Testing the variation of pressure with volume Fig 1 Testing the variation of pressure with volume Gas under pressure Pressure gauge (Pa) To foot pump

Gas under pressure Pressure gauge (Pa) To foot pump

(80, 40) (160, 20)

When a gas expands slowly (inside a closed container ) the pressure decreases quickly at first then less quickly. Pressure is inversely proportional to the volume P = ( constant ) x 1 V or P x V = (constant ) P1 V1 = P2 V2

a. P1E = m c θ = 0.20 x 900 x (40-15) P1 x V1 = P2 x V2 2a. Spot : Pressure halves so the volume doubles ! 0.00040 P1 x V1 = P2 x V2 2a. 100,000 x 0.00020 = 50,000 x V2 20 = 50,000 x V2 20 = V2 50,000 0.00040 m3 = V2

a. P1E = m c θ = 0.20 x 900 x (40-15) P1 x V1 = P2 x V2 2b. Spot : volume halves so the pressure doubles ! 0.00040 200,000 P1 x V1 = P2 x V2 2b. 100,000 x 0.00030 = P2 x 0.00015 30 = P2 x 0.00015 30 = P2 0.00015 200,000 Pa = P2

0.00040 Spot : not so easy this time! 200,000 0.00050 P1 x V1 = P2 x V2 2c. 120,000 x V1 = 100,000 x 0.00060 120,000 x V1 = 60 V1 = 60 120,000 V1 = 0.0005 m3

So pressure has decreased by 1/3 Spot : Volume increases by 3x So pressure has decreased by 1/3 0.00040 200,000 0.00050 180,000 P1 x V1 = P2 x V2 2d. P1 x 0.00015 = 60,000 x 0.00045 P1 x 0.00015 = 27 P1 = 27 0.00015 P1 = 180,000 Pa

a. P1E = m c θ = 0.20 x 900 x (40-15) P1 x V1 = P2 x V2 3. P1 V1 = P2V2 V1 = P2 V2 P1 120 = P2 100 100 P2 = 120 kPa P1 x V1 = P2 x V2 3. 100 x (20 + 100) = P2 x 100 12,000 = P2 x 100 12,000 = P2 100 120 kPa = P2

When thermal energy has time to leave ( or enter) the system these pressure and volume changes are called isothermal changes, because the gas temperature remains constant which is a condition for Boyles law.