using what equation?

I cant do this beacuse the ns dont drop out is there a way to relate the Va to Vc
20(Va) = nR(800)
2(Vc) = nR(500)

sorry if there is something that im just blatantly missing

nR = (20)*Va/800

2*Vc = ((20)*Va/800)*500

Im sorry I feel like a moron for missing the point
solving that you get
Vc = 6.25(Va) but what good does that do for me theres no other relationship between Vc and Va where knowing that helps me?

I was going to plug them into
20(Va) = nR(800)
2(Vc) = nR(500)

but that will just end up confirming Vc=Vc=6.25(Va)

Hm, if it's simpler, just use this equation P^(1-x)*T^(x) instead of the one with volume.

So doing this for points B and C.

Pb^(1-x)*(800)^x = 2^(1-x)*500^x

Pd^(1-x)*(500)^x = 20^(1-x)*800^x

The whole process relies on the fact that B -> C and D->A are reversible adiabatic steps for an ideal gas, so you use these equations:

PV^x = Constant
P^(1-x)T^x = Constant
TV^(x-1) = Constant

Where x = 5/3 or 7/5 depending on if it's a monatomic or diatomic gas.

This post was edited by RzChaos on Apr 5 2016 07:25pm

finally get it thanks a lot man