Final answer:
To find the final pressure of the gas mixture, the ideal gas law is used. With both compartments initially containing v moles of ideal gas at temperature T and pressure p bar, after removing the partition, the final pressure is p / (1+b/2) bar.
Step-by-step explanation:
To calculate the final pressure of the gas mixture in terms of p bar after the partition is removed from a thermally insulated container with one compartment volume b times larger than the other, we can use the ideal gas law. The final pressure can be determined by considering the total volume and the amount of gas present after the partition is removed.
Initially, each compartment has v moles of an ideal gas at temperature T and pressure p bar. Upon removing the partition, the total volume V becomes V + bV = (1+b)V because the right compartment is b times larger. Since both compartments were at the same initial pressure and temperature, they have the same number of moles (v moles). Therefore, the total moles of gas after removal is 2v moles.
The ideal gas law states PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the temperature T and gas constant R don't change, and the total number of moles is now 2v, we have:
P_final(V_final) = (2v)RT
Plugging in the final volume, we get:
P_final(1+b)V = (2v)RT
Solving for P_final, we find:
P_final = (2vRT) / (1+b)V
Given that the initial pressure is p bar, we can also write the initial state as P_initialV = vRT which implies P_initial = p. Hence:
P_final = (2 * p * V) / (1+b)V = p / (1+b/2)
So, the final pressure of the gas mixture in terms of p bar is p / (1+b/2).