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Specific Heat Capacities of Gases — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsSpecific Heat Capacities of Gases5 Marks
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 $(\gamma=1.5)$ 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.
✓

Answer



Given,
$\gamma=1.5$
We know fro adiabatic process $\text{TV}^{\gamma-1}=\text{Const}.$
So, $\text{T}_1\text{V}_1^{\gamma-1}=\text{T}_2\text{V}_2^{\gamma-1}\ \dots(1)$
As, it is an adiabatic process and all the other conditions are same. Hence the above equation can be applied.
So, $\text{T}_1\times\Big(\frac{3\text{V}}{4}\Big)^{1.5-1}=\text{T}_2\times\Big(\frac{\text{V}}{4}\Big)^{1.5-1}$
$\Rightarrow\text{T}_1\times\Big(\frac{3\text{V}}{4}\Big)^{0.5}=\text{T}_2\times\Big(\frac{\text{V}}{4}\Big)^{0.5}$
$\Rightarrow\frac{\text{T}_1}{\text{T}_2}=\Big(\frac{\text{V}}{4}\Big)^{0.5}\times\Big(\frac{4}{3\text{V}}\Big)^{0.5}=\frac{1}{\sqrt3}$
So, $\text{T}_1:\text{T}_2=1:\sqrt3$

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