Deriving First Law of Thermo Equations

[mortalapeman]

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

[tex] $ PV = RT $ [/tex]

[tex] $ C_{v} = C_{p} - R [/tex]

[tex] $ \Delta U = Q + W $ [/tex]

[tex] $ W = -PdV$ [/tex]

[tex] $ dQ = C_{v}dT + RTV^{-1}dV $ [/tex]

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

The Attempt at a Solution

I understand how to get to:

[tex] $ dQ = C_{v}dT + RTV^{-1}dV $ [/tex]

and there is another equation in my book that is:

[tex] $ dQ = C_{p}dT - RTP^{-1}dP $ [/tex]

with work equal to:

[tex] $ dW = -RdT + RTP^{-1}dP $ [/tex]

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

 

[Andrew Mason]

I understand how to get to:

[tex] $ dQ = C_{v}dT + RTV^{-1}dV $ [/tex]

and there is another equation in my book that is:

[tex] $ dQ = C_{p}dT - RTP^{-1}dP $ [/tex]

with work equal to:

[tex] $ dW = -RdT + RTP^{-1}dP $ [/tex]

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

 

[mortalapeman]

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

 

[Andrew Mason]

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

 

[rwisz]

First, let's start with the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

$ \Delta U = Q + W $

Now, let's consider an ideal gas in a closed system, where the only form of work is pressure-volume work. This means that the work done by the system is equal to the change in volume multiplied by the external pressure:

$ W = -PdV $

Substituting this into the first law equation, we get:

$ \Delta U = Q - PdV $

Next, we can use the ideal gas law, which states that the pressure and volume of an ideal gas are inversely proportional at constant temperature:

$ PV = RT $

Rearranging this equation, we get:

$ P = \frac{RT}{V} $

Substituting this into the first law equation, we get:

$ \Delta U = Q - \frac{RT}{V}dV $

Now, we can use the definition of heat capacity at constant volume ($ C_{v} $), which is the amount of heat required to raise the temperature of a substance by 1 degree at constant volume:

$ dQ = C_{v}dT $

Substituting this into the first law equation, we get:

$ \Delta U = C_{v}dT - \frac{RT}{V}dV $

Finally, we can use the definition of heat capacity at constant pressure ($ C_{p} $), which is the amount of heat required to raise the temperature of a substance by 1 degree at constant pressure:

$ dQ = C_{p}dT $

Substituting this into the first law equation, we get:

$ \Delta U = C_{p}dT - \frac{RT}{V}dV $

Now, we have two equations for heat ($ dQ $) in terms of temperature and volume, and we can solve for work by equating these two equations:

$ C_{v}dT - \frac{RT}{V}dV = C_{p}dT - \frac{RT}{V}dV $

Simplifying, we get:

$ \frac{C_{v}-C_{p}}{V}dV = 0 $

Since this equation must hold true for all