A cylinder is closed at both ends and has insulating walls. It is divided into two compartments by an insulating piston that is perpendicular to the axis of the cylinder as shown in Figure a. Each compartment contains 1.00 mol of oxygen that behaves as an ideal gas with γ = 1.40. Initially, the two compartments have equal volumes and their temperatures are T1i = 445 K and T2i = 300 K. The piston is then allowed to move slowly parallel to the axis of the cylinder until it comes to rest at an equilibrium position. Find the final temperatures in the two compartments.
To find the final temperatures in the two compartments, we can use the principle of adiabatic expansion/compression.
In this case, the process is adiabatic because the cylinder has insulating walls, meaning that no heat can enter or leave the system.
We can use the equation for adiabatic processes:
P1^(γ-1) * T1 = P2^(γ-1) * T2
where P1 and P2 are the initial and final pressures, T1 and T2 are the initial and final temperatures, and γ is the specific heat ratio.
Since the cylinder is divided into two compartments with equal volumes and equal amounts of oxygen, the pressures in the two compartments will be equal at all times during the process.
Therefore, we can simplify the equation to:
T1 = T2^(γ-1) * T2
Now, let's substitute the given values:
T1i = 445 K
T2i = 300 K
γ = 1.40
Now, we can solve for the final temperatures using algebraic manipulation:
T1i = T2^(γ-1) * T2
445 = T2^0.4 * T2
445 = T2^1.4
Taking the 1.4th root of both sides:
T2 ≈ 445^(1/1.4)
T2 ≈ 349.84 K
Now, we can find the final temperature in the other compartment by using the fact that the sum of the initial and final temperatures in the two compartments is constant (since no heat is transferred between them). Therefore:
T1f + T2f = T1i + T2i
T1f = T1i + T2i - T2f
T1f = 445 + 300 - 349.84
T1f ≈ 395.16 K
Therefore, the final temperatures in the two compartments are approximately T1f ≈ 395.16 K and T2f ≈ 349.84 K.