A cubical vessel with side 1 cm contains a gas at pressure -P1- When the side of the vessel is made 1-2 cm- the pressure of gas becomes- Assume constanttemperature-A- PB- P-8C- 2 PD- 8 P

The correct option is D 8P Applying Chalre's Law, P_1/V_1 = P_2/V_2 P_1 1^3 = P_2 (1/2)^3 P_2 = 8P_1 ...

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Standard X

Chemistry

Gay Lussac's Law

A cubical ves...

Question

A cubical vessel with side $1 c m$ contains a gas at pressure 'P1'. When the side of the vessel is made $\frac{1}{2}$ cm, the pressure of gas becomes? Assume constant temperature.

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Solution

The correct option is D
8P

Applying Chalre's Law, $\frac{P_{1}}{V_{1}}$ = $\frac{P_{2}}{V_{2}}$

$P_{1} 1^{3}$ = $P_{2}$ ${(\frac{1}{2})}^{3}$

$P_{2}$ = $8 P_{1}$

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A cubical vessel with side 1 cm contains a gas at pressure 'P1'. When the side of the vessel is made 12 cm, the pressure of gas becomes? Assume constant temperature.

The correct option is D 8P Applying Chalre's Law, P1V1 = P2V2 P113 = P2 (12)3 P2 = 8P1

A cubical vessel with side $1 c m$ contains a gas at pressure 'P1'. When the side of the vessel is made $\frac{1}{2}$ cm, the pressure of gas becomes? Assume constant temperature.

The correct option is D
8P

Applying Chalre's Law, $\frac{P_{1}}{V_{1}}$ = $\frac{P_{2}}{V_{2}}$

$P_{1} 1^{3}$ = $P_{2}$ ${(\frac{1}{2})}^{3}$

$P_{2}$ = $8 P_{1}$