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Pressure and Temperature

P 2, V 2,T 1=...

Question

Two spheres - 1,2 with flexible walls are connected by a tube of negligible volume. Their initial and final conditions are given by $P_{1}, V_{1}, n_{1}$ and $P_{2}, V_{2}, n_{2}$ respectively. The radius of these spheres depends on the temperature as $4 π r^{2} d r = k d T$ . Initially, sphere 1 was at $100 K$ and sphere 2 was at $200 K$ . As the temperature of sphere 1 is increased to $300 K$ , find the final mole ratio $(\frac{n_{1}^{'}}{n_{2}^{'}})$ .

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Solution

Since we know, $n_{t o t a l} = constant$
$n_{1} + n_{2} = n_{1}^{'} + n_{2}^{'}$

$4 π r^{2} d r = k d T$
Integrating above equation, $\int 4 π r^{2} d r = \frac{4}{3} π r^{3} |_{r_{1}}^{r_{2}} = k T |_{T_{1}}^{T_{2}}$
Volume of the sphere, $V = \frac{4}{3} π r^{3}$
so $d V = 4 π r^{2} d r = k d T$ .
Hence we get, $\frac{d V}{d T} = constant$

Using ideal gas equation, $P V = n R T$
$\frac{n_{1}}{n_{2}} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2} V_{2}} = \frac{P_{1}}{P_{2}} ∵ \frac{V}{T} = constant$

When the stop cork is open, pressure equilises i.e. $P_{1}^{'} = P_{2}^{'}$
Hence $\frac{n_{1}^{'}}{n_{2}^{'}}$ becomes 1.

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