Recent content by CVB

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

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

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...

Hello again, here is the exercise,do you think is ok?. I ´m sorry to send it by attachment. I will try to solve my Latex problem.Thank you for your help.

V_T= \frac{N_A . R. T_f}{P_f}+ \frac{N_B .R .T_f}{P_f} 2388,14 = 93,75 x 0,082 x 406,88/P_f + 124,91 x 0,082 x 406,88/P_f Then P_f= 3,05 Bar and Final volume in B is V_B = 124,91 x 8,3143 x 406,88/305 = 1385,44 litres And Final volume in A is V_A =2388,14-1385,44 = 1002,7 litres...

To Calculate final pressure : V_A+V_B= V_T I calculate V_T=477,9+1910,24=2388,14 Litres Like the total volume is constant

Have a rigid and thermally insulated tank having a diatermano piston that divides in 2 separate parts. Part A contains oxygen and inside there is a resistance connected to 220 v which supplies a electric current of 3 amperes. The other part B contains helium. Initially the system is in thermal...

Hello Andrew,thank you for your reply.It`s right that before electricity is supplied to the resistance the system is in static equilibrium: \Delta Q = \Delta U = W = 0 I wanted to say \Delta U_{total} = -W_{elec} . If you don´t mind I`m going to write all exercise and like I solve it.

Hello,thank you for your reply. Yes the question is that in the total balance before turn on the resistance \Delta U = -W_{elec} because the system is adiabatic and Q=0. But to calculate Q in left chamber after turn on the resistance would be Q=N C_v (T_f-T_i)+W_{expansion}+W_{elec} Thank you

First, I want to say thank you for your quick reply, I am very grateful. I am very confused because I understand Q=W_{elec} but I am going to upload a book where W_{out}=W_{exp}+W_{elec} and the first law is \Delta U=Q-W_{out} you can see example 4.16. Than you

Hello, my name is Roberto and I am engineering student from Spain. I am a new member in the forum and I don´t know if this is the right site to present me but I want to use this moment to say hello to the comunity. I have a doubt with a thermodinamic exercise.I have an adiabatic cylinder with...