Manipulation of Heat and Work Equations

[Wilson95]

Dear all,

How do you get from the 1st equation to the 2nd?

dW = -RT dV/V

dW = -RdT + RT dP/P

Also, I’m trying to do the same for this additional pair below…

dQ = C_V dT + RTdV/V

dQ = C_P dT - RT dP/P

Thanks.

 

[Mapes]

Hi Wilson95, welcome to PF. It looks like you're assuming an ideal gas system; for the first set, note that

$$dV=d\left(\frac{RT}{P}\right)$$

I'm not immediately seeing how to handle the second set.

 

[Blueshift5]

Dear reader,

The equations you have provided are known as the first and second law of thermodynamics, respectively. The first law states that the change in internal energy of a system (dU) is equal to the heat added to the system (dQ) minus the work done by the system (dW). The second law states that the change in entropy of a system (dS) is equal to the heat added to the system (dQ) divided by the temperature (T).

To answer your first question, the two equations are related through the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature. By rearranging the ideal gas law, we can get dV/V = dT/T - dP/P. Substituting this into the first equation, we get dW = -RT dT + RT dP/P.

For the second question, the equations are related through the specific heat capacities of a gas at constant volume (C_V) and constant pressure (C_P). These values are related through the equation C_P = C_V + R, where R is the gas constant. Substituting this into the second equation, we get dQ = C_P dT - RT dP/P.

I hope this explanation helps clarify the relationship between these equations. If you have any further questions, please do not hesitate to ask.