Name: Specific Heat Capacity of Water
Uploaded: 2019-07-19
Duration: 6 min 44 s
Description: Due to a lack of electricity, sometimes, we keep our cold drinks in water to maintain their coldness. Why? Because it takes time to draw heat out of water. W...

Question

A heat-conducting piston can freely move inside a closed thermally insulated cylinder, filled with an ideal gas. In equilibrium, the piston divides the cylinder into two equal parts, the gas temperature being equal to

T_{0}

. The piston is slowly displaced. Find the gas temperature as a function of the ratio

η

of the volumes of the greater and the smaller sections. The adiabatic exponent of the gas is equal to

γ

.

Accepted Answer

Since the piston is conducting and it is moved slowly, the temperature on the two sides increases and is maintained at the same value.

Elementary work done by the agent $=$ Work done in compression $-$ Work done in expansion,

i.e., $d A = p_{2} d V - p_{1} d V = (p_{2} - p_{1}) d V$ ,
where, $p_{1}$ and $p_{2}$ are pressures at any instant of the gas, on expansion and compression side, respectively.

From the ideal gas law,
$p_{1} (V_{0} + S x) = v R T$
and
$p_{2} (V_{0} - S x) = v R T$
For each section $(x$ is the displacement of the piston towards section $2$ )

So, $p_{2} - p_{1} = v R T \frac{2 S x}{V_{0}^{2} - S^{2} x^{2}} = v R T \frac{2 V}{V_{0}^{2} - V^{2}}$ (as $S x = V$ )

Thus, $d A = v R T \frac{2 V}{V_{0}^{2} - V^{2}} d V$

Also, from the first law of thermodynamics,
$d A = - d U = - 2 v \frac{R}{γ - 1} d T$ (as $d Q = 0$ )

So, work done on the gas $= - d A = 2 v \frac{R}{γ - 1} d T$

Thus, $2 v \frac{R}{γ - 1} d T = v R T \frac{2 V d V}{V_{0}^{2} - V^{2}}$

or, $\frac{d T}{T} = γ - 1 \frac{V d V}{V_{0}^{2} - V^{2}}$
When the left end is $η$ times the volume of the right end,
$(V_{0} + V) = η (V_{0} - V)$ or $V = \frac{η - 1}{η + 1} V_{0}$

On integrating, we get, $\int_{T_{0}}^{T} \frac{d T}{T} = (γ - 1) \int_{0}^{V} \frac{V d V}{V_{0}^{2} - V^{2}}$

or $\ln \frac{T}{T_{0}} = (γ - 1) {[- \frac{1}{2} \ln (V_{0}^{2} - V^{2})]}_{0}^{V}$
$= - \frac{γ - 1}{2} [\ln (V_{0}^{2} - V^{2}) - \ln V_{0}^{2} - V^{2}) - \ln V_{0}^{2}]$

$= \frac{γ - 1}{2} [\ln V_{0}^{2} - \ln V_{0}^{2} \{1 - {(\frac{η - 1}{η + 1})}^{2}\}] = \frac{γ - 1}{2} \ln \frac{(η + 1)^{2}}{4 η}$

$T = T_{0} {(\frac{(η + 1)^{2}}{4 η})}^{\frac{γ - 1}{2}}$

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