)) = T (N_(2))`

`T (H_(2)) gt T (N_(2))`

`T (H_(2)) lt T (N_(2))`

`T (H_(2)) = sqrt7 T (N_(2))`

The r.m.s. velocity of hydrogen is sqrt 7 times the r.m.s. velocity of nitrogen. If T is temperature of the gas then

`T_((H_2)) = T_((N_2))`

`T_((H_2)) gt T_((N_2))`

`T_((H_2)) lt T_((N_2))`

`T_((H_2)) = sqrt7 " "T_((N_2))`

The r.m.s. velocity of hydrogen is sqrt(7) times the r.m.s. velocity 1.0 g cm^(-3) and that of wate vapours is 0.0006 g cm^(-3) , then the volume of water molecules in 1 L of steam at this temperature is

`T(H_(2)) = T(N_(2))`

`T(H_(2)) gt T(N_(2))`

`T(N_(2)) gt T(H_(2))`

`T(H_(2)) = sqrt(7)T(N_(2))`