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Chemistry

800 cm3 of gas is collected at 654 mm pressure. At what pressure would the volume of the gas reduce by 40% of its original volume, temperature remaining constant?

Gas Laws

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Answer

V1 = 800 cm3

P1 (Initial pressure) = 654 mm of Hg

V2 = reduces by 40% of it's initial value

= 800 - (40100\dfrac{40}{100} x 800)

= 800 - 320 = 480 cm3

P2 (Final pressure) = ?

By Boyle's Law:

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

Substituting the values :

654×800=P2×480P2=800×654480P2=1090 mm of Hg654 \times 800 = \text{P}2 \times 480 \\[1em] \text{P}2 = \dfrac{800 \times 654}{480} \\[1em] \text{P}_2 = 1090 \text{ mm of Hg}

∴ Final pressure of the gas = 1090 mm of Hg.

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