Take two containers, each with volume 0.1 m^3, containing ideal gases of monatom

Question

Take two containers, each with volume 0.1 m^3, containing ideal gases of monatomic He at T= 107 K and diatomic N_2 at T = 201 K, respectively. Each gas is at pressure p= 10^6 Pa. A valve is opened allowing these two gases to mix. They are kept thermally isolated from the outside. 1) What is the total initial internal energy? U_i = 2) What is the final internal energy? U_f = 3) What is the final temperature of the mixed gas? T = 4) Now we want to calculate the change in entropy during this process. What is the change of dimensionless entropy solely due to the fact that the gases have more volume accessible? Deltasigma_vol = 5) What is the change of dimensionless entropy solely due to the fact that the gasses' temperatures have changed? Deltasigma_temp = 6) What is then the total change of dimensionless entropy? Deltasigma = 7) What is the change of the standard entropy in this process? DeltaS =

Explanation / Answer

for mono atomic gas, internal energy(U1)= kinetic energy = (3kT/2)*N ; N-number of particles

Apply PV=nRT for Hydrogen, 106 *0.1= n*8.31*107;

n= 112.404 mol ; So, N= Na*n ;Na-avogadro's number ; N=6.769 x 1025

U1= 1.5 x 105 J

for N2 gas, internal energy= (5*k*T/2 )*N where N- number of molecules of N2

As before apply PV=nRT ; n=59.837 ; N=3.6 x 1025

U2= 2.5 x 105 J

Total internal energy= U1 +U2 = 4 x 105 J

2) No net work is done so that final internal energy= 4 x 105 J

3)both gases becomes to a same temperature (T)

as in the problem 1); Uf= 3*N1*k*T/2 + 5*k*T*N2/2

T*(3*N1*k/2 + 5*k*N2/2) = 4 x 105

T= 151.26 K

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