Gas undergoing isothermic compression

[Karol]

Homework Statement

10 liters of air at atmospheric pressure and temperature 300⁰K were isothermally compressed to a volume of 2[liter] and then freely expanded adiabatically to their original volume. what's the final temperature.

Homework Equations

Adiabatic process: $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$
γ of air=1.4

The Attempt at a Solution

I ignore the first isothermic stage and start from the end conditions of the first stage: $$V_1=2[liter], T_1=300^0K$$:
$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}\rightarrow 300\cdot 2^{0.4}=T_2\cdot 10^{0.4}$
$\rightarrow T_2=158^0K=-115^0C$
It should be -102⁰C

 

[TSny]

...and then freely expanded adiabatically ...

A "free" expansion generally denotes an irreversible process in which the gas is allowed to expand into a vacuum without doing any work. Does $TV^{\gamma - 1} = const$ apply to an irreversible process? If the expansion is adiabatic and free, what happens to the energy of the gas during the expansion? What does that tell you about the temperature change during the expansion (assuming ideal gas behavior)?

Anyway, I don't see how to get the answer of -102 ^oC.

 

[Chestermiller]

I think that by the term "freely expanded," they meant for you to assume that the adiabatic expansion took place against a resistance of 1 atm. That would explain why their answer of -102 is higher than your answer of -115. I realize that the term free expansion correctly refers to expansion against a vacuum, but I don't think that is what they meant here. Try the problem with the assumption of 1 atm resistance and see what you get.

Chet

 

[Karol]

Does $TV^{\gamma - 1} = const$ apply to an irreversible process?

I didn't learn about reversible or irreversible processes. do the equations $\frac{PV}{T}=Const,\ TV^{\gamma - 1} = const$ apply only to reversible processes? and what is an irreversible process? is, in our case of free expansion, the process irreversible because if we compress the gas back we must invest work while the expansion occur without work and it will contradict the conservation of energy?
In the free adiabatic expansion no work is done and no heat exchange occurs so internal energy is conserved and the temperature remains unchanged, so $TV^{\gamma - 1} = const$ doesn't apply. so how do we solve?

I think that by the term "freely expanded," they meant for you to assume that the adiabatic expansion took place against a resistance of 1 atm. That would explain why their answer of -102 is higher than your answer of -115

The first law: $Q\mbox{(heat exchange)}=U_2-U_1+W$, Q=0, W>0 (work is done against resistance) so U, the internal energy, in this case should be lower, not higher, than expansion into vacuum so the temperature in expansion against 1[atm] must be lower than -115⁰, not higher.
Now i will try to solve with the assumption of resistance.

 

[Chestermiller]

The first law: $Q\mbox{(heat exchange)}=U_2-U_1+W$, Q=0, W>0 (work is done against resistance) so U, the internal energy, in this case should be lower, not higher, than expansion into vacuum so the temperature in expansion against 1[atm] must be lower than -115⁰, not higher.
Now i will try to solve with the assumption of resistance.

When you solved for the -115, it wasn't for free expansion into vacuum. It was for adiabatic reversible expansion. U will be higher for the expansion at 1 atm resistance than for the case of reversible adiabatic expansion that you assumed.

Chet

 

[Karol]

U will be higher for the expansion at 1 atm resistance than for the case of reversible adiabatic expansion that you assumed.

I don't know nothing about this and about reversible/irreversible processes. maybe this topic will appear at the second law, otherwise i will have to learn it by myself.
But in any case, in adiabatic expansion into vacuum doesn't U conserve? there isn't any work and no exchange of heat, so why is it even lower than against resistance?

 

[Karol]

I quote from wikipedia, about irreversible process:
"By releasing pressure on a sample and thus allowing it to occupy a large space, the system and surroundings will have completely left equilibrium, and heat dissipation will be large compared to the little work done."
But i read that ideal gas that undergoes a process of choking doesn't change temperature, but real gases do. so does the article in wikipedia refer to ideal gases or not?

 

[Karol]

I understand that in free expansion the heat loss is the biggest thus how to calculate assuming resistance of 1[atm]? so far i studied only first law and the varios forms of the equation of state but i am curios. secondly, is it possible with my knowlegde to get to -102⁰C or i did right by calculating only according adiabatic expnsion?

