Thermodynamics - Calorific Capacity

In summary: But you're also assuming pressure constant in ##dV = \frac{∂V}{∂T}dT##.This is why you're having trouble showing ##CP - CV##.In summary, the conversation discusses finding the total differential of entropy and volume in a system with constant particles, and introduces the calorific capacity at constant volume. However, there is confusion over the meaning of different partial derivatives and assumptions of constant pressure in the equations.
  • #1
Jalo
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0

Homework Statement



Consider a system with a constant number of particles.
Write the total differential dS in terms of the derivarives [itex]\frac{∂S}{∂T}[/itex] and [itex]\frac{∂S}{∂V}[/itex]. Introduce CV (calorific capacity at constant volume).
Next write the total differential of the volume dV in terms of the parcial derivatives [itex]\frac{∂V}{∂T}[/itex] and [itex]\frac{∂V}{∂P}[/itex]. Assume that the pressure is constant. Show that the result comes in the form of:

CP-CV= Expression

Homework Equations



CP=T[itex]\frac{∂S}{∂T}[/itex] , P and N Constant
CV=T[itex]\frac{∂S}{∂T}[/itex] , V and N Constant


The Attempt at a Solution



First I wrote the differential of the entropy as asked:

dS = [itex]\frac{∂S}{∂T}[/itex]dT + [itex]\frac{∂S}{∂V}[/itex]dV

I know that [itex]\frac{∂S}{∂V}[/itex] = CV/T. Substituting I get:

dS = CV/T dT + [itex]\frac{∂S}{∂V}[/itex]dV

Next I found the differential of the volume:

dV = [itex]\frac{∂V}{∂T}[/itex]dT + [itex]\frac{∂V}{∂P}[/itex]dP

Since the pressure is constant it reduces to the form

dV = [itex]\frac{∂V}{∂T}[/itex]dT

Substituting in our dS expression we get:

dS = CV/T dT + [itex]\frac{∂S}{∂V}[/itex][itex]\frac{∂V}{∂T}[/itex]dT =
= CV/T dT + [itex]\frac{∂S}{∂T}[/itex]dT =
= CV/T dT + CP/T dT ⇔
⇔ T[itex]\frac{dS}{dT}[/itex] = CV + CP

I'm making some mistake. If anyone could point me in the right direction I'd appreciate.

Thanks!
 
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  • #2
I think your problem is a confusion over the meanings of different partial derivatives. Your use of ##\frac{∂S}{∂V}## refers to changing volume, keeping temperature constant.
That's inconsistent with later equating ##\frac{∂S}{∂V}\frac{∂V}{∂T} = \frac{∂S}{∂T}##, which, to be valid, assumes pressure constant in all terms.
 

What is thermodynamics?

Thermodynamics is the branch of physics that deals with the relationship between heat, work, temperature, and energy. It studies how these quantities are affected by changes in a system.

What is calorific capacity?

Calorific capacity, also known as heat capacity, is a measure of the amount of heat required to raise the temperature of a substance by one degree. It is usually expressed in units of Joules per Kelvin (J/K) or calories per degree Celsius (cal/°C).

How is calorific capacity measured?

Calorific capacity can be measured experimentally by heating a substance and measuring the change in temperature, and the amount of heat added. It can also be calculated from the mass, specific heat capacity, and temperature change of a substance using the formula Q=mCΔT, where Q is the heat added, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

What factors affect the calorific capacity of a substance?

The calorific capacity of a substance is affected by its mass, chemical composition, and physical state. Substances with higher mass and specific heat capacity have a higher calorific capacity. The chemical composition and physical state of a substance determine how much energy is required to raise its temperature.

Why is it important to understand calorific capacity in thermodynamics?

Calorific capacity is an essential concept in thermodynamics because it helps us understand how heat and energy are transferred and used in different systems. It allows us to calculate and predict the behavior of substances when they are heated or cooled and plays a crucial role in many industrial and technological processes.

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