# How Do You Derive P(V) for an Ideal Gas Transformation with dQ=0?

• fluidistic
In summary, the homework statement asks for the relation P=P(V) for a transformation dQ=0 in an ideal gas. Differentiating we get PdV+VdP =nRdT. Substituting dT into -PdV=CRndT does not help much, so the student is stuck. They are asked to find V on the other side and substitute c. They get PV^x=constant where x is gamma i.e C(P)/C(V). The constant(1/B) will depend on initial conditions.
fluidistic
Gold Member

## Homework Statement

Find the relation $P=P(V)$ for a transformation $dQ=0$ in an ideal gas ($PV=nRT$ and $U=CnRT$).

## Homework Equations

$dU=dQ-PdV$.

## The Attempt at a Solution

If I assume that C and R are constant I get $dU=CR \left [ \frac{\partial (nT)}{\partial n} dn + \frac{\partial (nT)}{\partial T } dT \right ] =-PdV$.
If I assume that n does not depend on T, this simplifies to $dU=CR(Tdn+ndT )=-PdV$. If I assume that n is constant, dn=0 and so $-PdV=CRndT$. But since the pressure can depend on the volume, I cannot just integrate this equation. I don't know how to find P(V). Also I think I assumed too many things that weren't stated in the problem statement. Any idea on how to proceed?

EDITed out. This is deriving the fundamental p-V relationship of an adiabatic process in an ideal gas ...

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Whenever, you come across such questions its best to assume that PV^x or PT^x is constant.(valid only when amount of gas i.e effective number of moles are not changing)

Now you have to find what the value of x and constant are.

Apply two laws.
Ideal gas law and 1st law of thermodynamics

However, since you have made a different approach(and you have dine quiet well) I would like you to continue with that.

You have used the 1st law of Thermodynamic.Now use the ideal gas law to substitute dT.

Thanks guys. Yeah I'm going to stick with my way of doing it. So you suggest me to use the relation $PV=nRT$ to get $dT$ and then replace in $-PdV=CRndT$.
So I get $dT=\left ( \frac{\partial T}{\partial P} \right ) dP+\left ( \frac{\partial T}{\partial V} \right ) dV$. Replacing into $-PdV=CRndT$ does not help much I think. I'm wondering if this is the way to go.

fluidistic said:
Thanks guys. Yeah I'm going to stick with my way of doing it. So you suggest me to use the relation $PV=nRT$ to get $dT$ and then replace in $-PdV=CRndT$.
So I get $dT=\left ( \frac{\partial T}{\partial P} \right ) dP+\left ( \frac{\partial T}{\partial V} \right ) dV$. Replacing into $-PdV=CRndT$ does not help much I think. I'm wondering if this is the way to go.

This is derived in every calculus-level physics and thermodynamics textbook you'll ever run into, guaranteed.

The ideal gas equation is PV=nRT

Differentiating we get PdV+VdP =nRdT

Sunstitute dT now.

Don't use partial derivatives

Thanks guys, I'd rather do it myself with your help :)
So $dT=\frac{PdV+VdP}{R}$.
$\Rightarrow c(PdV+VdP)=-PdV \Rightarrow -P(1+c)dV=cVdP \Rightarrow -(1+c) \ln V =c \ln P + K \Rightarrow P= \frac{(V^{-1-c})^{1/c}}{B}$ where B is a constant.
I have never seen that before, so I'm guessing I made error(s)...

You have got it correctly.:-)

Just get V on the other side and substitute c.
You will get PV^x=constant where x is gamma i.e C(P)/C(V).
The constant(1/B) will depend on initial conditions
:-)

p.s. (Avoiding using symbol B as constant.People may confuse it with Bulk Modulus in Thermodynamics.Strangely its better to rather use x,y,z :P

fluidistic said:
Thanks guys, I'd rather do it myself with your help :)
So $dT=\frac{PdV+VdP}{R}$.
$\Rightarrow c(PdV+VdP)=-PdV \Rightarrow -P(1+c)dV=cVdP \Rightarrow -(1+c) \ln V =c \ln P + K \Rightarrow P= \frac{(V^{-1-c})^{1/c}}{B}$ where B is a constant.
I have never seen that before, so I'm guessing I made error(s)...

Hate to crash the party, but that is incorrect. C is a heat capacity but as emailanmol alludes to in his last post we need a ratio of heat capacities, not just one heat capacity. You have been dealing with just one heat capacity so far if I'm not mistaken (which I often am ... )

rude man said:
Hate to crash the party, but that is incorrect. C is a heat capacity but as emailanmol alludes to in his last post we need a ratio of heat capacities, not just one heat capacity. You have been dealing with just one heat capacity so far if I'm not mistaken (which I often am ... )

Lol Rudeman :-)

Actually C is not heat capacity here.

From what i can see, C is degrees of freedom/2.

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emailanmol said:
C is not heat capacity here.

From what i can see c is degrees of freedom/2

OK, $U=CnRT$). That's from the original problem statement. So C here is actually CV/nR = cV/R, CV is total heat capacity at constant volume, cV is molar specific heat at constant volume, and n= no. of moles of the gas.

The relation dU = CVdT is a direct consequence of the definition of an ideal gas.

My point is he's missing another heat capacity, that at constant pressure. You obviously know the correct formula but he's not there himself. He has to derive it whereas you & I know it from our textbooks and/or instructors.

