Change in heat and internal energy

AI Thread Summary
The discussion centers on the calculation of heat transfer (ΔQ) and internal energy change (ΔU) in a thermodynamic cycle involving an ideal gas. It raises the question of whether to use specific heat at constant pressure (Cp) or constant volume (Cv) due to changes in both pressure and volume. The internal energy change is determined by ΔU = mCv(T2-T3), emphasizing that for an ideal gas, internal energy depends solely on temperature. The area under the curve in a pressure-volume diagram represents the work done by the gas, which can be calculated using the ideal gas law (PV = nRT). The importance of adhering to sign conventions in calculations is also highlighted, ensuring accurate results.
Ashshahril
Messages
4
Reaction score
1
Homework Statement
What is the change in heat and internal energy in this PV diagram (attached link) from T2 to T3 (T represents temperature)?
Relevant Equations
ΔQ=mC(T3-T1)
Δu=ΔQ + Δw
ΔQ=mC(T3-T1)

But, will this C be Cp or Cv. Both pressure and volume changes. So, neither of them can be.

Feeling so confused
 

Attachments

  • Screenshot_20210622-213430_Chrome.jpg
    Screenshot_20210622-213430_Chrome.jpg
    34.3 KB · Views: 157
Physics news on Phys.org
Calculate the work from 2-3 which will be the area under the line from 2-3( trapezium area) and you know the change in internal energy= Cv(T2-T3) and u will get ∆Q
 
Note this is a full cycle. You start at point 1 and finish at point 1. So the system's final state is the same as its initial state. What can you say about the overall change in the system's internal energy, ΔU?

The area gives you the work done by the gas. ΔW:
https://www.ux1.eiu.edu/~cfadd/1150/14Thermo/Images/Diag10.gif
If you are dealing with n moles of an ideal gas you can use PV = nRT to get the value of V at each vertex and then derive a formula for ΔW in terms of the P's, T's, n and R.

Then you can find ΔQ.

Make sure the signs (+ or -) of your answers match the sign-convention you are using.

Edit: poor wording corrected.
 
Last edited:
For an ideal gas, internal energy is a function only of temperature. By definition, Cv is given by $$mC_v=\left(\frac{\partial U}{\partial T}\right)_V$$But, since U is dependent only on T for an ideal gas, $$\Delta U=mC_v\Delta T$$
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top