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 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) 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 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 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 What volume is needed to store 0.050 moles of helium gas at 202.6kPa and 400K? P = 202.6 kPa n = 0.050 mol T = 400K V = ? L R = 8.314 J K1 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) What pressure will be exerted by 20.16g hydrogen gas in a 7.5L cylinder at 20oC? P = ? kPa V = 7.5L n = mass ÷ MM › mass = 20.16g › MM(H2) = 2 x 1.008 = 2.016g/mol n = 20.16 ÷ 2.016 = 10mol T = 20o = 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 A 50L cylinder is filled with argon gas to a pressure of 10130.0kPa at 30oC. How many moles of argon gas are in the cylinder? P = 10130.0kPa V = 50L n = ? mol R = 8.314 J K1 mol-1 T = 30oC = 30 + 273 = 303K 10130.0 x 50 = n x 8.314 x 303 506500 = n x 2519.142 n = 506500 ÷ 2519.142 = 201.1mol 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? 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 Problem A hydrogen gas thermometer is found to have a volume of 100.0 cm3 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 cm3. What is the temperature of the boiling point of chlorine? 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: P1 = P, V1 = 100 cm3, n1 = n, T1 = 0 + 273 = 273 K PV1 = nRT1 Finally: P2 = P, V2 = 87.2 cm3, n2 = n, T2 = ? PV2 = nRT2 Note that P, n, and R are the same. Therefore, the equations may be rewritten: P/nR = T1/V1 = T2/V2 and T2 = V2T1/V1 Plugging in the values we know: T2 = 87.2 cm3 x 273 K / 100.0 cm3 T2 = 238 K Answer 238 K (which could also be written as -35°C) http://www.ausetute.com.au/idealgas.h tml http://chemistry.about.com/od/chemistr yglossary/a/idealgaslawdef.htm