As shown in the figure, a cylindrical tube possesses adiabatic walls and is equipped with an adiabatic separator. An external mechanism can be used to slide the separator within the tube. An ideal gas ( γ = 1.5 ) is

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Correct answer: B. 3 : 1

Let the initial volume of each part of the tube be

V_0

. The total volume of the tube is

2V_0

.

The separator is shifted such that it divides the tube in the ratio

1 : 3

.
Let the final volumes of the two parts be

V_1

and

V_2

.

V_1 = \dfrac{1}{4} \times 2V_0 = \dfrac{V_0}{2}

V_2 = \dfrac{3}{4} \times 2V_0 = \dfrac{3V_0}{2}

Since the walls and the separator are adiabatic, the gas in both parts undergoes a reversible adiabatic process.
For an adiabatic process,

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

.

For the first part:

T_1 V_1^{\gamma - 1} = T_0 V_0^{\gamma - 1}

T_1 = T_0 \left( \dfrac{V_0}{V_1} \right)^{\gamma - 1} = T_0 \left( \dfrac{V_0}{V_0 / 2} \right)^{1.5 - 1} = T_0 (2)^{0.5} = \sqrt{2} T_0

For the second part:

T_2 V_2^{\gamma - 1} = T_0 V_0^{\gamma - 1}

T_2 = T_0 \left( \dfrac{V_0}{V_2} \right)^{\gamma - 1} = T_0 \left( \dfrac{V_0}{3V_0 / 2} \right)^{1.5 - 1} = T_0 \left( \dfrac{2}{3} \right)^{0.5} = \sqrt{\dfrac{2}{3}} T_0

The ratio of the temperatures in the two parts is:

\dfrac{T_1}{T_2} = \dfrac{\sqrt{2} T_0}{\sqrt{2/3} T_0} = \sqrt{3}

Therefore,

T_1 : T_2 = \sqrt{3} : 1

.

As shown in the figure, a cylindrical tube possesses adiabatic...

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