An ideal gas is initially at temperature T and volume V. It's volume increases by `DeltaV` due to an increase in temperature of `DeltaT`, pressure remaining constant. The quantity `mu = (DeltaV)/(V DeltaT)` varies with temperature as -

Options

A.
B.
C.
D.

Answer

D

Explanation

`V = KT rArr DeltaV = KDeltaT`
`therefore (DeltaV)/(DeltaT) = K`…….(i)
`rArr mu = (DeltaV)/(V DeltaT) rArr mualpha 1/V rArr mu alpha 1/T`
So graph between u & T will be rectangular hyperbola.