Consider two containers, adjacent to each other, each containing an ideal gas wi
ID: 982005 • Letter: C
Question
Consider two containers, adjacent to each other, each containing an ideal gas with N particles kept at the same temperature t, but at different pressures p_1 and p_2. By making use of the equation of state for an ideal gas, determine the pressure p that arises after an opening is made in the wall that separates those containers, allowing for a free flow of gas in-between. Combine the Sackur-Tetrode formula with the ideal-gas equation of state and express the entropy of an ideal gas in terms of temperature, number of particles and pressure. Find the entropy for the gas in the containers above, before and after the connection, and compute the change of entropy in the process.Explanation / Answer
a) Each container contains 'N' no of molecules.
Presssures in respective containers are P1 and P2 respectively.
Temperature of the system = T
Ideal Gas equation => PV = nRT
After an opening is made,
Let the resultant pressure be P
V1 = (NRT)/(No P1) [ Here n (moles) = N/No , Where No is avagadros number ]
V2 = (NRT/(No P2)
Now the final volume will become = V1 + V2
Therfore the final presure will be P = nRT/V ---- 1
Here n = (N+N)/No = 2N/No ---- 2
V = V1 + V2 = NRT/No{(1/P1)+(1/P2)} --- 3
SUBSTITUTING 2 AND 3 IN 1
Solving we get,
P = { (2P1P2) / (P1 + P2) } ----------- ANS
c) Entropy of mixing gases = -nR [x1 ln(x1) + x2 ln(x2) ]
Where x1 and x2 are the mole fraction of gases in both containers.
x1 = P1 / P ; x2 = P2 / P
Therefore ,
Entropy of mixture S = {-(2NR/No)[(P1/P) ln(P1/P) + (P2/P) ln(P2/P) ]] } ----- ANS
Where { P = 2P1P2 / (P1 + P2) }