# Cannot understand formula for molar heat capacities of an ideal gas

• I
• vcsharp2003
In summary: When you plug in ##pV^{x}=constant## for very large x, the second term on the right-hand side becomes zero, and the equation becomes true for an adiabatic process.
vcsharp2003
Homework Statement:: I am trying to understand a formula given in our book for determining molar heat capacity of an ideal gas under different thermodynamic processes using a single formula, but it is confusing. The exact formula for different processes is in the screenshots below. Can someone please explain how ##x## is calculated for different processes? The power of x is confusing me.
Relevant Equations:: As given in screenshot

To apply the formula for molar heat capacity using the equation mentioned above, one needs to first determine ##x## value and plug it into the molar heat capacity formula. But I am hitting a roadblock in finding ##x## for certain processes like isochoric, isothermal and adiabatic processes. I do get it for isobaric process.

The equation ##PV^x = constant ## doesn't make sense to me for isochoric process, in which ##V## is constant. To satisfy the above equation. ##x## would need to take a value so the left side of the equation is constant i.e. as ##P## increases, ##V^x## should decrease making ##x## take a variable set of multiple values. If, we can conclude that ##x= \infty ## then the second term on right hand side will be ##\frac {R}{1- \infty} = 0##, then the formula makes sense for isochoric process.

For isobaric process I get it. Since ##P## is already constant, so we need to only make sure that ##V^x## is constant, which is only true if ##x=0##.

Last edited:
The generic ideal gas equation of state is PV=nRT. Assuming that the amount of gas does not change, you can write it as PV=kT. The isothermal, isobaric and isochroric EOS’s should be derived from in directly. Your book should have the adiabatic derivation of PVϒ=constant.

vcsharp2003
Frabjous said:
The generic ideal gas equation of state is PV=nRT. Assuming that the amount of gas does not change, you can write it as PV=kT. The isothermal, isobaric and isochroric EOS’s should be derived from in directly. Your book should have the adiabatic derivation of PVϒ=constant.
I get the adiabatic part that you mentioned which is based on derivation, but the book uses the value of x for other processes. I was hoping to know how the x value is reached.

My bad. The book says that it derives the equation later. Is that derivation consistent with V=constant?

vcsharp2003
If ##pV^x=constant## for very large x then as p varies V hardly changes. In the limit, it doesn’t change at all.
I presume that just before the extract you show there was some other equation defining C. Would you post that please?

vcsharp2003
haruspex said:
If ##pV^x=constant## for very large x then as p varies V hardly changes. In the limit, it doesn’t change at all.
I presume that just before the extract you show there was some other equation defining C. Would you post that please?
I assume that C is the molar heat capacity for a given process, ##C = {\lim_{\Delta T \rightarrow 0}}{\frac {\Delta Q} {n \Delta T}}##. I am guessing that since there is no work involved in a constant volume process, the derivation assumes that there is some which is why the infinity appears.

Last edited:
haruspex said:
If ##pV^x=constant## for very large x then as p varies V hardly changes. In the limit, it doesn’t change at all.
I presume that just before the extract you show there was some other equation defining C. Would you post that please?
Below are screenshots of full page from where the original post screenshots were extracted.

Screenshot 1

Screenshot 2

Frabjous said:
My bad. The book says that it derives the equation later. Is that derivation consistent with V=constant?
I think for adiabatic process of ideal gas we know that ##PV^{\gamma} = constant ## is always true ( we don't care about the derivation of this equation for now since the goal is determining ##x## for different thermodynamic processes). Then we compare above equation with ##PV^{x} = constant ## and we get ##x= \gamma##. So, I am now also clear how ##x## is obtained for an adiabatic process.

Last edited:
vcsharp2003 said:
I think for adiabatic process we know that for an ideal gas ##PV^{gamma} = constant ##. Then we compare above equation with ##PV^{x} = constant ## and we get ##x= \gamma##. So, I am now also clear how ##x## is obtained for an adiabatic process. My problem is only that I want to know how ##x## value is arrived at.
For a constant volume process, ##\Delta Q = \Delta U##. I believe that when you look at the derivation of the equation, you will see that they using ##\Delta Q = \Delta U + W##. The infinity is essentially showing up to drive W to zero. They say “we will derive it later“ on page 208 after the boxed C equation.

Last edited:
haruspex said:
If ##pV^x=constant## for very large x then as p varies V hardly changes. In the limit, it doesn’t change at all.
Think about what @haruspex is saying here. It is not the common way we think about infinities in an equation.
You can write (P+δP)(V+δV)x=constant. Since the limit of δV is zero, the only way it can influence the equation is if the limit of x is infinity and you get some zero-infinity action.

Last edited:
haruspex said:
If ##pV^x=constant## for very large x then as p varies V hardly changes. In the limit, it doesn’t change at all.
I presume that just before the extract you show there was some other equation defining C. Would you post that please?

I think the following may be an apt way of getting the value of ##x## in an isochoric process.

We need to satisfy the equation ##PV^{x} = constant## when volume ##V## remains constant. We know for an isochoric process, the pressure ##P## must change and the only way ##PV^{x} = constant## is if ##V^x =0## else if we take finite value of ##x## then ##V^x## will be some constant number causing ##PV^{x} ## to not remain constant as ##P## varies . Thus ##V^x= 0 \implies x=-\infty##.