 

[Chestermiller]

I quote from wikipedia, about irreversible process:
"By releasing pressure on a sample and thus allowing it to occupy a large space, the system and surroundings will have completely left equilibrium, and heat dissipation will be large compared to the little work done."
But i read that ideal gas that undergoes a process of choking doesn't change temperature, but real gases do. so does the article in wikipedia refer to ideal gases or not?

Choking (aka throttling) is a process that takes place in an open system with gas flowing into and out of the system. To analyze that, you need to use the open system version of the first law, which you probably haven't learned yet. In any case, choking is not relevant to your problem which involves a closed system.

 

[Chestermiller]

I don't know nothing about this and about reversible/irreversible processes. maybe this topic will appear at the second law, otherwise i will have to learn it by myself.

If you wanted the expansion to occur reversibly, you would back off very gradually on the pressure you apply to the piston (assumed massless), rather than releasing it suddenly. Your gamma equations apply to the case where you back off very gradually.

But in any case, in adiabatic expansion into vacuum doesn't U conserve?

Yes.

there isn't any work and no exchange of heat, so why is it even lower than against resistance?

That's why I said in response #3 that maybe they didn't really mean perfectly free expansion (which is expansion against a vacuum). Maybe they meant expansion against a pressure of 1 atm., and they mistakenly used the term free expansion. As you correctly indicated, if it were truly free expansion, the final temperature would be the same as the initial temperature.

Chet

 

[Chestermiller]

I understand that in free expansion the heat loss is the biggest

Actually, there is no heat loss in adiabatic free expansion and there is no work done.

thus how to calculate assuming resistance of 1[atm]? so far i studied only first law and the varios forms of the equation of state but i am curios.

The work done against a constant resistance of 1 atm is W = 1 (10 - 2) liter-atm, which is equal to 800 J. So ΔU would be -800 J.

secondly, is it possible with my knowledge to get to -102⁰C or i did right by calculating only according adiabatic expnsion?

I solved the problem assuming expansion against a constant pressure of 1 atm., but got a final temperature of 204K (= -69 C). I can't account for how they got -102 C. To get that result, they would have had to expand against a constant pressure of about 1.34 atm.

Chet

 

[Karol]

I am totally confused, i don't understand. i saw your result for calculating against 1[atm] but i want to try myself.
If i insert 1[atm] and 10[liter], with the original number of moles n into the equation of state PV=nRT, i get the same temperature like in the beginning: 300⁰K.
I don't understand when to use what.
If we assume expansion against 1[atm] then it's also P, the final pressure, and the volume is 10[liter] so i don't have a choice and i have to use these numbers in the equation of state. and the mass of the gas remains unchanged, so i have to use the same number of mols n also.

 

[Chestermiller]

I am totally confused, i don't understand. i saw your result for calculating against 1[atm] but i want to try myself.
If i insert 1[atm] and 10[liter], with the original number of moles n into the equation of state PV=nRT, i get the same temperature like in the beginning: 300⁰K.
I don't understand when to use what.
If we assume expansion against 1[atm] then it's also P, the final pressure, and the volume is 10[liter] so i don't have a choice and i have to use these numbers in the equation of state. and the mass of the gas remains unchanged, so i have to use the same number of mols n also.

I completely understand your source of confusion. What you are missing is that, once the piston reaches the final location where the volume is 10 liters (and then stops moving), the gas within the cylinder has not yet achieved thermodynamic equilibrium. There are still spatial variations in gas temperature and pressure within the cylinder. No more work will be done, and no heat will enter or leave the gas, but it will take some additional time before the gas achieves thermodynamic equilibrium. At the end of this additional time, the temperature and pressure of the gas will become uniform at the final thermodynamic equilibrium values. The final temperature will be less than 300 C (because the gas has done adiabatic work) and the final pressure will be less than 1 atm.

Chet

 

[Karol]

How did you calculate with 1[atm] resistance, or should i wait until i learn it later?
And how is it possible that the pressure will go down below i[atm] if the gas expands against resistance of 1[atm]?

 

[Chestermiller]

How did you calculate with 1[atm] resistance, or should i wait until i learn it later?

See post # 11.

And how is it possible that the pressure will go down below i[atm] if the gas expands against resistance of 1[atm]?

After the expansion is complete, the pressure decreases as the system equilibrates further at constant volume.

