Figure (24-EI) shows a cylindrical tube with adiabatic walls and fitted with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected into the two sides at equal pressures and equal temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio of 1: 3. Find the ratio of the pressures in the two parts of the vessel.


We know ideal gas equation


PV=nRT


Where V= volume of gas


R=gas constant


T=temperature


n=number of moles of gas


P=pressure of gas


Given


Volume of first part=V


Volume of second part=3V


Initially separator had divided cylinder in two equal parts so, number of moles in both the parts will be same.


n1=n2=n


Since the walls of separator is diathermic, the temperature of both the parts will always be same.


T1=T2=T


Pressure of part 1


P1=


Pressure of part 2


P2==


Diving P1 and P2, we get



P1:P2 =3:1


The ratio of the pressures in the two parts of the vessel is 3:1.


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