How Does Heat Addition Affect Gas Behavior in a Dual-Chamber Cylinder?

[CVB]

Homework Statement

-We have a cylinder with two chambers separated by an adiabatic piston. The external walls are adiabatics but left wall is diatermana. Friction produced by the piston moving is absorbed by the right system. Inicially gas conditions are the same for both ( in the left chamber CO2, 500 litres, 1 Bar, 27 ºC, and in right chamber N2 with the same conditions).
We add heat (by an exterior focus with constant temperature of 900ºK) through diatemana wall and piston moving until pressure in right chamber is 2 Bar. Left system is a polytropic process with n = -2. We assume gases are perfect.
Calculate:
a) P,T,V in final equilibrium in both chambers
b) ΔU , heat and work in left chamber
c) ΔU , heat and work in right chamber
d) Expansion work and friction work in right chamber

Homework Equations

The Attempt at a Solution

a) First I find moles for both gases and It`s the same 20.32 moles.
Final pressure in both chambers is 2 Bar. To calculate final temperature in left chamber I used T_2=T_1(P_2/P_1)^{n-1/n}and it is T_2=848,52 ºK.
For the volume P_1(V_1)^n=P_2(V_2)^n and it is V_2= 707,1 Litres
For the right chamber: Total volume of the cylinder is 1000 litres and doesn´t change, so V_t=V_{1f}-V_{2f}, 1000=707,1-V_{2f}, V_{2f}=292,9 litres now I used PV=NRT to calculate final temperature in right chamber T=351,57 ºK.
b) To calculate ΔU, I use ΔU= N Cv(Tf-Ti) and ΔU=229.0788 Kj after
W_{exp}=NR(T_f-T_i)/1-n and W_{exp}=30.8901 Kj Now I calculate Q
Q= ΔU+W_{exp}=198.1887 Kj

I don´t know if it´s ok?
Thank you

 

[Studiot]

You have followed it through well.

The only thing that bothers me about this calculation is the value of 2 for n.
Where did this come from?

I have
For carbon dioxide n=1.2
For nitrogen n=1.4

Incidentally I apologise for confusing Italian with Spanish before.
The English word is diathermanous.

go well

 

[CVB]

Hello Studiot,thank you for your reply.The n value is an exercise data,is given by the teacher,I suppose it`s invented. Thank you to correct diathermanous word.
Best regards

 

[Studiot]

Seems a bit pointless stating the gas, then supplying the wrong value of the ratio of specific heats.

Further, I presume you realize that the stated conditions are instantaneous?
With the heat source specified the gas in the left chamber would continue to heat up until its temperature was the same as the heat source.

I normally expect to see given information used (needed) in some way in a question. Does your teacher not do this?

 

[CVB]

Hello, unfortunaly we have a very bad teacher for this subject.
Thank you