Question

. A cubical vessel has a side with 1 cm length
contained a gas at a pressure of'p'. when the side
of the vessel is made 1/2 cm, the pressure of the
gas becomes
1) P
2)P/8
3) 2P
4) 8P​

Answer 1

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• Given

=> pressure of gas is P when length of cube is 1 cm

• To Find

=> pressure of gas when length of cube is 1/2 cm

• Formula

=> volume of cube is given by

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \boxed{ \pink{ \bold{V = {l}^{3}}}}$

=> relation between pressure and volume is given by

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \boxed{ \pink{ \bold{P \propto \: \frac{1}{V} }}}$

• Calculation

$\implies \sf \: volume \: of \: cube \: in \: first \: case \\ \\ \implies \sf \: </strong><strong>V</strong><strong>1 = {1}^{3} = 1 {cm}^{3} \\ \\ \implies \sf \: volume \: of \: cube \: in \: second \: case \\ \\ \implies \sf \: </strong><strong>V</strong><strong>2 = {</strong><strong>(</strong><strong> \frac</strong><strong>{1}{2}</strong><strong>)</strong><strong> }^{3} = \frac{1}{8} {cm}^{3} \\ \\ \implies \sf \: now \: \: \frac{</strong><strong>P</strong><strong>1}{</strong><strong>P</strong><strong>2} = \frac{</strong><strong>V</strong><strong>2}{</strong><strong>V</strong><strong>1} \\ \\ \implies \sf \: </strong><strong>P</strong><strong>2 = \frac{</strong><strong>P</strong><strong>1 \times </strong><strong>V</strong><strong>1}{</strong><strong>V</strong><strong>2} = \frac{</strong><strong>P</strong><strong> \times 1}{ \frac{1}{8} } = 8</strong><strong>P</strong><strong> \\ \\ \boxed{ \sf{ \red{ \bold{</strong><strong>P</strong><strong>2 = 8</strong><strong>P</strong><strong>}}}}$