Calculating Heat Transfer to Ideal Gas During Isochoric Process

  • Thread starter Thread starter rammer
  • Start date Start date
  • Tags Tags
    Gas Ideal gas
AI Thread Summary
To calculate heat transfer to an ideal gas during an isochoric process, start with the relationship dU = dQ, where no work is done. Use the ideal gas equation PV = nRT to find the change in temperature, ΔT, which can be expressed as ΔT = Δp*V / (n*R). By substituting ΔT into the equation for heat transfer, you arrive at Q = n*cv*ΔT. The final equation can be simplified to Q = (Cv/R) V Δp, applicable to any ideal gas regardless of atomicity. This method effectively eliminates the need to know the temperature directly.
rammer
Messages
22
Reaction score
0
How to calculate heat transfer to an ideal gas during isochoric process? I only know initial and final pressures and volume. (Do I have to know whole cycle (closed loop)?)

Here, no work is done so:

dU = dQ
n*c*dT=dQ

But T and its change is unknown, so what would be the next step?
 
Physics news on Phys.org
You should specify which C you are using Cp or Cv, as each ahs a spcific value.

Also have you heard of the ideal gas equation, PV=nRT?
That is the hint to get you started to the next step.
 
As molar heat capacity I use "cv". Let's say an ideal gas is monoatomic, so "cv"=3/2 R.

I involved equation pv=nRT as you suggested and figured something out:

dU = dQ
n*cv*dT=dQ
∫n*cv*dT=∫dQ
n*cv*ΔT = Q

------
p*V = n*R*T
dp*V + p*dV = n*R*dT (isobaric p. dV=0)
∫dp*V = ∫n*R*dT
Δp*V = n*R*ΔT
ΔT = Δp*V / (n*R)
------

After substituting ΔT into the first integrated equation:

Q = 3/2 *VΔp

I got rid out of T and it seems correct, is it?
 
Perfect firugring out !
But, you are missing an R in your final equation

In any case, you could just leave as:
Q = ( Cv/R ) V Δp
and that would work for any ideal gas, momoatomic, diatomic, ... polyatomic
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...

Similar threads

Replies
14
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
8
Views
3K
Replies
3
Views
1K
Back
Top