$561 d m^{3}$ of a gas at S.T.P. is filled in a $748 d m^{3}$ container. If temperature is constant, calculate the percentage change in pressure required.

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Solution

$P_{1} V_{1} = P_{2} V_{2}$
$P_{1} = 1 a t m, V_{1} = 561 d m^{3}, V_{2} = 748 d m^{3}$
$P_{2} = \frac{1 \times 561}{748} = 0.75 a t m$
%Change in pressure $(0.75 - 1) \times 100 = - 25$ % decrease.

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561 dm3 of a gas at S.T.P. is filled in a 748 dm3 container. If temperature is constant, calculate the percentage change in pressure required.

P1V1=P2V2P1=1atm,V1=561dm3,V2=748dm3P2=1×561748=0.75 atm%Change in pressure (0.75−1)×100=−25% decrease.

$561 d m^{3}$ of a gas at S.T.P. is filled in a $748 d m^{3}$ container. If temperature is constant, calculate the percentage change in pressure required.

$P_{1} V_{1} = P_{2} V_{2}$
$P_{1} = 1 a t m, V_{1} = 561 d m^{3}, V_{2} = 748 d m^{3}$
$P_{2} = \frac{1 \times 561}{748} = 0.75 a t m$
%Change in pressure $(0.75 - 1) \times 100 = - 25$ % decrease.