Is This PV Diagram Calculation Correct?

[zaf]

Hello forum members

I have a HW problem on ploting a P-V digram, I worked out a solution..can someone have a look at it and let me know if I am correct

the Question is as follows

2 moles of a monatomic ideal gas initially occupies a volum of 100 liters at a pressure of 100 KPa. The gas then expands iobarically to 200 liters (leg 1->2), followed by an isometric pressure increase to 300 KPa (leg 2-3), and ending in an isobaric expansion to 300 liters (leg 3->4).

Find change in U (internal energy) for each leg
Find W (work) for each leg
Find Q (heat exchange) for each leg


here is part of the solution

T1 = (100000 * .10) / (8.314 * 2) = 601 K
U1 = (3/2)*2*8.31*601 = 14982 J

T2 = (100000 * .20) / (8.31 * 2) = 1203 K
U2 = (3/2)*2*8.31*1203 = 29990 joules

Therefore
U for leg 1->2 is +15008.79 joules

W for leg 1-2 would be 10000 joules, which is P times change in volume

Am i correct so far...if that is so...then i can solve the rest


Zaf

 

[S.P.P]

hmmm

Its a while since i did this, but i think that's correct. I got dU by using dU = nCdT = 2 * 12.47 * 602 = 15014 J which is close enough.

Your method for working out the work is correct i think, i also got 10,000 J

What is isometric? i know of isobaric (const pressure), isochoric (const volume), isothermic (const temp), but isometric I've not heard of..

I presume isometric means isochoric?

 

[M98Ranger]

ar

Hello Zafar,

Your solution looks correct so far. The change in internal energy for leg 1-2 is +15008.79 joules. The work done by the gas for this leg would be 10000 joules, as you have calculated. To find the heat exchange for this leg, you would need to use the first law of thermodynamics, which states that the change in internal energy (ΔU) equals the heat exchange (Q) minus the work done (W). So for leg 1-2, the heat exchange would be 25008.79 joules (15008.79 + 10000). You can continue this approach for the remaining legs and solve the problem. Good job on your solution so far!