Thermodynamics: Finding the molar heat capacity

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SUMMARY

The discussion focuses on calculating the molar heat capacity \(C_V\) of an ideal gas as a function of its volume \(V\) during a process defined by the equation \(T = T_0 e^{\alpha V}\). Participants explore the integration of work done by the gas, using the ideal gas law \(PV = nRT\) and the relationship \(Q = W + \Delta U\). Key insights include the necessity of integrating with respect to volume and temperature, and the clarification that the molar heat capacity can depend on both \(V\) and \(T\). The final expression derived is \(C = C_V + \frac{nR}{\alpha V}\).

PREREQUISITES
  • Understanding of ideal gas laws and equations.
  • Familiarity with thermodynamic concepts such as work, internal energy, and heat capacity.
  • Knowledge of calculus, particularly integration techniques.
  • Basic principles of thermodynamics, including the First Law of Thermodynamics.
NEXT STEPS
  • Study the derivation of the First Law of Thermodynamics and its applications.
  • Learn about the Exponential Integral and its relevance in thermodynamic calculations.
  • Explore the concept of polytropic processes in thermodynamics.
  • Investigate the relationship between heat capacity and state variables in different thermodynamic processes.
USEFUL FOR

This discussion is beneficial for students and professionals in physics and engineering, particularly those focusing on thermodynamics, heat transfer, and gas laws. It is especially relevant for those tackling advanced problems in thermodynamic systems.

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Homework Statement


An ideal gas has a molar heat capacity ##C_V## at constant volume. Find the molar heat capacity of this gas as a function of its volume ##V##, if the gas undergoes the following process: ##T=T_0e^{\alpha V}## here ##T_0## and ##\alpha## are constants.

Homework Equations


The Attempt at a Solution


Work done by gas: ##\displaystyle W=\int PdV##
From ideal gas equation: ##PV=nRT=nRT_0e^{\alpha V}##. Hence
W=\int_{V_1}^{V_2} \frac{nRT_0 e^{\alpha V}}{V}dV
But I cannot integrate this. :confused:
 
Last edited:
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no work is done if the volume is kept constant - sorry I jumped too quickly on this.
 
Basic_Physics said:
no work is done if the volume is kept constant

Yes but the question does not state that the volume is constant during the process.
 
It might help to integrate by parts.
 
Ask yourself: What is heat capacity?

ehild
 
ehild said:
Ask yourself: What is heat capacity?

ehild

Heat required to change the temperature of a body by a given amount.

I approached the problem the same way as I have done for the polytropic processes. I calculate the work done in those processes through integration in terms of ##\Delta T##, change in temperature. I then write the change in internal energy in terms of ##\Delta T## and plug them in ##Q=W+\Delta U##. Rewriting Q in terms of heat capacity, I obtain the answer. Is it incorrect to use the same approach here?
 
Pranav-Arora said:
Heat required to change the temperature of a body by a given amount.

Change by what amount? Can you formulate the definition of heat capacity mathematically?

Pranav-Arora said:
I approached the problem the same way as I have done for the polytropic processes. I calculate the work done in those processes through integration in terms of ##\Delta T##, change in temperature. I then write the change in internal energy in terms of ##\Delta T## and plug them in ##Q=W+\Delta U##. Rewriting Q in terms of heat capacity, I obtain the answer. Is it incorrect to use the same approach here?

How do you write Q (the total amount of heat during a process) in terms of heat capacity, when it changes during the process? Q depends on V and you integrate with respect to V.

Have you solved the problem? What is your solution?

ehild
 
ehild said:
Change by what amount? Can you formulate the definition of heat capacity mathematically?
Sorry ehild, but I am less familiar with the technical terms. :redface:

C=\frac{\Delta Q}{n\Delta T}

How do you write Q (the total amount of heat during a process) in terms of heat capacity, when it changes during the process? Q depends on V and you integrate with respect to V.

Is it wrong to write ##Q=nC\Delta T## where C is the molar heat capacity?
 
  • #10
C can depend on other state variables. Here it depends on V, and also on T as T is function of V. The total heat Q taken in a process is not necessarily proportional to ΔT. The heat taken by the system depends also on the process. How do your lecture notes or books define the molar heat capacity? Certainly not as you said that "Heat required to change the temperature of a body by a given amount." If the given amount is 218 K, and you need 1000 J heat to change the temperature by that amount than the heat capacity is 1000 J?


ehild
 
  • #11
ehild said:
C can depend on other state variables. Here it depends on V, and also on T as T is function of V. The total heat Q taken in a process is not necessarily proportional to ΔT. The heat taken by the system depends also on the process. How do your lecture notes or books define the molar heat capacity? Certainly not as you said that "Heat required to change the temperature of a body by a given amount." If the given amount is 218 K, and you need 1000 J heat to change the temperature by that amount than the heat capacity is 1000 J?


ehild

Sorry, I just looked up heat capacity in my book. It defines heat capacity as C=Q/ΔT.

I am still not sure how to begin with the problem. :confused:
 
  • #12
The same way as you define speed as ds/dt, you can define heat capacity at a given state as small amount of heat divided by small change of temperature C=δQ/dt . The First Law of Thermodynamics says that dU=δQ+δW, that is, δQ=dU-δW. The small amount of heat taken is equal to the sum of the infinitesimal change of internal energy + infinitesimal amount of work.

dU=CvdT, δW=-pdV - write this in terms of V and dT. ehild
 
  • #13
ehild said:
δW=-pdV - write this in terms of V and dT.
I can replace P with ##nRT/V=nRT_0e^{\alpha V}/V## and ##dV## with ##dT/(T_0 \alpha e^{\alpha V}##

$$dW=\frac{nRdT}{\alpha V}$$
 
  • #14
Very good, plug into the equation δQ=dU-δW and find C=δQ/dt. (Recall that it is an ideal gas, so dU=Cvdt.)

ehild
 
  • #15
ehild said:
Very good, plug into the equation δQ=dU-δW and find C=δQ/dt. (Recall that it is an ideal gas, so dU=Cvdt.)

ehild

CdT=C_VdT+\frac{nRdT}{\alpha V}
I can cancel dT but what about n? :confused:
 
  • #16
The molar heat capacity is the heat capacity of one mole gas. n=1.
 
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  • #17
ehild said:
The molar heat capacity is the heat capacity of one mole gas. n=1.

Completely missed that, thanks! :smile:
 

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