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Chemistry

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.

Gas Laws

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Answer

V1 = 561 dm3

P1 = P

V2 = 748 dm3

P2 = ?

By Boyle's Law:

P1V1=P2V2\text{P}1 \text{V}1 = \text{P}2 \text{V}2

Substituting the values :

561×P=748×P2P2=561×P748P2=0.75P561 \times \text{P} = 748 \times \text{P}2 \\[1em] \text{P}2 = \dfrac{561\times \text{P}}{748} \\[1em] \text{P}_2 = 0.75 \text{P}

Change in pressure = P - 0.75P = 0.25P decrease

Percentage decrease in pressure = DecreaseOriginal\dfrac{\text{Decrease}}{\text{Original}} x 100 = 0.25PP\dfrac{\text{0.25P}}{\text{P}} x 100 = 25%

∴ Percentage decrease in the pressure of the gas = 25%

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