Question

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 13. Find the ratio of the pressures in the two parts of the vessel.
Figure

Answer 1

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

Explanation:

Step 1:

Since the separator initially splits the cylinder in equal measure, the number of moles of gas in both sections is identical. Accordingly,

$n_1 = n_2= n$

first part of the volume  = V

second part of the volume  = 3V

The walls are given as being diathermic. Therefore, the two pieces are equal in temperature. Accordingly,

$T_1 = T_2 = T$

Let the first- and second-parts pressure be P1 and P2, respectively.

First part: We obtain state equation

$P_1V =nRT$…………….eqn  (1)          

Second part: Applying State equation,

we get

$P_2 (3V)=nRT$…………..eqn (2)

Step 2:

Dividing equation 1 by equation 2, we get

$\begin{aligned}&\frac{p_{1} V}{p_{2}(3 V)}=1\\&\frac{P_{1}}{P_{2}}=\frac{3 V}{V}\\&\frac{p_{1}}{p_{2}}=\frac{3}{1}\\&P_{2}: P_{2}=3: 1\end{aligned}$

Thus the ratio of the pressures is 3:1

Answer 2

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The ratio will be 3:1