BTW I have problems with that statement to begin with. U = CnRT assumes validity down to absolute zero which is by no means a fait accompli at the OP's stage of learning (or mine, for that matter). It's a lot safer to just say dU = CnRdT or as I prefer it, dU = CVdT.

rude man said:
OK, $U=CnRT$). That's from the original problem statement. So C here is actually CV/nR = cV/R, CV is total heat capacity at constant volume, cV is molar specific heat at constant volume, and n= no. of moles of the gas.

The relation dU = CVdT is a direct consequence of the definition of an ideal gas.

My point is he's missing another heat capacity, that at constant pressure. You obviously know the correct formula but he's not there himself. He has to derive it whereas you & I know it from our textbooks and/or instructors.

BTW I have problems with that statement to begin with. U = CnRT assumes validity down to absolute zero which is by no means a fait accompli at the OP's stage of learning (or mine, for that matter). It's a lot safer to just say dU = CnRdT or as I prefer it, dU = CVdT.

As you said C= C(V)/R and C(V)/R is degrees of freedom/2.

Your other missing heat capacity is the term c+1 in power of V. ( what is (c+1)R ?)

As for the last point, i think the OP made a typo by writing U=CnRT .
It should be dU=CnR(dT).

He has used the correct formula in this post

fluidistic said:
Thanks guys. Yeah I'm going to stick with my way of doing it. So you suggest me to use the relation $PV=nRT$ to get $dT$ and then replace in $-PdV=CRndT$.
So I get $dT=\left ( \frac{\partial T}{\partial P} \right ) dP+\left ( \frac{\partial T}{\partial V} \right ) dV$. Replacing into $-PdV=CRndT$ does not help much I think. I'm wondering if this is the way to go.

His answer is perfect . :-)

As i mentioned the only thing he needs to work on is to avoid using standard variable as different ones .(which is causing this whole confusion over c)
But if he does he should declare them in advance as he did for B. :-)

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I didn't make a typo when writing $U=CnRT$.
I reached the formula $-PdV=CRndT$ because $dU=dQ-PdV$ and $U=CnRT$. I calculated $dU$ assuming n, R and C as constants, yielding $dU=CnRdT$.

Anyway I did what you said (try to get PV in one side), reaching $PV^{1+1/c}=K_2$ where $K_2$ is a constant.
I hope this is right. Actually I prefer to derive $PV^\gamma = \text{constant}$ rather than getting a "useless" P(V) as they asked me. Well I should get both.

Hey. U =CnRT is not right.

dU =CnRdT is right.

U=CnRT+ U(0)

Which goes away on differentiating :-)

Also you do realize that 1+c/c is gamma.Right ?

( there is a typo in your previous post in power of V where you have written 1+1/c)

emailanmol said:
Hey. U =CnRT is not right.

dU =CnRdT is right.

U=CnRT+ U(0)

Which goes away on differentiating :-)

Also you do realize that 1+c/c is gamma.Right ?

( there is a typo in your previous post in power of V where you have written 1+1/c)

Sorry for being so late (too busy with other courses, it's hard not to fall behind in all courses!).
I can see U=CnRT in wikipedia: http://en.wikipedia.org/wiki/Ideal_gas, are you sure it's wrong? Basically the problem consider an ideal gas whose internal energy tends to 0 when the temperature tends to 0. I don't know if it's relevant for this problem though, but I do think the equation is right.
Do you mean $\gamma=\frac {1+c}{c}$? In which case it's worth 1/c+1 which is what I found. I don't see my typo in post #13.

Taking U as CnRT + U(0) is a good practice, as defining energy requires us to define a reference point.No one knows what will happen at 0K :-)

Practically and Logically it makes no difference to skip writing U(0) but it is a good practice to write it. :-)

Yes
y(gamma)=(1/c) +1

The typo (which isn't there.Sorry !) in your post is that
you wrote PV^1+1/c=K2

I thought you were writing (1+1)/c (which is really stupid of me :P).
It is actually fine :-)

fluidistic said:
$PV^{1+1/c}=K_2$ where $K_2$ is a constant.

Okay, thank you very much. So basically the problem is solved, right?

Yes, it is :-)

## 1. What is the ideal gas law?

The ideal gas law is a fundamental equation in thermodynamics that describes the relationship between the pressure, volume, and temperature of an ideal gas. It is written as PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.

## 2. How is the ideal gas law derived?

The ideal gas law is derived from the combination of three other gas laws: Boyle's law, Charles's law, and Avogadro's law. These laws describe the relationships between pressure and volume, volume and temperature, and volume and number of moles, respectively. When combined, these laws result in the ideal gas law.

## 3. What is the significance of the ideal gas law?

The ideal gas law is important in understanding the behavior of gases in different conditions. It can be used to predict the behavior of gases at different temperatures, pressures, and volumes. It also helps in determining the number of moles of a gas present in a given volume. The ideal gas law is also the basis for many thermodynamic processes and calculations.

## 4. Is the ideal gas law always accurate?

No, the ideal gas law is only accurate for ideal gases, which are hypothetical gases that have no intermolecular forces and occupy no volume. Real gases do not behave exactly according to the ideal gas law, especially at high pressures and low temperatures. However, the ideal gas law is a good approximation for many real gases under normal conditions.

## 5. How is the ideal gas law used in real-life applications?

The ideal gas law is used in various real-life applications, such as in the design of gas storage tanks and pipelines, the calculation of the volume and pressure of gases in chemical reactions, and the determination of the number of moles of a gas in a given sample. It is also used in the study of atmospheric gases and in the development of new technologies, such as fuel cells and gas turbines.

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