I have started this problem, but I'm getting a weird answer. Ithink I made a mistake when calculating the pressure to getM (molar mass). The problem is this: A 1.00-liter vessel contains 5.00 grams of Nitrogen(N2) and 1.50 grams of water at 25oC. a) Determine the RMS speed of nitrogen molecules and watermolecules in the vapor phase at 25oC. b) Calculate the average kinetic energy per molecule ofnitrogen and of water vapor at 25oC. I have started this problem, but I'm getting a weird answer. Ithink I made a mistake when calculating the pressure to getM (molar mass). The problem is this: A 1.00-liter vessel contains 5.00 grams of Nitrogen(N2) and 1.50 grams of water at 25oC. a) Determine the RMS speed of nitrogen molecules and watermolecules in the vapor phase at 25oC. b) Calculate the average kinetic energy per molecule ofnitrogen and of water vapor at 25oC.
Explanation / Answer
A 1.00-liter vessel contains 5.00 grams of Nitrogen(N2) and 1.50 grams of water at 25oC. a)rms speed is a measure of the speed of the particles in agas given by : Vrms = (3*R*T/M) where Vrms is root mean square speed. M= molar mass of gas(in kg) R= universal gas constant i.e 8.314 J/K/mol T =temperature in kelvin so, here RMS speed of nitrogen molecules molar mass of nitrogen molecule = 2*14 = 28 gm = 0.028 kg = (3*8.314*(25+273.15)/0.028) =515.35 m/s and water molecules in the vapor phase at25oC. molar mass of oxygen molecule = 2*16 = 32 gm = 0.032 Kg = (3*8.314*(273.15+25)/0.032) = 482.067 m/s b)now average kinetic energy per molecule is given by: E = (3/2)RT/N where n = no of moles and N = avogadro number = 6.023 x1023 so average kinetic energy for nitrogen = average kineticenergy for water vapour in gaseous state. = 3/2(8.314/6.023*1023)*(273.15 + 25)) = 6.171 x 10 -21 (ans) so average kinetic energy for nitrogen = average kineticenergy for water vapour in gaseous state. = 3/2(8.314/6.023*1023)*(273.15 + 25)) = 6.171 x 10 -21 (ans)