IDEAL GAS LAW .pptx

Bshaueiga

1. IDEAL GAS LAW

2. Key Concepts An Ideal Gas (perfect gas) is one which obeys  Boyle's Law  and  Charles' Law  exactly. An Ideal Gas obeys the Ideal Gas Law (General gas equation): PV = nRT

3. where  P = pressure V = volume n = moles  of gas T = temperature R = gas constant (dependent on the units of pressure, temperature and volume) R = 8.314 J K -1  mol -1  if Pressure is in kilopascals( kPa ) Volume is in litres(L) Temperature is in  kelvin (K) R = 0.0821 L atm K -1  mol -1  if Pressure is in atmospheres( atm ) Volume is in litres(L) Temperature is in  kelvin (K) PV = nRT

4. Key Concepts An Ideal Gas is modelled on the  Kinetic Theory of Gases  which has 4 basic postulates: Gases consist of small particles (molecules) which are in continuous random motion

5. The volume of the molecules present is negligible compared to the total volume occupied by the gas Intermolecular forces are negligible Pressure is due to the gas molecules colliding with the walls of the container

6. Key Concepts Real Gases deviate from Ideal Gas Behaviour because: at low temperatures the gas molecules have less kinetic energy (move around less) so they  do  attract each other at high pressures the gas molecules are forced closer together so that the volume of the gas molecules becomes significant compared to the volume the gas occupies

7. Key Concepts Under ordinary conditions, deviations from Ideal Gas behaviour are so slight that they can be neglected A gas which deviates from Ideal Gas behaviour is called a  non-ideal gas .

8. Ideal Gas Law Calculations

9. Calculating Volume of Ideal Gas: V = ( nRT ) ÷ P What volume is needed to store 0.050 moles of helium gas at 202.6kPa and 400K?

10. PV = nRT P = 202.6 kPa n = 0.050 mol T = 400K V = ? L R = 8.314 J K -1  mol -1 202.6(V) = 0.050 x 8.314 x 400  202.6(V)= 166.28  V = 166.28/202.6  V = 0.821 L (821mL)

11. Calculating Pressure of Ideal Gas: P = ( nRT ) ÷ V What pressure will be exerted by 20.16g hydrogen gas in a 7.5L cylinder at 20 o C?

12. PV = nRT P = ? kPa   V = 7.5L  n = mass ÷ MM mass = 20.16g MM(H 2 ) = 2 x 1.008 = 2.016g/mol  n = 20.16 ÷ 2.016 = 10mol  T = 20 o  = 20 + 273 = 293K R = 8.314 J K -1  mol -1 P x 7.5 = 10 x 8.314 x 293 P x 7.5 = 24360.02 P = 24360.02 ÷ 7.5 P = 3248kPa

13. Calculating moles of gas: n = (PV) ÷ (RT) A 50L cylinder is filled with argon gas to a pressure of 10130.0kPa at 30 o C. How many moles of argon gas are in the cylinder?

14. PV = nRT P = 10130.0kPa  V = 50L  n = ? mol  R = 8.314 J K -1  mol -1   T = 30 o C = 30 + 273 = 303K 10130.0 x 50 = n x 8.314 x 303 506500 = n x 2519.142  n = 506500 ÷ 2519.142 = 201.1mol

15. Calculating gas temperature: T = (PV) ÷ ( nR ) To what temperature does a 250mL cylinder containing 0.40g helium gas need to be cooled in order for the pressure to be 253.25kPa?

16. PV = nRT P = 253.25kPa V = 250mL = 250 ÷ 1000 = 0.250L n = mass ÷ MM mass = 0.40g MM(He) = 4.003g/mol  n = 0.40 ÷ 4.003 = 0.10mol R = 8.314 J K mol -1   T = ? K 253.25 x 0.250 = 0.10 x 8.314 x T  63.3125 = 0.8314 x T T = 63.3125 ÷ 0.8314 T = 76.15K

17. Other sample problem: Problem A hydrogen gas thermometer is found to have a volume of 100.0 cm 3  when placed in an ice-water bath at 0°C. When the same thermometer is immersed in boiling liquid chlorine, the volume of hydrogen at the same pressure is found to be 87.2 cm 3 . What is the temperature of the boiling point of chlorine?

18. Solution For hydrogen, PV = nRT , where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature. Initially: P 1  = P, V 1  = 100 cm 3 , n 1  = n, T 1  = 0 + 273 = 273 K PV 1  = nRT 1 Finally: P 2  = P, V 2  = 87.2 cm 3 , n 2  = n, T 2  = ? PV 2  = nRT 2

19. Note that P, n, and R are the  same . Therefore, the equations may be rewritten: P/ nR = T 1 /V 1  = T 2 /V 2 and T 2  = V 2 T 1 /V 1 Plugging in the values we know: T 2  = 87.2 cm 3  x 273 K / 100.0 cm 3 T 2  = 238 K Answer 238 K (which could also be written as -35°C)