5E-3 : | Expansion of an Ideal Gas to Fill an Evacuated Chamber | 4 pts |
|||||||

An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 6 kg of an ideal gas at 800 kPa and 50°C and the other part is evacuated. The partition is now removed and the gas expands into the entire tank. Determine the final temperature and pressure in the tank. | |||||||||

Read : | The most important thing to recognize in this problem is that removing the partition is equivalent to allowing the partition to move the to the right (in our diagram) until the ideal gas fills the entire tank. The resisting force in this expansion process is zero because there is a vacuum in the right chamber of the tank. Since the resisting force is zero, the work done by the expanding gas is also zero. If we take the ideal gas to be our system, there is no heat transfer during the expansion either, because the tank is insulated. The 1st Law tells us that ΔU = 0 when no work or heat transfer occur during a process on a closed system. Also, because U is a function of T only for an ideal gas, T2 = T1. Then, all we need to do is apply the IG EOS to determine P2. | ||||||||

Given : | m | 6 |
kg | Find: | P2 | ??? |
kPa | ||

P1 | 800 |
kPa | |||||||

T1 | 50 |
°C | |||||||

Assumptions: | - The gas behaves as an ideal gas | ||||||||

- Changes in kinetic and potential energy are negligible. | |||||||||

- The tank is perfectly insulated, so the process is adiabatic: Q = 0. | |||||||||

Equations: | The ideal gas is the system that we will analyze. This is a closed system because no mass crosses the boundary during the expansion process. | ||||||||

The 1st Law for a closed system is: | Eqn 1 | ||||||||

Since the restraining force overcome during the expansion is zero, the boundary work for the expansion is also zero. Combining this fact with the 2nd and 3rd assumptions listed above, allows us to simplify Eqn 1 to Eqn 2 : | |||||||||

or: | Eqn 2 | ||||||||

For an ideal gas, internal energy depends only on the temperature of the gas. If the internal energy is the same in state 2 as in state 1, then the temperature in state 2 must also be the same as the temperature in state 1 ! | |||||||||

Eqn 3 | |||||||||

Now that we know T2, we can apply the Ideal Gas EOS to both states 1 and 2 to determine P2. | |||||||||

Eqn 4 | Eqn 5 | ||||||||

Solve : | Divide Eqn 5 by Eqn 4 and cancel like terms : | ||||||||

Eqn 6 | |||||||||

But, we know that T1 = T2 and because the system is closed, n1 = n2. Also, because the left and right chambers of the tank are equal in size, V2 = 2 V1. Therefore, Eqn 6 can be simplified as follows: | |||||||||

Eqn 7 | or: | Eqn 8 | |||||||

Solving Eqn 8 for P2 yields : | Eqn 9 | ||||||||

Putting values into Eqn 9 gives us the answer : | P2 | 400 | kPa | ||||||

Verify Assumptions: | None of the assumptions can be verified. | ||||||||

Answer Questions: | P2 | 400 | kPa |