Air at atmospheric pressure (760 mm Hg) is trapped inside a container. When the piston is pulled out slowly so that the volume is increased by 20% the temperature remaining constant the pressure of the air becomes?
According to the Boyle's law, given a constant temperature, the pressure and volume of an ideal gas are inversely related. In other words:
P `alpha` 1/V
or, PV = constant
where, P is the pressure of the gas and V is its volume. This means that as long as the temperature stays constant, an increase in either the pressure or the volume will decrease the other quantity. Similarly, a decrease in one of them will cause the other property to rise.
Here, the original pressure is 760 mm Hg and the volume is increased by 20%.
that is , P1 = 760 mm, V2 = 120% of V1 = 1.2 V1
Using the Boyle's law: P1V1 = P2V2
or, P2 = P1V1/V2 = 760 x V1 / (1.2 V1) = 760/1.2 = 633.3 mm Hg.
Thus, an increase of 20% in volume caused 20% reduction in the pressure of the air, at constant temperature.
Hope this helps.