Question

1. You have two identical containers, one containing gas A and the other containing gas B. Both gases are under the same pressure and are at 5.0 ?C. The molecular masses are mA = 3.29 × 10?27 kg and mB = 6.12 × 10?26 kg.

(a) (1 point) Which gas has greater translational kinetic energy per molecule?

(b) (1 point) Which gas has greater rms speed?

(c) (1 point) Assuming you can only change one of the containers, the temperature of which gas should be raised so that both gases will have the same rms speed?

(d) (2 points) What temperature will do the job in (c)?

(e) (1 point) Now that the molecules of the two gases have the same rms speed, which gas’s molecules have greater average kinetic energy?

(f) (2 points) Suppose that instead of changing only one in (c), you could change both; the energy to raise the temperature of one of the gases will be removed from the other, both in constant-volume processes. In terms of the molar heat capacities of these gases at constant volume, CA and CB, express the ratio of the final temperature TA of gas A to that of gas B TB such that the rms speeds are again the same.

Answer #1

1. (a) KE depends only on temp.

both have same temp hence both have same KE.

(B) v_rms = sqrt(2KE/m)

so inversly related to molecular mass.

less the mass, greater is the speed.

Ans: GAS A

(c) gas B have less rms speed so we need to raise the temp of gas
B.

Ans: B

(D) v is propotional to sqrt(T / m).

T/m = constant

(T) / (6.12 x 10^-26) = (273+5) / (3.29 x 10^-27)

T = 5171 K

delta(T) = 5171-5 = 5166 deg C

(E) B is at greater temp so B will have greater KE.

(F) TA/TB = mA/mB = 0.054

Please show work and give detailed
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