A 10 g of a gas at atmospheric pressure is cooled from 273- C to 0- C keeping the volume constant- its pressure would becomeA- - atm B- 1-273 atmC- 2 atmD- 273 atm

The correct option is A atm T_1 = 273^∘ C = 273 + 273^∘ K = 546^∘ K T_2 = 0^∘ C = 273 + 0^∘ C = 273^∘ K P_1 = 1; P_2 = ? According to Gay Lussac's law P_1/T_ ...

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Standard XII

Chemistry

Henry's Law

A 10 g of a g...

Question

A 10 g of a gas at atmospheric pressure is cooled from $273^{\circ} C t o 0^{\circ} C$ keeping the volume constant, its pressure would become

atm

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Solution

The correct option is A
atm

$T_{1} = 273^{\circ} C = 273 + 273^{\circ} K = 546^{\circ} K$

$T_{2} = 0^{\circ} C = 273 + 0^{\circ} C = 273^{\circ} K$

$P_{1} = 1; P_{2} = ?$

According to Gay-Lussac's law

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} ∴ P_{2} = \frac{P_{1} T_{2}}{T_{1}} = \frac{1 \times 273^{\circ} K}{546^{\circ} K} a t m; \frac{1}{2} a t m .$

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A 10 g of a gas at atmospheric pressure is cooled from 273 ∘ C to 0∘ C keeping the volume constant, its pressure would become

The correct option is A atm T1=273∘C=273+273∘K=546∘K T2=0∘C=273+0∘C=273∘K P1=1;P2=? According to Gay-Lussac's law P1T1=P2T2∴P2=P1T2T1=1×273∘K546∘Katm;12atm.

A 10 g of a gas at atmospheric pressure is cooled from $273^{\circ} C t o 0^{\circ} C$ keeping the volume constant, its pressure would become