Ideal Gas Cycle: N2 Undergoing abcd w/ Varying Pressures

[Luis2101]

Two moles of N2 gas undergo the cycle abcd. The pressure of the gas in each state is
pa = pd = 5500 Pa;

pb = pc = 1500 Pa.


Note that 5500 Pa > 1500 Pa. The volume of the gas in each state is
Va = Vb = 1.80m^3 ;

Vc = Vd = 8.70m^3.

Note that 8.70 m^3 > 1.80 m^3. The gas may be treated as ideal.

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Part A asks: Find the magnitude of the total work done on (or by) the gas in the complete cycle.

I tried to approach the problem in the following manner and got an incorrect answer for Part A.

I treated the process from A->B, as well as from C->D as Isochoric since Volume is constant, and said no work was done.

I then calculated the work from B->C:
Using: W = p(deltaV) = 1500 (8.7-1.8) = 10,350J.

I said this was the total work done, but it's not...
Any help would be greatly appreciated.

-Luis

 

[Luis2101]

Nevermind. I found a link in the archives and figured it out from there.
Here it is if anyone needs it:
https://www.physicsforums.com/showthread.php?t=55853

 

[kyle8921]

I would first like to clarify that the ideal gas law, which states that PV = nRT, cannot be used to analyze this cycle as the volume is not constant. Instead, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system. In this case, ΔU = Q - W.

To find the total work done on (or by) the gas in the complete cycle, we need to calculate the work done in each process and then sum them up. Let's break down the cycle into its four processes:

Process A-B: Since the volume is constant (Va = Vb), this process is isochoric and no work is done (W = 0).

Process B-C: In this process, the volume increases from Vb to Vc while the pressure decreases from Pb to Pc. This is an isobaric process, and the work done can be calculated as W = pΔV = Pb(Vc - Vb) = 1500(8.7 - 1.8) = 10,350 J. Note that the sign of the work is positive as the gas is expanding.

Process C-D: Similar to process A-B, this process is also isochoric and no work is done (W = 0).

Process D-A: In this process, the volume decreases from Vd to Va while the pressure increases from Pd to Pa. This is also an isobaric process, and the work done can be calculated as W = pΔV = Pd(Va - Vd) = 5500(1.8 - 8.7) = -38,500 J. The negative sign indicates that the work is done by the gas as it is being compressed.

Now, to find the total work done in the cycle, we simply sum up the work done in each process: Wtotal = 0 + 10,350 + 0 - 38,500 = -28,150 J. Note that the negative sign indicates that the gas has done work on its surroundings.

In conclusion, the magnitude of the total work done on (or by) the gas in the complete cycle is 28,150 J. It is important to note that the work done by the gas is