Question

Figure shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas (γ = 1.5) is injected in the two sides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1 3. Find the ratio of the temperatures in the two parts of the vessel.
Figure

Answer 1

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

Explanation:

The tube is in the ratio 1:3

Volume of one part of tube = V/4

Volume of another part of tube = 3V/4

From question, the process is adiabatic process,

$P V^{\gamma}=K$

Where,

P = Pressure

V = Volume

$\gamma=\frac{C_{p}}{C_{v}}$

For ideal gas, $P V=n R T$

Where,

P = Pressure

V = Volume

T = Temperature

R = Gas constant

n = Number of moles of gas

Now, the equation becomes,

$T V^{\gamma-1}=K'$

$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}$

$T_{1}\left(\frac{V}{2}\right)^{(1.5-1)}=T_{2}\left(\frac{V}{4}\right)^{(1.5-1)}$

$T_{1}\left(\frac{V}{2}\right)^{(0.5)}=T_{2}\left(\frac{V}{4}\right)^{(0.5)}$

$T_{1}\left(\frac{1}{2}\right)^{(0.5)}=T_{2}\left(\frac{1}{4}\right)^{(0.5)}$

$T_{1}\left(\frac{1}{2}\right)^{(0.5)}=T_{2}\left(\frac{1}{2}\right)$

$\therefore T_{2}=\sqrt{2} T_{1}$

Now, the other part of the cylinder is:

$T_{1}^{\prime}\left(\frac{V}{2}\right)^{(1.5-1)}=T_{2}^{\prime}\left(\frac{3 V}{4}\right)^{(1.5-1)}$

$T_{1}^{\prime}\left(\frac{V}{2}\right)^{(0.5)}=T_{2}^{\prime}\left(\frac{3 V}{4}\right)^{(0.5)}$

$T_{1}^{\prime}\left(\frac{V}{2}\right)^{(0.5)}=T_{2}^{\prime}\left(\frac{3V}{4}\right)^{(0.5)}$

$\therefore T_{2}^{\prime}=\sqrt{\frac{2}{3}} T_{1}^{\prime}$

Since, the pressure and volume are same, then, temperature is also same.

$T_{1}=T_{1}^{\prime}$

$T_{2}=T_{2}^{\prime}=\sqrt{2}: \sqrt{\frac{2}{3}}=\sqrt{3}: 1$

Answer 2

The ratio of the temperatures in the two parts of the vessel is $\frac{T_{1}}{T_{2}}=\frac{\sqrt{3}}{1}$

Explanation:

Given  Data

γ = 1.5

In case of adiabatic method,

$\mathrm{TV}^{\gamma-1}=\text { constant }$

$\mathrm{T}_{1} \mathrm{V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{V}_{2}^{\gamma-1}$

The above equation can be applied because it is an adiabatic process and all other conditions are the same. The slid divides the tube in the ratio 1:3 at the new position.

So if the total volume is V, then one side occupies $\frac{3}{4} v$ and the other side occupies $\frac{V}{4}$

So,  $\mathrm{T}_{1}\left(\frac{3 \mathrm{v}}{4}\right)^{\gamma-1}=\mathrm{T}_{2}\left(\frac{\mathrm{v}}{4}\right)^{\gamma-1}$

$\mathrm{T}_{1}\left(\frac{3 \mathrm{V}}{4}\right)^{1.5-1}=\mathrm{T}_{2}\left(\frac{\mathrm{V}}{4}\right)^{1.5-1}$

$\frac{T_{1}}{T_{2}}=\frac{\sqrt{3}}{1}$

The ratio of the temperatures in the two parts of the vessel with adiabatic wall and adiabatic separator is $\frac{T_{1}}{T_{2}}=\frac{\sqrt{3}}{1}$ when the slid moved to a new position and divides the tube in the ratio of 1 : 3. Also  an ideal gas is injected in the slides with equal pressure and temperatures, but the separator remains equilibrium in the middle.