Figure 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.

Use R=8.314J K^{-1} mol^{-1}

#### Solution

Since the separator initially divides the cylinder equally, the number of moles of gas are equal in the two parts. Thus,

n_{1 }= n_{2}= n

Volume of the first part = V

Volume of the second part =3V

It is given that the walls are diathermic. So, temperature of the two parts is equal. Thus,

T_{1} = T_{2} = T

Let pressure of first and second parts be P_{1} and P_{2}, respectively.

For first part:- Applying equation of state, we get

\[ P_1 V = nRT ..........\left( 1 \right)\]

For second part:- Applying equation of state, we get

\[ P_2 \left( 3V \right) = nRT .............\left( 2 \right) \]

Dividing eq. (1) by eq. (2), we get

\[\frac{P_1 V}{P_2 \left( 3V \right)} = 1\]

\[ \Rightarrow \frac{P_1}{P_2} = \frac{3}{1}\]

\[ \Rightarrow P_1 : P_2 = 3: 1\]