Chet[/QUOTE]

 

[Chestermiller]

Here's the full calculation:

$n=\frac{pv}{RT}=\frac{(1)(10)}{300R}$

where R = 0.082 (liter-atm)/(degree-mole)

So $ΔU=nC_v(T-300)=-W=-PΔV=-(1)(8)$

So $\frac{(1)(10)}{300R}C_v(T-300)=-8$ liter-atm

So $T-300=-300(0.8)\frac{R}{C_v}=-300(0.8)(γ-1)$

Chet

 

[Karol]

Why C_v for the adiabatic expansion? it's not at fixed volume

 

[Chestermiller]

Why C_v for the adiabatic expansion? it's not at fixed volume

C_v is defined more generally by the equation:

C_v=\left(\frac{\partial U}{\partial T}\right)_V

where U is the internal energy per unit mass. For a process at constant volume, this is consistent with C_vdT=dU=dQ. For an ideal gas, both U and C_v are functions only of temperature.

Chet

 

[Karol]

this is consistent with C_vdT=dU=dQ.

You meant C_vdT=dU=dQ, the dash was a typing mistake, right?
So what is C_p then? and why is C_v suitable for a changing volume? C_v is the temperature rise for a unit mass for 1⁰ at constant volume, that's what i learned

 

[Chestermiller]

You meant C_vdT=dU=dQ, the dash was a typing mistake, right?

Yes. Sometimes I mistakenly use LaTex symbology when I'm applying the regular text symbology.

So what is C_p then?

C_p=\left(\frac{\partial H}{\partial T}\right)_P
where H is the enthalpy U + PV.

and why is C_v suitable for a changing volume? C_v is the temperature rise for a unit mass for 1⁰ at constant volume, that's what i learned

This is a problem with the way they teach this stuff in many of the books and at schools. It's because they start out relating the heat capacities to Q, which is a characteristic of the process, rather than of the material being processed. Cv and Cp are properties of the material, not the process. Yes, in a constant volume process, dQ = Cv dT and dU = CvdT for an ideal gas. In a non-constant volume process, dQ is not equal to CvdT, but, because Cv is a material property, dU is still equal to CvdT for an ideal gas. This has been an unending source of confusion for students over the ages. You are not alone.

If you still have doubts about what I'm saying, look up the derivations of your "gamma equations" for adiabatic reversible processes. These are not constant volume process nor constant pressure processes, but they still use Cv in their derivations for relating pressures, volumes, and temperatures.

Chet

 

[Chestermiller]

Oh yes. One more thing. For an ideal gas, the internal energy (and enthalpy) are functions only of temperature, independent of pressure and volume. So, even if the volume changes, dU is still equal to CvdT.

Chet

 

[Karol]

In a non-constant volume process, dQ is not equal to CvdT, but, because Cv is a material property, dU is still equal to CvdT for an ideal gas.

I combine this with what i read in Wikipedia. So C_v isn't $\frac{\mbox{invested heat}}{\mbox{temperature rise}}$ it's:
$C_v=\left(\frac{\partial U}{\partial T}\right)=\frac{\mbox{invested heat}}{\mbox{internal energy rise}}$, right?
But is it $\left(\frac{\partial U}{\partial T}\right)$ or $\left(\frac{\partial U}{\partial T}\right)_v$? i mean is C_v the internal energy rise in any case and any condition or just in constant volume? from your answer i guess it's only $C_v=\left(\frac{\partial U}{\partial T}\right)$, in any process, not necessarily at constant volume, right?

Just now i saw your last post. i didn't learn yet about enthalpy but i read it's equal to U+PV, so it depends not only on temperature, no?

 

[Chestermiller]

I combine this with what i read in Wikipedia. So C_v isn't $\frac{\mbox{invested heat}}{\mbox{temperature rise}}$ it's:
$C_v=\left(\frac{\partial U}{\partial T}\right)=\frac{\mbox{invested heat}}{\mbox{internal energy rise}}$, right?
But is it $\left(\frac{\partial U}{\partial T}\right)$ or $\left(\frac{\partial U}{\partial T}\right)_v$? i mean is C_v the internal energy rise in any case and any condition or just in constant volume? from your answer i guess it's only $C_v=\left(\frac{\partial U}{\partial T}\right)$, in any process, not necessarily at constant volume, right?

No. Only for an ideal gas is it dU/dT, since, for an ideal gas, U is not a function of V. For non-ideal (real) gases (and other materials), $dU=C_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV$

Chet

 

[Chestermiller]

No. Only for an ideal gas is it dU/dT, since, for an ideal gas, U is not a function of V. For non-ideal (real) gases (and other materials), $dU=C_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV$

When you are thinking about U, you should not be thinking about the process. U is a function of state, independent of process path, and depends only on the initial and final state of the material (T,V).