Now if we plug in ##x=-\infty## into the equation ##C = \frac {R}{\gamma -1}. + \frac {R}{1 -x}## then we get the second term on right hand side as ##\frac {R}{1 - (-\infty)} = \frac {R}{\infty} = 0##. That gets us the correct formula for C in isochoric process using the framework mentioned in my original post. We get ##C = \frac {R}{\gamma -1} + 0## when volume ##V## is held constant.

Last edited:
TSny
If you rewrite ##PV^x=C## as $$P^{\frac{1}{x}}V=C^{'}$$you can easily see how ##x=\infty## gives constant V.

Frabjous, PhDeezNutz and vcsharp2003
Chestermiller said:
If you rewrite ##PV^x=C## as $$P^{\frac{1}{x}}V=C^{'}$$you can easily see how ##x=\infty## gives constant V.
But then we will get in the left hand side ##C^{\frac {1}{x}}##, which will not be a constant.

@Chestermiller
Is this formula for C used in practice (i.e., several other values of x that we care about)?

vcsharp2003 said:
But then we will get in the left hand side ##C^{\frac {1}{x}}##, which will not be a constant.
The actual equation should read $$PV^x=P_0V_0^x$$What then?

vcsharp2003
Frabjous said:
@Chestermiller
Is this formula for C used in practice (i.e., several other values of x that we care about)?
No. In my judgment, polytropic processes are used for pedagogical purposes only. In addition, I am strongly opposed to expressing heat capacity (which is supposed to be a function of state) in terms of Q (which is a function of path). lr is very confusing to students.

Frabjous
Chestermiller said:
The actual equation should read $$PV^x=P_0V_0^x$$What then?
I see, so right hand side is a constant if we raise both sides to ##\frac{1}{x} ##. Still we get ##P_o^{\frac{1}{x}} V_o##, which should be a constant inspite of power involving x?

Last edited:
vcsharp2003 said:
I see, so right hand side is a constant if we raise both sides to ##\frac{1}{x} ##. Still we get ##P_o^{\frac{1}{x}} V_o##, which should be a constant inspite of power involving x?
We rewrite it as $$\frac{V}{V_0}=\left(\frac{P_0}{P}\right)^{\frac{1}{x}}$$

vcsharp2003
Chestermiller said:
We rewrite it as $$\frac{V}{V_0}=\left(\frac{P_0}{P}\right)^{\frac{1}{x}}$$
Ok. So LHS would be a constant for isochoric process. This would means the RHS is also the same constant, which is only possible if ## x = -\infty##.

vcsharp2003 said:
Ok. So LHS would be a constant for isochoric process. This would means the RHS is also the same constant, which is only possible if ## x = -\infty##.
Or ##x=\infty##

vcsharp2003
Chestermiller said:
polytropic processes are used for pedagogical purposes only.
You are too young. I remember reading reports that reported gamma for various types of rock for simulation in hydrocodes.

Frabjous said:
You are too young. I remember reading reports that reported gamma for various types of rock for simulation in hydrocodes.
I’m 81.

Frabjous
Chestermiller said:
Or ##x=\infty##
Yes

## 1. What is the formula for calculating molar heat capacities of an ideal gas?

The formula for calculating molar heat capacities of an ideal gas is Cp = (5/2)R for a monatomic gas and Cp = (7/2)R for a diatomic gas, where R is the gas constant.

## 2. Why is it difficult to understand the formula for molar heat capacities of an ideal gas?

The formula for molar heat capacities of an ideal gas can be difficult to understand because it involves complex concepts such as thermodynamics and statistical mechanics. Additionally, the formula may differ depending on the type of gas being studied.

## 3. How is the formula for molar heat capacities of an ideal gas derived?

The formula for molar heat capacities of an ideal gas is derived using the first law of thermodynamics and the equipartition theorem, which states that each degree of freedom of a molecule contributes (1/2)RT to the total energy of the gas.

## 4. Can the formula for molar heat capacities of an ideal gas be applied to real gases?

The formula for molar heat capacities of an ideal gas is an approximation and may not accurately predict the heat capacities of real gases. Real gases have more complex behavior due to intermolecular forces and other factors, so the formula may need to be modified for more accurate results.

## 5. What are some factors that can affect the molar heat capacities of an ideal gas?

The molar heat capacities of an ideal gas can be affected by factors such as temperature, pressure, and the type of gas. Additionally, the number of atoms or molecules in a gas can also impact its heat capacity.

### Similar threads

• Thermodynamics
Replies
2
Views
451
• Thermodynamics
Replies
1
Views
917
• Thermodynamics
Replies
32
Views
2K
• Thermodynamics
Replies
23
Views
1K
• Thermodynamics
Replies
5
Views
1K
• Thermodynamics
Replies
22
Views
2K
• Thermodynamics
Replies
1
Views
604
• Thermodynamics
Replies
4
Views
2K
• Thermodynamics
Replies
12
Views
1K
• Thermodynamics
Replies
3
Views
1K