Why Does My Ideal Gas Formula Derivation Not Include Moles?

[PeterPoPS]

I'm trying to show a formula for an ideal gas, but I don't get the right results.

Homework Statement

For an ideal gas PV = nRT where n is the number of momles. Show that the heat transferred in an infinitesimal quasistatic process of an ideal gas can be written as

$$dQ = \frac{C_V}{nR}VdP + \frac{C_P}{nR}PdV$$

Homework Equations

$$dU = dQ + dW$$

$$C_P = C_V + R$$

$$dU = nC_VdT$$

$$dW = -PdV$$

The Attempt at a Solution

I differented the formula for the ideal gas PV = nRT so it becomes

PdV + VdP = nRdT

$$dT = \frac{PdV + VdP}{nR}$$

$$dU = C_V\frac{PdV + VdP}{R}$$

$$dQ = C_V\frac{PdV + VdP}{R} + PdV = \left(\frac{C_V}{R} + 1\right)PdV + \frac{C_V}{R}VdP = \frac{C_P}{R}PdV + \frac{C_V}{R}VdP$$

What have I done wrong? There is no dependens on n in my final equation.
I know there should be bars on dW and dQ but i didn't got it to work in latex :/

 

[Andrew Mason]

The Attempt at a Solution

I differented the formula for the ideal gas PV = nRT so it becomes

PdV + VdP = nRdT

$$dT = \frac{PdV + VdP}{nR}$$

$$dU = C_V\frac{PdV + VdP}{R}$$

$$dQ = C_V\frac{PdV + VdP}{R} + PdV = \left(\frac{C_V}{R} + 1\right)PdV + \frac{C_V}{R}VdP = \frac{C_P}{R}PdV + \frac{C_V}{R}VdP$$

What have I done wrong? There is no dependens on n in my final equation.
I know there should be bars on dW and dQ but i didn't got it to work in latex :/

Your answer is correct. The solution posed by the question is wrong. There is no "n" in the denominator. dQ must have the same dimensions as VdP or PdV, which has dimensions of energy. C_v/R is dimensionless.

AM

 

[phjw]

I have just come across the same problem in an exercise book (no solution unfortunately). Its very unlikely two sources are incorrect?

 

[Andrew Mason]

I have just come across the same problem in an exercise book (no solution unfortunately). Its very unlikely two sources are incorrect?

How do you know they are two different sources?

The dimensions of the suggested answer are dimensions of energy per mole. If the dQ was the specific heat flow per mole the suggested answer would be correct, which is maybe what the OP was saying.Consider an expansion at constant pressure. By definition:

(1) $$dQ = nC_pdT$$

where $C_p$ is the molar heat capacity at constant pressure.

The solution of the OP gives:

(2) $$dQ = \frac{C_P}{R}PdV$$

since VdP = 0 (constant pressure).

You can see that (2) is equivalent to (1) if:

$$nRdT = PdV$$

This, of course, follows from the ideal gas law for a constant pressure process.

AM