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?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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. 

 

Approved by eNotes Editorial Team

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial