Thermodynamics - 4-4 Ideal Gas Specific Heat example 4

Calculating U (internal energy) and boundary work for the conservation of energy equation.
Ideal Gas. Like and subscribe! And get the notes here:
Thermodynamics: drive.google.c...
Mechanics of Materials: drive.google.c...
Statics: drive.google.c...
Dynamics: drive.google.c...
And now you can purchase my Thermodynamics notes completely filled in (184 pages of notes and problems) for $50 here: sowl.co/ksG5e

[@chrissmith2034]

This helped so much with me so much with one of my assignments. Thanks dude great explanation.

[@vidhyamakwana1430]

A well insulated frictionless piston-cylinder assembly contains 0.5 kg of air initially at 75℃ and 300 kPa. An electric-resistance heating element inside the cylinder is energized and causes the air temperature to reach 150 ℃. The pressure of the air is maintainded constant throughout the process. Determine the work for the process and the amount of electrical work.

[@hakanavc8665]

Hi professor, thank you so much for helpful videos. Process 1-2 is a constant volume, then 2-3 is constant pressure. Why dont we use cp ? Why do we use only cv ? I guess we should use cv for the process1-2 and cp for the process 2-3. If we use boundary work, then we dont need cv. We can directly solve it with table A-17. But we use only cv. I think this problem should be divided into the two parts. Then we can use cv and cp together.

[@davidcohen6791]

Hi Professor, In all other of your videos, you say "expansion" is work out (+) and "compression" is work in (-)... I don't understand why you take Wb as (-) even though it is "expansion" (+). If you don't mind, would you be able to clarify this confusion? Thanks in advance😁

[@emirbfitness]

expansion is negative, the system is doing work

[@sunmichoi6888]

Thank you so much!!!!!!!!!!!!!!!!!!!!

[@Justin-yc3eb]

How come you used Cv instead of Cp? The piston is moving while the heat is added right, therefore pressure is constant and volume is not. Right?

[@engineeringdeciphered]

We used Cv because we needed to find u, not h. The reason we needed to u is because the whole process (from state 1 to 3) is not just constant pressure or constant volume, its kind of a combination. So for those problems you calculate u AND boundary work.

[@qanimfarah1796]

Hey professor, i would appreciate if you explain why do we have to calculate T_av to find Cv? what if we use one of the Ts?

[@mecha_engineer]

Since it's not a constant pressure process.. Shouldnt the boundary work have been delta p times delta v???

[@solounomas0]

It is better to see it as 2 processes in one. And so from 1 to 2, V is constant so boundary work is 0 . From 2 to 3, volume doubles while constant pressure, there you calculate the boubdary work. I think it is better drawing the graph P vs V, and get the area under the curve as it is an integral really.

[@sannafinsrud3685]

how can you tell that the pressure is constant from 2-3 ?

[@nirajjha6028]

From 2-3 there isn't any Material or surface which hinders the flow of heat. Air molecules or water molecules can move freely as there isn't any Stopper or Sth. Like that. So the pressure remains constant

[@sevgipnar5261]

ın 10:06 why we take Tavg? the formula is deltaT. why we didnt write T2-T1?

[@danialraja]

Hi professor, i was stuck on this question and i tried as hard as i could but i cannot do it. Can you provide a solution for this questions please? A piston-cylinder device initially contains air at 250 kPa and 500 0C and total enclosed volume is 400 liters. Due to the heat loss, the piston moves downwards to a position where it sits on a pair of stops and its final enclosed volume is 200 liters. Further heat loss leads to the pressure reduced to 200 kPa Determine (a) the final temperature, (b) the work done by the air, and (c) the total heat loss to the environment. Assuming: 1) air is a perfect gas; 2) there is no friction between piston and cylinder walls, and 3) ignore the volume occupied by the stops; 5) there is no air leakage; 4) C_p=1.013 kJ/kgK; C_v=0.726 kJ/kgK; 5) use 273 for the conversion between Celsius and Kelvin degrees.

[@afinanadiyah]

why suddenly Q-140kJ? wasn't the formula is Q+W+Wb=dU?

[@munerimukhethwa4487]

Expansion is work out and compression is work in. Remember in is positive and out is negative. The boundary work in this problem is in expansion. Hence q-140

[@solounomas0]

@@munerimukhethwa4487 Also there is no other work than the boundary work, just why it is just Q and negative boundary work(as it is work out)

[@syarizahnorhashimah833]

u calculated 1400-300 and divide 2 for the average.. which equals to 550.. shouldnt we find Cv at 550K instead?

[@qanimfarah1796]

its plus. 1400+300/2.