5.19 Boyle's Law · 5.19 use the relationship between the pressure and volume of a fixed mass of gas at constant temperature: p1V1 = p2V2 p1 = Pressure at the beginning [kPa, bar or atm] V1 = Volume at the beginning [m3 or cm3] p2 = Pressure at the end [kPa, bar or atm] V2 = Volume at the end [m3 or cm3] (Note: can use any units for V and p as long as they are the same at the beginning and end)
5.19 Boyle's Law demos
Fun with the vacuum pump!
· Marshmellows
· Food colouring in pipettes
· Surgical gloves 5.19 Ideal graph and conclusion
5.19 Questions PFY, p.36, Q.1a, 3 and 4
1) Boyle's Law: for a "constant" mas of gas, at constant "temperature", "pressure" x "volume" is constant. Pressure is "inversely" proportional to "volume" 3) P(1)T(1) = P(2)V(2)
4 x 1 = 1 x V(2)
V(2) = 4cm^3 4) P(1)V(1) = P(2)V(2)
1 x 60 = P(2) x 40
P(2) = 1.5 atm Extension: PFY, p.36, Q.5.
5) P(1)V(1) = P(2)V(2)
2.5 x 1000 = 1 x V(2)
V(2) = 2500 cm^3 Pressure in tyre and pump = 1atm
Volume of Tyre + pump = 1000+100 = 1.100cm^3
Volume of Tyre (without the pump) = 1,000 cm^3 1,100 x 1 = 1,000 x P(2)
P(2) = 1.1 atm
5.19 Boyle's Law demos
Fun with the vacuum pump!
· Marshmellows
· Food colouring in pipettes
· Surgical gloves 5.19 Ideal graph and conclusion
5.19 Questions PFY, p.36, Q.1a, 3 and 4
1) Boyle's Law: for a "constant" mas of gas, at constant "temperature", "pressure" x "volume" is constant. Pressure is "inversely" proportional to "volume" 3) P(1)T(1) = P(2)V(2)
4 x 1 = 1 x V(2)
V(2) = 4cm^3 4) P(1)V(1) = P(2)V(2)
1 x 60 = P(2) x 40
P(2) = 1.5 atm Extension: PFY, p.36, Q.5.
5) P(1)V(1) = P(2)V(2)
2.5 x 1000 = 1 x V(2)
V(2) = 2500 cm^3 Pressure in tyre and pump = 1atm
Volume of Tyre + pump = 1000+100 = 1.100cm^3
Volume of Tyre (without the pump) = 1,000 cm^3 1,100 x 1 = 1,000 x P(2)
P(2) = 1.1 atm