Reversible cycle and adiabatic expansion

[dancergirlie]

Homework Statement

One mole of a monatomic ideal gas is taken through the reversible cycle shown below. Process bc is an adiabatic expansion, with Pb = 10.0 atm and Vb = 1.00 x 10-3 m3. Find (a) the energy added to the gas as heat (b) the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.

I attached the diagram of the cycle:

Homework Equations

Ideal gas law:
pv=nRT

dQ=dW + dU

The Attempt at a Solution

For the first part I need to find the Q into the system which would be from dQ from a to b.

so dQ= dU + dW
dW=0 since there is constant volume. Meaning,
dQ=dU

Since the gas is a monotomic ideal gas, that means
dU= (3/2)nR(Tb-Ta), since n=1 that means:
dU=(3/2)R(Tb-Ta) and thus
dQ=(3/2)R(Tb-Ta)

However, I don't know how to find out the temperatures. I tried figuring them out using the adiabatic equations assuming gamma=1.67 (for an ideal gas)

where PV^gamma=constant
and V^(gamma-1)T= constant
but I'm getting values like 15 K, and I know that is unrealistic, especially for a gas. Any help/tips would be appreciated!

 

[anathema]

Thank you for your question. I can help you with finding the temperatures in this problem.

First, let's take a closer look at the adiabatic expansion process from b to c. In this process, the gas is expanding without any heat exchange with the surroundings. This means that the change in internal energy (dU) is equal to the work done by the gas (dW). We can use the ideal gas law to find the initial and final temperatures for this process.

Since we know the initial pressure (Pb) and volume (Vb), we can use the ideal gas law to find the initial temperature (Tb):

Tb = PbVb/nR

Next, we can use the fact that the final pressure (Pc) and volume (Vc) are equal to the initial pressure and volume, but with different values. This is because the gas is expanding adiabatically, so the product of pressure and volume (PV) remains constant. This gives us:

PcVc = PbVb

We can rearrange this equation to find the final volume (Vc):

Vc = (PbVb)/Pc

Now, we can use the ideal gas law again to find the final temperature (Tc):

Tc = PcVc/nR

Substituting in the values for Pc and Vc, we get:

Tc = (PbVb)/(nR/Pc)

We know that Pc is equal to Pb/10, so we can substitute that in:

Tc = (PbVb)/(nR/(Pb/10))

Simplifying, we get:

Tc = (Vb/10)nR

Now, we have the initial and final temperatures for the adiabatic expansion process. We can use these temperatures to find the change in internal energy (dU) and the heat added to the system (dQ).

dU = (3/2)nR(Tc - Tb) = (3/2)nR[(Vb/10)nR - Tb]

dQ = dU + dW = (3/2)nR[(Vb/10)nR - Tb] + 0

So, to summarize, the temperatures for the adiabatic expansion process are:

Tb = PbVb/nR

Tc = (Vb/10)nR

 

[tomcue]

Your approach is correct in using the ideal gas law and the first law of thermodynamics to find the heat added to the system. However, the temperatures cannot be found using the adiabatic equations as the process is not adiabatic. The process bc is an isobaric process, meaning the pressure is constant. Therefore, the temperature can be found using the ideal gas law, where PV=nRT. Since the number of moles and gas constant are both constant, the temperature can be found by taking the ratio of the initial and final volumes. Use this temperature to find the heat added to the system and the efficiency of the cycle.