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Standard XII

Physics

1st Law of Thermodynamics

If 250 mL o...

Question

If $250 m L$ of $N_{2}$ over water at $30^{\circ} C$ and a total pressure of $740 t o r r$ is mixed with $300 m L$ of $N e$ over water at $25^{\circ} C$ and a total pressure of $780 t o r r$ , what will be the total pressure if the mixture is in a $500 m L$ vessel over water at $35^{\circ} C$ .
(Given : Vapour pressure (Aqueous tension) of $H_{2} O$ at $25^{\circ} C, 30^{\circ} C$ and $35^{\circ} C$ are $23.8, 31.8$ and $42.2 t o r r$ respectively. Assume volume of $H_{2} O (l)$ is negligible in final vessel)

760 t o r r

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828.4 t o r r

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807.6 t o r r

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870.6 t o r r

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Solution

The correct option is D $870.6 t o r r$
$n_{N_{2}} = \frac{(\frac{708.2}{760} \times 0.25)}{0.0821 \times 303} = 9.36 \times 10^{- 3}$
$n_{O_{2}} = \frac{(\frac{756.2}{760}) \times 0.3}{(0.0821) \times 298} = 0.0122$
$n_{t o t a l} m o l e s = 0.02156$

In final vessel $= P$

$\frac{(n_{t o t a l}) R T}{V} = \frac{0.02156 \times 0.0821 \times 308}{0.5}$

$P = 1.09 a t m$ or $828.4 t o r r$

$P_{t o t a l} = P_{(O_{2} + N_{2})} + V . p r . o f H_{2} O = 828.4 + 42.2 = 870.6 t o r r$ .

So, the correct option is $D$

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The correct option is D 870.6torrnN2=(708.2760×0.25)0.0821×303=9.36×10−3nO2=(756.2760)×0.3(0.0821)×298=0.0122ntotalmoles=0.02156In final vessel=P(ntotal)RTV=0.02156×0.0821×3080.5P=1.09atm or 828.4torrPtotal=P(O2+N2)+V.pr.ofH2O=828.4+42.2=870.6torr.So, the correct option is D