# PV = 1/3 Nmc2 = 1/3 nMc2 (n = number of moles, N = number of

PV = 1/3 Nmc2 = 1/3 nMc2 (n = number of moles, N = number of molecules; M = molar mass, m = mass of molecule)Heat, Gas and KT questionsPV = nRT = NkT where N = n x NAR = 8.31 J/K mol; k = 1.38 x 10-23 J/K; NA = 6.02 x 10231. A reaction produces 0.3 litres (what’s that in m3?) of CO2 (Molar mass 44 g/mol) at 105 Pa and 27 oC.a) How many moles of gas is this?b) What mass of gas is this?c) If this gas were collected at 3 x 105 Pa and -13 oC, what volume would it occupy?2. 1 kg of steam (molar mass = 0.018 kg / mol) is at 105 Pa and 200 oC. a) How many moles of steam is this? b) So what volume does it occupy? c) What is the kinetic energy of an average water molecule in this steam? d) What is the total K.E. in the steam? The steam is now heated to 300 oC. The volume is kept constant. d) What is the new pressure?e) What is the total K.E. contained in the steam now? f) How much energy has the steam gained? How much energy would it take to raise the temperature of 1 kg of steam by 1 oC? (This is called the ‘specific heat capacity’ of steam (at constant volume).) g) If the steam were allowed to expand as it was heated, so that the pressure, rather than the volume, remained constant, would you expect the amount of energy required to raise the temperature to be more than, less than, or the same as the amount of energy you found above, and why? (Hint: remember why the gas in a bike pump, or the piston of a BB gun, gets hotter on compression? Same idea…)3. a) Your lungs have a capacity of 4 litres (4 x 10-3 m3). How many molecules do they contain at a pressure of 105 Pa and a temperature of 300 K? b) If, at the top of Everest, the temperature is only 250 K and your 4 litre lungs hold 3.6 x 1022 molecules, what is the pressure there?4. Abdul the Iranian is refining uranium. The process involves diffusion of uranium hexafluoride, UF6, a gas of RMM 352 if the uranium isotope is U-238, and RMM 349 if it is U-235. A chamber contains UF6 gas at a temperature of 350 K. a) What is the RMS speed of a UF6 molecule if the uranium atom is U-238? b) The isotope separation process relies on the different rates of diffusion of the two different isotopes, which depends on their speeds. What is the ratio of RMS speeds of a U235F6 molecule and a U238F6 5. a) A 1.0 mole sample of hydrogen has a mass of 2.0 x 10-3 kg and occupies 2.26 x 10-2 m3 at 273 K and 1.0 x 105 Pa. What is the rms speed of a hydrogen molecule at this temperature? b) The molar mass of radon is 0.22 kg mol-1. What is the rms speed of a radon atom at 273 K?6. The escape velocity from the earth is 11 km/s. This means that any gas molecule at the top of the atmosphere whose vertical velocity exceeds 11 km/s will probably be lost from the earth. a) At what temperature is the rms speed of hydrogen molecules equal to 11 km/s? The molar mass of hydrogen is 2.0 x 10-3 kg mol-1. b) In practice hydrogen will gradually escape from the earth even if the temperature at the top of the atmosphere is much less than your answer to part a. Explain why. c) It has been shown that if the rms speed of molecules in a gas is more than about 1/5 of the escape velocity, the gas will escape almost completely in a billion years. If the top of the earth’s atmosphere has been at 1000 K for the past billion years, find the molar mass below which you would not expect to find any gases in Earth’s atmosphere.7. A further demonstration of diffusion involves releasing bromine molecules (molar mass = 0.16 kg/mol) into an evacuated tube, when the colour moves to the top of the tube very fast indeed, and then repeating it but with air in the tube, when the colour of the bromine spreads much more slowly. This is because of the frequent collisions with air molecules that mean that the path of the bromine molecules becomes a ‘drunkard’s walk’. There is a result that says that the average displacement from your start point after N steps, each of length l in a random direction, is N l. Thus, if the bromine molecules make n collisions per second, each separated by a ‘mean free path’ of length l, their average speed is nl, while their average velocity will be a) Find the average kinetic energy, and hence rms (root mean square) speed of bromine molecules at room temp. (20 oC.) b) Bromine in air is observed to diffuse at an average rate of 6 x 10-3 m/s. Use this information, along with the information already given / calculated, to find the mean free path in air, l, and the number of collisions