Question

A 90.0cm3 box contains helium at a pressure of 1.90atm and a temperature of 100?C. It is placed in thermal contact with a 220cm3 box containing argon at a pressure of 4.30atm and a temperature of 400?C.

A. What is the initial thermal energy of each gas?

B. What is the final thermal energy of each gas?

C. How much heat energy is transferred, and in which direction?

D. What is the final temperature?

E. What is the final pressure in each box?

Answer 1

Since Both the gases are ideal so degree of freedom is One.

thermal energy = 1/2 K*T, K=boltzman constant, so thermal energy of helium = 1/2*K*373=2.5737e-21 J for each atom, overall energy = 0.000085324J.

thermal energy of the argon=1/2*K*673=4.6437e-21 J for each atom, overall =0.00046864J

using PV=NRT you can solve all the subsequent question

(B) Final Thermal Energy of each gas atom = 1/2 K 600 =4.14e-21 J per atom., overall energy = 0.0005590754 J

(c) N1= 1.9*90*10^-6/ 8.314*373=3.3211625e+16 atom , N2= 4.3 * 220 *10^-6/ 8.314*673 = 1.0183074e+17

total energy lost= 0.0000052014 J

(D) T' = (n??T? + n??T?) / (n? + n?)

T' = (p??V? + p??V?) / (p??V?/T? + p??V?/T?)

T= 599.3 Kelvin.

(E) Gas A:

initial volume = 90
initial pressure = 1.9 atm

final volume = 90 + 220 = 310

Using Boyles' Law, the final partial pressure of gas A after the valve is released is 1.9*90 / 310 =0.55 atm

For gas B, the final partial pressure = 4.3*220 / 310 = 3.05 atm

So the total pressure = 0.55 + 3.05 = 3.6 atm