SOLVED - READ BOTTOM QUESTION.
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What is the work done from B to A if n=4 mol?
A: -18000 J
I got -12000 J... HUGE difference.

Question

kittiwitti1 · Answer

Information (V is volume and P is pressure - originally given in kPa)$$V_B=10 m^3,~P_B=1\times10^3 Pa$$$$V_A=4m^3,~P_A=5\times10^3Pa$$***
Formulas$$\Delta U=\frac{3}{2}P\Delta  V=\frac{3}{2}nRT$$$$P\Delta V=nRT$$

agent0smith · Answer

Can't you solve the very last equation for T?

Then plug it into the prior equation.

kittiwitti1 · Answer

For T?

kittiwitti1 · Answer

But doesn't$$W=P\Delta V$$

kittiwitti1 · Answer

o.o?

agent0smith · Answer

^wouldn't that be for constant pressure...? Idk, i'm too tired to think about it

anonymous · Answer

"PΔV=nRT" 
that equation is definitely wrong :-/

anonymous · Answer

you have only provided the intial and final states..and not mentioned the type of process 

work done depends upon the PATH taken (the process u used to go from state A to state B).. so unless u mention that, u can't calculate work done!

anonymous · Answer

$$W = \int\limits_{A}^{B}PdV$$

kittiwitti1 · Answer

@Mashy this is what we were taught...

kittiwitti1 · Answer

I believe the process is .... adiabatic?

anonymous · Answer

u believe? :P.. or you know? cause different processes can give u different answers..

do you know what is the formula for the work done in case of adiabatic process ?

kittiwitti1 · Answer

Well it's a triangle, and B>A is the hypotenuse.
Since it's a pressure-volume graph work would be PV or area under the curve.

anonymous · Answer

yea.. work done is area under the curve.. the graph is given to you? mentioned in the question itself?

kittiwitti1 · Answer

It is given but I cannot post it.
B is the bottom corner....|dw:1392184132736:dw|

kittiwitti1 · Answer

That's the best I can do....

anonymous · Answer

u did not mention this :P.. without mentioning the graph how do u expect anyone to solve it? :D :D 

The graph itself indicates what process it is.. (its not adiabatic.. not important now what the process is anyways.. cause the graph is known)

ok .. lemme see if i can draw it better!

raffle_snaffle · Answer

This looks exactly like one of my hw problems.....

raffle_snaffle · Answer

the work done from B to A is w=1/2*P*V

raffle_snaffle · Answer

No work is being done from A to C

anonymous · Answer

|dw:1392184330165:dw|

now the area equals work done under curve = shaded region !