Question

560 cm3 of air is at a pressure of 76 cm of mercury. Find the pressure of air (assuming temperature remains constant), when its volume is : [4]
(i) 210 cm3 (ii) Volume increases by 25%

Answer 1

Answer:

Given,

Volume 1 (V1) = 560 cm³

Pressure (P1) = 76 cm of Hg

Temperature is constant,

1. Volume 2 (V2) = 210 cm³

According to the Boyle's law,

P1V1 = P2V2

(76*560) = 210P2

202.666 cm of Hg = P2

2. Volume 2 (V2) = V1 + (25% of V1)

= 560 + (25/100 * 560) cm³

= 560 + 140 cm³

= 700 cm³

According to the Boyle's law,

P1V1 = P2V2

(76*560) = 700P2

60.8 cm of Hg = P2

I hope this helps.