A cubical vessel has a side with '1' cm length contains a gas at a pressure of 'P'. When the side of the vessel is made 0.5 cm, the pressure of the gas becomes:

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Solution

The correct option is D 8P
According to Boyle's law,
$P V$ is constant.
As a cubical vessel is considered, its volume comes after be $l^{3}$ .
So, $P_{1} V_{1}$ is equal to $P_{2} V_{2}$
$P l^{3} = P_{2} (l^{3} / 8)$
And then $P_{2}$ comes up to be $8 P$ .

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A cubical vessel has a side with '1' cm length contains a gas at a pressure of 'P'. When the side of the vessel is made 0.5 cm, the pressure of the gas becomes:

The correct option is D 8PAccording to Boyle's law,PV is constant.As a cubical vessel is considered, its volume comes after be l3.So, P1V1 is equal to P2V2Pl3=P2(l3/8)And then P2 comes up to be 8P.

The correct option is D 8P
According to Boyle's law,
$P V$ is constant.
As a cubical vessel is considered, its volume comes after be $l^{3}$ .
So, $P_{1} V_{1}$ is equal to $P_{2} V_{2}$
$P l^{3} = P_{2} (l^{3} / 8)$
And then $P_{2}$ comes up to be $8 P$ .