Problem 20.37
You build a heat engine that takes 1.20 mol of an ideal diatomic gas through the cycle shown in the figure pv diagram starts at pa=2.00*10^(5) m3= 0.01 for the first cycle. The second cycle is 4*10^5 m3= 0.005. third cycle i 2*10^5 m3= 0.001.

Part A
Part complete
During which segment(s) of the cycle is heat absorbed by the gas?

only ab
only bc
only ca
ab and ca
Previous Answers
b is the Correct answer:
Part B
Part complete
During which segment(s) is heat rejected?

only ab
only bc
only ca
ab and ca
Previous Answers
a and c is Correct
Part C
Calculate the temperature at points a, b, and c.
Enter your answers using two significant figures separated by commas.


Ta, Tb, Tc =
K
Previous AnswersRequest Answer
Incorrect; Try Again;
With this question I used pv=nrt and convert to t=vp/nr. This didn't work, thought it may be with the volume and pressure being potentially negative, that wasn't it. NEED HELP HELP here figuring out what I need to calculate. PLEASE DON'T USE THE MATH EDDITOR ON HERE. PLEASE JUST WRITE OUT EQUATIONS FOR THE LOVE AND MERCY OF GOD. I am blind and can't see/read pictures with my screen readers. Part D
Calculate the net heat exchanged with the surroundings.
Express your answer using two significant figures.
with this one I would suspect to use du=q-w for this one but unsure too.

Qnet =
J
Previous AnswersRequest Answer
Incorrect; Try Again; 4 attempts remaining
Part E
Calculate the net work done by the engine in one cycle.
Express your answer using two significant figures.


Wnet =
J

Just haven't gotten to this one yet, attempted to see if this was 0 just incase but neiter of the previous answers are 0. Would love to get some help here, WITHOUT MATH equation IMAGES please and thank you.