13-2) A certain gas mixture at a temperature of 1000 K consists of two moles of
ID: 1786539 • Letter: 1
Question
13-2) A certain gas mixture at a temperature of 1000 K consists of two moles of N2 and one mole of CH4. Data on these gases can be obtained from the graph in the online text. a) Explain which type of molecule, if either, has a greater average translational KE at 1000 K. b) Explain how you know which molecule, if either, has the greater number of active modes at 1000 K. c) Explain which type of molecule, if either, has a greater average rotational KE at 1000 K. d) Explain which type of molecule, if either, has a greater average translational speed at 1000 K e) If you added the same small amount of heat to one mole of N2 and to one mole of CHa, which would you expect to have the greater temperature change? Assume that both gases are at the same high temperature. Explain using the relevant concepts of the Particle Model of Thermal Energy, not by doing any calculations.?Explanation / Answer
given
2 moles of N2, one mole of CH4
molar mass of N2 = 14 g
molar mass of CH4 = 16 g
a. now KEav = 3kT/2
hence it is independent of the mass of the molecule, hence both the types of molecules have same average kinetic energy at 1000 K
but as CH4 has more degrees of freedom, is has energy partitioned into more modes and hence N2 is left with more average translational KE
b. as there are more atoms in CH4, it has more modes active at 1000 K as N2 is just a linear molecule
c. rotatinoal KE is given by E = L^2/2I
now, L = Iw ( w being angular speed)
E = (I^2w^2)/2I = Iw^2/2
hence
for CH4 the I is greater ( moment of inertia)
hence
CH4 has more rotational Kinetic energy
d. now, average translational speed of a gas molecule os given by
v = sqrppt(8*k*T/pi*m)
hence it is inversly proportional to mass of molecule
hence
N2 has greater average translational speed
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