Why Does My Ideal Gas Formula Derivation Not Include Moles?

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SUMMARY

The discussion centers on the derivation of the heat transferred in an infinitesimal quasistatic process of an ideal gas, represented by the equation dQ = (C_V/R)PdV + (C_V/R)VdP. The original poster (OP) expresses confusion regarding the absence of the number of moles (n) in their final equation. Participants clarify that the dimensions of dQ must align with energy dimensions, and the suggested formula is dimensionally consistent only if interpreted as specific heat flow per mole. The OP's derivation aligns with the ideal gas law, confirming the relationship between dQ and the molar heat capacity at constant pressure (C_P).

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV = nRT)
  • Familiarity with thermodynamic concepts such as heat transfer (dQ) and work (dW)
  • Knowledge of specific heat capacities (C_V and C_P)
  • Basic proficiency in calculus for differentiation
NEXT STEPS
  • Study the derivation of the Ideal Gas Law and its applications in thermodynamics
  • Learn about the relationship between specific heat capacities (C_P and C_V) and their implications in thermodynamic processes
  • Explore the concept of quasistatic processes and their significance in thermodynamics
  • Investigate dimensional analysis in thermodynamic equations to ensure consistency
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This discussion is beneficial for students and professionals in physics and engineering, particularly those focusing on thermodynamics and heat transfer principles.

PeterPoPS
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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


<br /> dU = dQ + dW<br />

<br /> C_P = C_V + R<br />

<br /> dU = nC_VdT<br />

<br /> dW = -PdV<br />


The Attempt at a Solution



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

PdV + VdP = nRdT

<br /> dT = \frac{PdV + VdP}{nR}<br />

<br /> dU = C_V\frac{PdV + VdP}{R}<br />

<br /> 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<br />

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 :/
 
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PeterPoPS said:

The Attempt at a Solution



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

PdV + VdP = nRdT

<br /> dT = \frac{PdV + VdP}{nR}<br />

<br /> dU = C_V\frac{PdV + VdP}{R}<br />

<br /> 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<br />

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

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