Deriving First Law of Thermo Equations

AI Thread Summary
The discussion focuses on deriving the first law of thermodynamics equations in terms of pressures P1 and P2 for an ideal gas. The user seeks clarity on transitioning from PdV to a function of dP, using key equations like PV = RT and the relationships involving internal energy, heat, and work. A differentiation approach is suggested, starting from the ideal gas law and applying the product rule to relate changes in volume and pressure. The conversation highlights the importance of consistent definitions for work done by or on the gas to avoid confusion. Overall, the thread emphasizes the mathematical relationships that underpin thermodynamic principles.
mortalapeman
Messages
2
Reaction score
0

Homework Statement



This is really just a question that I can't seem to find a good solution for in my book. Basically I'm trying to understand for the first law of thermodynamics how you can derive the equation in term of P1 and P2. I don't understand how to go from PdV to (something)dP. This assuming we are dealing with an ideal gas.

Homework Equations



$ PV = RT $

$ C_{v} = C_{p} - R

$ \Delta U = Q + W $

$ W = -PdV$

$ dQ = C_{v}dT + RTV^{-1}dV $

$ W = -PdV = -RTV^{-1}dV = -RT \ln \left ( V_{1}/V_{2} \right ) = RT \ln \left ( P_{2}/ P_{1} \right ) $

The Attempt at a Solution



I understand how to get to:

$ dQ = C_{v}dT + RTV^{-1}dV $

and there is another equation in my book that is:

$ dQ = C_{p}dT - RTP^{-1}dP $

with work equal to:

$ dW = -RdT + RTP^{-1}dP $

And its getting to those equations that i don't understand how to do. Any help in the right direction would be appreciated :)
 
Last edited:
Physics news on Phys.org
mortalapeman said:
I understand how to get to:

$ dQ = C_{v}dT + RTV^{-1}dV $

and there is another equation in my book that is:

$ dQ = C_{p}dT - RTP^{-1}dP $

with work equal to:

$ dW = -RdT + RTP^{-1}dP $

And its getting to those equations that i don't understand how to do. Any help in the right direction would be appreciated :)
Start with:

PV = RT (we will assume n=1)

Differentiate with respect to T:

d(PV)/dT = R = P(dV/dT) + V(dP/dT)

So: RdT = PdV + VdP = dW + VdP

Which means that: dW = RdT - VdP = RdT - RTdP/P

AM
 
Andrew Mason said:
Start with:

PV = RT (we will assume n=1)

Differentiate with respect to T:

d(PV)/dT = R = P(dV/dT) + V(dP/dT)

So: RdT = PdV + VdP = dW + VdP

Which means that: dW = RdT - VdP = RdT - RTdP/P

AM

Thanks so much for clearing that up for me. These things always seem to be so simple, can't believe I didn't see that xD
 
mortalapeman said:
Thanks so much for clearing that up for me. These things always seem to be so simple, can't believe I didn't see that xD
Note: I am using dW as the work done BY the gas. I see you are using dW is the work done ON the gas, so dW = -PdV. Most texts now use dW = PdV. It is much less confusing.

AM
 
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