Chet

 

[Chestermiller]

I combine this with what i read in Wikipedia. So C_v isn't $\frac{\mbox{invested heat}}{\mbox{temperature rise}}$ it's:
$C_v=\left(\frac{\partial U}{\partial T}\right)=\frac{\mbox{invested heat}}{\mbox{internal energy rise}}$, right?

For an ideal gas, it's just the rate of change of the internal energy of the material with respect to temperature. The internal energy of a monoatomic ideal gas is just the sum of the kinetic energies of the individual molecules. For a diatomic gas, it also includes the vibrational energy. These energies both increase with temperature.

Just now i saw your last post. i didn't learn yet about enthalpy but i read it's equal to U+PV, so it depends not only on temperature, no?

Yes. Good observation.

Chet

 

[Karol]

so C_v is measured in constant volume but is used in all processes? there must be an explanation to the constant volume notation, other than the effect on ideal gas that you described in post #20: "Yes, in a constant volume process, dQ = Cv dT and dU = CvdT for an ideal gas". so far i didn't find one.
$$\left(\frac{\partial U}{\partial V}\right)_TdV$$ change of volume against pressure is work, does this member account for the work? if yes, why is it missing in ideal gas?
If not please give me an example since i don't know how just a change in volume changes internal energy

 

[Chestermiller]

so C_v is measured in constant volume but is used in all processes?

Yes.

there must be an explanation to the constant volume notation, other than the effect on ideal gas that you described in post #20: "Yes, in a constant volume process, dQ = Cv dT and dU = CvdT for an ideal gas". so far i didn't find one.

Actually, dU=CvdT applies generally to any single phase constant composition system at constant volume.

$$\left(\frac{\partial U}{\partial V}\right)_TdV$$ change of volume against pressure is work, does this member account for the work?

Not completely. The reversible work -PdV is one term that contributes to $$\left(\frac{\partial U}{\partial V}\right)_TdV$$. But that's not the only term. When you study the second law of thermo, you will find that there is also another term that contributes, related to the effect of volume on entropy. The complete expression is

$$\left(\frac{\partial U}{\partial V}\right)_TdV=\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)dV$$

So, the full expression for dU in terms of changes in temperature and specific volume is:

$dU=C_vdT+\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)dV$

if yes, why is it missing in ideal gas?

What is the term in parenthesis equal to for an ideal gas?

If not please give me an example since i don't know how just a change in volume changes internal energy

For an ideal gas, it doesn't. For a real gas, the term in parenthesis is not zero.

Chet

 

[Karol]

For ideal gas:
$$PV=nRT\rightarrow \left(\frac{\partial P}{\partial T}\right)_V=\frac{nR}{T}$$
$$\left(\frac{\partial P}{\partial T}\right)_V=T\frac{nR}{T}=P$$
$$\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)=0$$

 

[Chestermiller]

For ideal gas:
$$PV=nRT\rightarrow \left(\frac{\partial P}{\partial T}\right)_V=\frac{nR}{T}$$
$$\left(\frac{\partial P}{\partial T}\right)_V=T\frac{nR}{T}=P$$
$$\left(T\left(\frac{\partial P}{\partial T}\right)_V-P\right)=0$$

So you now kind of see how the internal energy and constant volume heat capacity of an ideal gas can be a function only of temperature? I know that the explanation is not totally satisfying, since you haven't learned about the 2nd law and entropy yet. Another way we know that the internal energy is not a function of volume is from experimental evidence, where, if we carry out an isothermal expansion or compression of a gas in the ideal gas region of low pressures, we find observationally that the heat added is equal to the work done, and thus, the change in internal energy is zero.

Chet

 

[Karol]

I made a mistake in the last post:
$$PV=nRT\rightarrow \left(\frac{\partial P}{\partial T}\right)_V=\frac{nR}{V}$$
$$\left(\frac{\partial P}{\partial T}\right)_V=T\frac{nR}{V}=P$$
Internal energy of non-ideal gas is also the potential energy between the molecules or atoms? is that the reason why U in non-ideal gas is a function of more than temperature?
And also we didn't talk at all about the dependence of the specific heat, C_v, on temperature:

So you now kind of see how the internal energy and constant volume heat capacity of an ideal gas can be a function only of temperature?