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Standard XII

Physics

Ideal Gas Equation from KTG

Equal masses ...

Question

Equal masses of air are sealed in two vessels, one of volume $V_{0}$ and the other of volume 2 $V_{0}$ . If the first vessel is maintained at a temperature 300 K and the other at 600 K, find the ratio of the pressures in the two vessels.

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Solution

Since mass is same

$n_{1} = n_{2} = n$

$P_{1} = \frac{n R \times 600}{V_{0}}$

$P_{2} = \frac{n R \times 600}{2 V_{0}}$

$\frac{P_{1}}{P_{2}} = \frac{n R \times 300}{V_{0}} \times \frac{2 V_{0}}{n R \times 600}$ = 1:1

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Equal masses of air are sealed in two vessels, one of volume V0 and the other of volume 2V0. If the first vessel is maintained at a temperature 300 K and the other at 600 K, find the ratio of the pressures in the two vessels.

Since mass is same n1=n2=n P1=nR×600V0 P2=nR×6002V0 P1P2=nR×300V0×2V0nR×600 = 1:1

Equal masses of air are sealed in two vessels, one of volume $V_{0}$ and the other of volume 2 $V_{0}$ . If the first vessel is maintained at a temperature 300 K and the other at 600 K, find the ratio of the pressures in the two vessels.

Since mass is same

$n_{1} = n_{2} = n$

$P_{1} = \frac{n R \times 600}{V_{0}}$

$P_{2} = \frac{n R \times 600}{2 V_{0}}$

$\frac{P_{1}}{P_{2}} = \frac{n R \times 300}{V_{0}} \times \frac{2 V_{0}}{n R \times 600}$ = 1:1