Ah yes they do, thanks a lot much appretiated.
Out of interest, is TVγ-1 = constant applicable to this question? When I tried just there for T1, T2 and V1 to find V2 it didnt work, I got V2 = 0.0365 m3
Thanks so much for the help! Sorry, awkwardness of timezones means this late reply.
OK so,
Vi = nRTi/Pi = (2.5*8.314*325)/(250000) ≈ 0.027 m3
and
Vf = nRTf/Pf = (2.5*8.314*Tf)/(125000) ≈ 1.66*10-4*Tf
and so using nCv(Tf - Ti) = -Pext(Vf - Vi)
2.5*12.47*(Tf - 353) = -100000(1.66*10-4*Tf -...
Thanks for explaining what's actually physically going on, I had no idea.
OK so,
ΔU = Q - W, adiabatic => Q = 0
ΔU = -W
ΔU = nCVΔT = -W = -Pext(Vf - Vi)
so nCv(Tf - Ti) = -Pext(Vf - Vi)
Would that be correct?
Sure,
this is an image of it, it was a past exam question. http://i.imgur.com/KT8XOOx.png
sorry I definitely wasn't clear enough in my initial post, I see that now
Homework Statement
[/B]
Find the final temperature, Q, ΔU, ΔH given the following
Initial state of gas
Ti = 353K
Pi = 250000Pa
2.5mols of gas
Cv = 12.47Jmol-1
Final pressure = 125000Pa
Homework Equations
PV = nRT
W = -PΔV
ΔH = ΔU + Δ(PV)
PVγ = constant
The Attempt at a Solution
Cv / R ≈3/2...
Thanks for the info! So, for part 3,4 and 5 I want to fine Δλ = λ/(2*5000*3.5) ? Then, λ +/- Δλ is the shortest/longest above/below I can resolve and in turn this λ range is indistinguishable from the iron line?
Homework Statement
One diffraction grating has 9600 lines uniformly spaced over a width of 3cm, it is illuminated by light from a mercury vapour discharge.
The other has 5000 lines/cm and is a 3.5cm grating, this one is used in the second order to resolve spectral lines close to 587.8002 nm...