5.18 Gay-lussac's law

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 Experiment

07 November 2011

14:32

<<Ideal Gas - Gay Lussac's law.xlsx>>

·        Change the temperature of a fixed mass of gas at a constant volume

·        Measure the pressure

 

 

5.18 Ideal graph and conclusion

09 November 2011

15:15

 

 

5.18 Question

07 November 2011

15:08

Collins, p.116

 Degrees Celsiusàkelvin

Tk=ToC+270

Tk=20+270

Tk=290k

290k=T1

ToC=Tk -270

ToC =55-270

ToC =-215k

-215k=T2

P1/T1=P2/T2

P1/T1*T2=P2

3/290*-215

=-2.2bar

a.              If we cool the gas in a rigid, sealed tin can, what happens to the pressure inside the can? (1 mark)

Decreases

b.             Explain your answer to part a. by using the Kinetic Theory (4 marks)

-Pressure is caused by the gas particles hitting the walls of the container.

As the temperature is decreasing, the average speed of the gas particles also decreases.

This means that there will be less collisions between the rigid walls of the container and the gas particles

Therefore decreasing the pressure.