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 vertical cylinder closed from both ends is equipped with an easily moving piston dividing the volume into two parts, each containing one mole of air. In equilibrium at $T_{0} = 300 K$ the volume of the upper part is $η = 4.0$ times greater than that of the lower part. At what temperature will the ratio of these volumes be equal to $η^{'} = 3.0 ?$

Accepted Answer

The initial state of the system is as shown in the figure,

We have taken pressures as $p_{1}$ and $η = 4.0$ and volumes as $V_{1}$ and $V_{2}$ in the upper and lower part of the cylinder respectively at the initial temperature of $T_{0}$ . Let the mass of the piston be $m$ .

According to the question, $V_{1} = η V_{2}$

Also using the equation of state, we have

$p V = n R T \Rightarrow p = \frac{R T_{0}}{V} . . . (1)$

[as number of moles of the gas is one]

As the piston is in equilibrium, so we have

$p_{1} S + m g = p_{2} S$ , (where, $S$ is cross-section area of piston)

$\Rightarrow p_{1} + \frac{m g}{S} = p_{2}$

$\Rightarrow \frac{R T_{o}}{V_{1}} + \frac{m g}{S} = \frac{R T_{0}}{V_{2}}$ [using $η^{'} = 3.0 ?$ ]

$\Rightarrow \frac{R T_{o}}{η V_{2}} + \frac{m g}{S} = \frac{R T_{0}}{V_{2}} [∵ V_{1} = η V_{2}]$

$\Rightarrow \frac{m g}{S} = \frac{R T_{0}}{V_{2}} (1 - \frac{1}{η}) . . . (2)$

The final state of the system is as shown in the figure,

Let $T$ be the final temperature and volume of the lower part be $V,$ then according to the question the volume of the upper part becomes $η^{'} V$ .

Hence, using equation $(2)$ , we can write equation for final state as

$\frac{m g}{S} = \frac{R T}{V} (1 - \frac{1}{η^{'}}) . . . (3)$

Equating right-hand sides of equations $(2) & (3)$ , we get,

$\frac{R T_{0}}{V_{2}} (1 - \frac{1}{η}) = \frac{R T}{V} (1 - \frac{1}{η^{'}})$

$\Rightarrow T = \frac{T_{0} (1 - \frac{1}{η}) V}{(1 - \frac{1}{η^{'}}) V_{2}} . . . (4)$

Now, we have to find the relation between $V & V_{2}$ . As, the total volume in initial and final states must be constant, so we have

$V_{2} + η V_{2} = V + η^{'} V \Rightarrow V_{2} (1 + η) = V (1 + η^{'}) \Rightarrow \frac{V}{V_{2}} = \frac{1 + η}{1 + η^{'}} . . . (5)$

Putting $(5)$ into $(4)$ , we get,

$T = \frac{T_{0} (1 - \frac{1}{η})}{(1 - \frac{1}{η^{'}})} \frac{1 + η}{1 + η^{'}}$

$\Rightarrow T = \frac{T_{0} (η^{2} - 1) η^{'}}{(η^{'} {}^{2}- 1) η} \Rightarrow T = \frac{300 \times (4^{2} - 1) \times 3}{(3 {}^{2}- 1) \times 4} \Rightarrow T = 421.87 K \Rightarrow T = 0.42 kK .$

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