Cloud formation
· Place a little water in the bottom of a 1½ litre plastic bottle
· Squeeze a few times
· Introduce a small amount of smoke
· Squeeze and release several times
· When you squeeze, the cloud disappears; when you release, the cloud reforms Explanation
· When the pressure increases the temperature increases and vica versa
· The smoke particles are nucleating sites on which the water can condense 5.18 Gay-lussac's law 11:11
· 5.18 use the relationship between the pressure and Kelvin temperature of a fixed mass of gas at constant volume: p1 / T1 = p2 / T2 p1 = Pressure at the beginning [kPa, bar or atm ] T1 = Absolute temperature at the beginning [K] p2 = Pressure at the end [kPa, bar or atm] T2 = Absolute temperature at the end [K] (Note: the units of temperature must be Kelvin, not oC! The units of pressure can be any, as long as the same at the beginning and the end) 5.18 Ideal graph and conclusion 09 November 2011 15:15
[cid:image001.png@01CCA464.D69055D0] 5.18 Question
Collins, p.116 [cid:image002.jpg@01CCA464.D69055D0] p1 p2
------ = ------
T1 T2 p1 = 3
T1 = 293
T2 = 328
p2 = 3.36 Bar a. If we cool the gas in a rigid, sealed tin can, what happens to the pressure inside the can? (1 mark) It will Decrease
b. Explain your answer to part a. by using the Kinetic Theory (4 marks) As the Temperature decreases in the container and volume stays the same, the pressure will decrease as the particles will move slower so they will not hit the walls of the container as hard
· Place a little water in the bottom of a 1½ litre plastic bottle
· Squeeze a few times
· Introduce a small amount of smoke
· Squeeze and release several times
· When you squeeze, the cloud disappears; when you release, the cloud reforms Explanation
· When the pressure increases the temperature increases and vica versa
· The smoke particles are nucleating sites on which the water can condense 5.18 Gay-lussac's law 11:11
· 5.18 use the relationship between the pressure and Kelvin temperature of a fixed mass of gas at constant volume: p1 / T1 = p2 / T2 p1 = Pressure at the beginning [kPa, bar or atm ] T1 = Absolute temperature at the beginning [K] p2 = Pressure at the end [kPa, bar or atm] T2 = Absolute temperature at the end [K] (Note: the units of temperature must be Kelvin, not oC! The units of pressure can be any, as long as the same at the beginning and the end) 5.18 Ideal graph and conclusion 09 November 2011 15:15
[cid:image001.png@01CCA464.D69055D0] 5.18 Question
Collins, p.116 [cid:image002.jpg@01CCA464.D69055D0] p1 p2
------ = ------
T1 T2 p1 = 3
T1 = 293
T2 = 328
p2 = 3.36 Bar a. If we cool the gas in a rigid, sealed tin can, what happens to the pressure inside the can? (1 mark) It will Decrease
b. Explain your answer to part a. by using the Kinetic Theory (4 marks) As the Temperature decreases in the container and volume stays the same, the pressure will decrease as the particles will move slower so they will not hit the walls of the container as hard
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