In an adiabatic process, volume is doubled then find the ratio of final average relaxation time & initial relaxation time. Given `C_p/C_v = gamma`

Resistivity of a material of conductor in terms of relaxation time is given by

The volume of an ideal gas (gamma 1.5 ) is changed adiabatically from 4.00 liters to 3.00 liters . Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.

A gas is compressed adiabatically till its temperature is doubled. The ratio of its final volume to initial volume will be

During an adiabatic process , the pressure p of a fixed mass of an ideal gas change by Deltap and its volume V change by DeltaV . If gamma=C_(p)//C_(v) then DeltaV//V is given by

One mole of perfect gas, initially at a pressure and temperture of 10^(5)Nm^(-2) and 300 K, respectively, expands isothermally until its volume is doubled and then adiabatically until its volume is again doubled. Find the final pressure of the gas. Given gamma=1.4 .

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from tau_(1) to tau_(2) . If (C_(p))/(C_(v))=gamma for this gas then a good estimate for (tau_(2))/(tau_(1)) is given by :

The pressure of a gas is doubled keeping its temperature constant. Find the ratio of the final volume of the gas to its initial volume.