At constant temperature, if the pressure of a gas of volume V in a container is doubled then the change in its volume is __________.

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\frac{V}{2}

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\frac{V}{4}

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Solution

The correct option is B $\frac{V}{2}$
We know PV=constant at constant temperature
according to question
$V_{1} = V, P_{1} = P$ and $P_{2} = 2 P, V_{2} = ?$

$\frac{P_{1} V_{1}}{P_{2} V_{2}} =$ constant
$\frac{P V}{2 P V_{2}} =$ constant
$V_{2} = V / 2$

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At constant temperature, if the pressure of a gas of volume V in a container is doubled then the change in its volume is __________.

The correct option is B V2We know PV=constant at constant temperatureaccording to question V1=V,P1=P and P2=2P,V2=?P1V1P2V2=constantPV2PV2=constantV2=V/2

The correct option is B $\frac{V}{2}$
We know PV=constant at constant temperature
according to question
$V_{1} = V, P_{1} = P$ and $P_{2} = 2 P, V_{2} = ?$

$\frac{P_{1} V_{1}}{P_{2} V_{2}} =$ constant
$\frac{P V}{2 P V_{2}} =$ constant
$V_{2} = V / 2$