Question

Figure represents a closed cycle for a gas. The changes in the internal energy of the gas as it moves from a to c along the path abc is -200J must be transferred to it as heat. An additional transfer of 80J to it as heat is needed as it moves from d to a. How much work is done on the gas as it moves from c to d?

Answer 1

@douglaswinslowcooper

Answer 2

Do you see the graph I posted?

Answer 3

still nothing bro

Answer 4

ugh

Answer 5

@douglaswinslowcooper

Answer 6

@roadjester

Answer 7

check the attachement

Answer 8

ok, so what about it?

Answer 9

So can you help me with this question? The attachment is the picture I was trying to draw earlier.

Answer 10

this is going to take me awhile, these tables are annoying to draw

Answer 11

Take your time man. No rush.

Answer 12

I know there isn't any work going on from bc and ad

Answer 13

put simply though
adiabatic expansion: Q=0; deltaE = W
nC_v dT= - PdV

isovolumetric: W=0; deltaE=Q

isobaric
W=-P delta V
delta E= Q-P D=delta V

Answer 14

delta E interal=Einternal,f-E internal,i=Q-W

Answer 15

oops forgot to also mention the arrows are point from right to left

Answer 16

what arrows? you didn't draw any, so i don't know the direction of the cycle

Answer 17

as it moves from c to d ,180J ust be transferred

Answer 18

i forgot to put that in the question

Answer 19

the cycles it from right to left <-----

Answer 20

ok, is it clockwise or counterclockwise?

Answer 21

that's easier to understand

Answer 22

So E=Q-W so the total energy from c to a is 260J

Answer 23

counter clockwise

Answer 24

do you have the pressure, temperature, and volume numbers?

Answer 25

No, I don't

Answer 26

Q=heat?

Answer 27

yes

Answer 28

do you at least know the coordinates on the graph?

Answer 29

nope

Answer 30

*yawn* I am getting tired bro

Answer 31

well i am out, don't worry about it. I will ask my instructor tomorrow.

Answer 32

sorry, it's just I can't get a numeric value; i keep getting one variable, either deltaE or -P deltaV

Answer 33

@douglaswinslowcooper

Answer 34

In a cycle; the net work done by/on the gas is equal to the area enclosed in the PV graph. If the cycle is clockwise; net work is done by the gas and vice-versa.