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Ideal Gas. Review Q = Energy transfer in form of heat SI unit : J ( Joule) What is a Joule ?

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Presentation on theme: "Ideal Gas. Review Q = Energy transfer in form of heat SI unit : J ( Joule) What is a Joule ?"— Presentation transcript:

1 Ideal Gas

2 Review Q = Energy transfer in form of heat SI unit : J ( Joule) What is a Joule ?

3 Joule <- base SI unit named after a dead guy English physicist James Prescott Joule (1818–1889 J= (Kg * m 2 )/ s 2 = N *m = Pa *m 3

4 Ideal Gas Objective: Relationship among pressure, heat, volume and the amount of gas Red Bull Stratos  Felix Baumgartner. On 14 October 2012

5 Ideal Gas Law PV = nRT Macro scopic Microscopic P- Pressure V- Volume T- Temperature Atoms and Molecules Number of moles Mass Velocity (KE) – Kinetic Energy

6 Kinetic theory of Pressure Change in momentum Δ P = m (V f –V i ) I = I = Δ P Ft = Δ P F = Δ P/Δ T

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9 Sample gas = Large number of particles Force is constant in a time interval - Particles move randomly in all direction -pressure same on all the walls

10 Pressure ( P)  force( F) per unit surface area (A) P=F/A Si unit 1 Pa = 1 N/m 2 = 1 Kg/ms 2 1 Pa = 1 N/m 2 = 1 Kg/ms 2

11 Pressure ( P ) Standard pressure – average pressure of the atmosphere at sea level K Pa

12 Boyle’s law Boyle’s law – fixed sample of gas, at constant temperature- --volume( V )varies inversely with the pressure( P ) When 2 variables are inversely related – product is constant ( PV=Constant)

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14 A sample gas is held in a 2.6 m 3 volume at 226 kPa. The temperature is kept constant while the volume is decreased until pressure is 565 kPa. What is the new volume of the gas. Given : P i = 226 kPa P i = 226 kPa V i = 2.6 m 3 P f = 565 kPa Unknown: V f ? Relationship – P i V i = P f V f P i V i = P f V f

15 A sample gas is held in a 2.6 m 3 volume at 226 kPa. The temperature is kept constant while the volume is decreased until pressure is 565 kPa. What is the new volume of the gas Solution : P i V i = P f V f V f = P i V i / P f (226 kPa ) (2.6 m 3 )/565 kPa = 1.0 m 3 Pressured (P) went up – Volume (V) went down

16 Charles’s Law also known as the law of volumes What did Charles observed? Increased Temperature 0 C°  1 C° Volume increased by : 1/ 273 of original Volume Increase T by 2 C° V increased by 2/273 If Increased T by 273C° V would of doubled

17 Charles’s Law also known as the law of volumes What about when cooling the gas? Volume sharked by 1/273 for decreasing each C° Max cooling point -20 C° Extended the graph to its lower limit. Extrapolated that the lowest -273 C° At -273C° volume is zero. Now called Absolute zero ( Kelvin)

18 Charles’s Law Gases tend to expand when heated Volumes of gas depends on temperature Volume of fixed amount of gas at a constant pressure depends linearly on the temperature. Constant Pressure V i /T i = V f /T f = Constant V i /T i = V f /T f V i /T i = V f /T f

19 A container of.22 m 3 of nitrogen gas at 20.0 ° C is heated under constant pressure to ° C What is its new volume? Given: V i =.22 m 3 T i = 20.0 ° C T f = 167 ° C Unknown: V 2 V i /T i = V f /T f

20 A container of.22 m 3 of nitrogen gas at 20.0 ° C is heated under constant pressure to ° C What is its new volume? First switch ° C to K. Use Kelvin for gas law T i = 20.0 ° C = 293 K T f = 167 ° C = 440 K V i /T i = V f /T f V f = ( V i T f )/ T i Plug and chug ((.22 m 3 )(440K))/ (293K) =.33 m 3

21 Boyle’s law Vs. Charles's Law

22 Gay-Lussac's law The combined gas law Boyle’s law and Charles's law combined For a fixed amount of gas: (P i V i )/ T i = Constant = (P f V f )/ T f (P i V i )/ T i = (P f V f )/ T f

23 A 20.0-L sample of argon gas at 273 K is at atmospheric pressure, kPa. The temperature is lowered to 77 K, and the pressure is increased to 145 kPa. What is the new volume of the argon sample ? Given: V i = 20.0-L P 1 = kPa T i = 273 K P f = 1.45 kPa T f = 77 K Unknown : V f (P i V i )/ T i = (P f V f )/ T f 1 Litre (L) =.001 m 3

24 A 20.0-L sample of argon gas at 273 K is at atmospheric pressure, kPa. The temperature is lowered to 77 K, and the pressure is increased to 145 kPa. What is the new volume of the argon sample? Solution (P i V i )/ T i = (P f V f )/ T f V f = (P i V i T f )/ P f T i Plug and chug ((101.3kPa)( 20.0-L)(77 K))/( (145 kPa)(273 K)) =3.9 L = 3.9 x10 -3 m 3 = 3.9 x10 -3 m 3


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