The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times.As a result, the diffusion coefficient of this gas increases `x` times. The value of `x` is........

Answer

4

Explanation

`"Diffusion coefficient " prop lambdaxx C_("mean")`
`"Mean speed, C"_("mean")prop sqrtT`
`"Diffusion coefficient " prop (T)/(p) xx sqrtT`
`prop (T^((3)/(2)))/(p)`
`=K(T^((3)/(2)))/(p)`
At temperature `T_(1)` and pressure `p_(1)`
`D_(1)=(K(T_(1))^((3)/(2)))/(p_(1))`
If `T_(1)` is increased 4 times and pressure in increased 2 times,
`D_(2)=(K(4T_(1))^((3)/(2)))/(2p_(a))`
`(D_(2))/(D_(1))=((4)^((3)/(2)))/(2)=(2^(3))/(2)=4`