Thermodynamics temperature question

[Meowzers]

Homework Statement

The pressure-volume graph shows an engine cycle with four processes: two isobaric, and two constant-volume. The engine uses 0.800 mol of an ideal monatomic gas as its working substance. For one engine cycle, calculate (a) the heat added to the gas, (b) the heat removed from the gas (stated as a positive number), and (c) the net work done by the engine.

http://www.slowbie.com/pressure-volume.gif

Homework Equations

Q=nCp(delta T), Cp=Cv+R
Q=heat, n=moles, Cp=molar specific heat, constant pressure, Cv=molar specific heat, constant volume, R=gas constant, T=temperature

The Attempt at a Solution

For part A, I used the molar specific heat-constant pressure equation to try to solve for the heat added to the gas.

Q=nCp(delta T), Cp=Cv+R
Cp=12.5+8.31
Cp=20.81
Then, I read the temperature difference from A to B.
Q=(.800)(20.81)(568K-425K)
Q=2380.064 J (Where am I making the error?)

For part B, is the amount of heat removed from the gas the same thing as part A but with temperature readings between C
and D?
With that, the equations would look like Q=(0.800)(20.81)(318-425)
Q=-1781.336 J

Thanks in advance!

 

[Andrew Mason]

The Attempt at a Solution

For part A, I used the molar specific heat-constant pressure equation to try to solve for the heat added to the gas.

Q=nCp(delta T), Cp=Cv+R
Cp=12.5+8.31
Cp=20.81
Then, I read the temperature difference from A to B.
Q=(.800)(20.81)(568K-425K)
Q=2380.064 J (Where am I making the error?)

For part B, is the amount of heat removed from the gas the same thing as part A but with temperature readings between C
and D?
With that, the equations would look like Q=(0.800)(20.81)(318-425)
Q=-1781.336 J

Thanks in advance!

Heat is being added from D to A and from A to B. Heat is being removed from B to C and C to D. From D to A, $Q+ = nC_v\Delta T$ and from A to B, $Q+ = nC_p\Delta T$. The total is Qh. Figure that out to get the total heat added. I think you can similarly determine the heat removed (Qc).

To find the work, use the first law: $\Delta Q = Qh - Qc = \Delta U + W$. What is $\Delta U$ in a full cycle?

AM

 

[Meowzers]

Delta U is 0 in a full cycle since it returns to its initial state. So the work done is equal to the change in heat.

Thanks so much for your help! :)