Need clarification on adiabatic process and heat capacity

[Chestermiller]

Ah, the dissipative process is the cause of why the ΔSsurrounding is less than ΔS.

Who said anything about the surroundings? Our focus here is on the system. There are many possibilities for what can be happening with the surroundings, and how the surroundings are controlled. But, if we only focus on the system, we will find that, no matter what is done with the surroundings, the entropy change for the system will always be greater or equal to dq/T_I. That's why, in my judgement, bringing the surroundings into the picture only leads to un-necessary confusion.

So in a reversible process, the dissipative process is negligible because the oscillation is extremely small?

If the process is done reversibly, there is no oscillation whatsoever, because we are controlling the motion of the piston so that it moves with exceedingly low speed. Also, it is not only the oscillation that causes the dissipation. It is also the very rapid deformation of the gas that contributes to the viscous dissipation.

And that allows the system to do the most work on the surrounding.

Yes. If, since there is no dissipation.

Chet

 

[gracy]

for an ideal gas, Delta_H = Cp Delta_T.

Is this always applicable.

Yes,

Even when pressure is not constant during the process?

 

[Chestermiller]

Even when pressure is not constant during the process?

Yes. But, please, if you want to pursue this, start a new thread. I would like to keep this present thread on topic.

Chet

 

[gracy]

, if you want to pursue this, start a new thread. I would like to keep this present thread on topic.

Ok.

 

[worryingchem]

Starting with the following equations:
$V_f = \frac{nRTA}{Mg}$

$V_i = \frac{nRTA}{Mg + mg}$

$W_f = \frac{Mg(V_f-V_i)}{A}$

Then substituting and simplifying.
$W_f = Mg(\frac{nRT\frac{A}{Mg} - nRT\frac{A}{Mg + mg}} {A})$

$W_f = Mg(\frac{nRT}{Mg} - \frac{nRT}{Mg + mg})$

And got
$W_f = nRT - \frac{nRTM}{M+m}$

Since it's an isothermal process dU = 0 and therefore, dq = -dw.
The total change in heat is
$Q_f = -nRT + \frac{nRTM}{M+m}$

 

[Chestermiller]

$W_f = nRT - \frac{nRTM}{M+m}$

This result is correct. But, if we reduce the terms to a common denominator, we obtain:
$W_f = nRT \frac{m}{(M+m)}$

Since it's an isothermal process dU = 0 and therefore, dq = -dw.

In the calculations were have done here, we have treated W_f as the work done by the system (the gas) is the surroundings. With this sign, convention ΔU=Q-W, so that
$Q =W_f = nRT \frac{m}{(M+m)}$

The last two items on our to do list for this version of the problem are to determine:
4. The change in Entropy of the gas between the initial and final equilibrium states
5. The heat transferred Q divided by the temperature T_I at the boundary where the heat transfer takes place during the irreversible expansion process

Item 5 is trivial to do at this point, because the boundary of the system is maintained at the temperature T for then entire process, so

$\int \frac{dQ}{T_I}=\frac{Q}{T}=nR \frac{m}{(M+m)}$

Now for determining the change in entropy of the system between the same initial and final thermodynamic equilibrium states. Any ideas on how to get the change in entropy ΔS (so that we can compare it with the above integral)?

Chet

 

[worryingchem]

Since entropy is a state variable, we can find the ΔS for the reversible and it would still apply for the irreversible.
$ΔS = ∫\frac {dQ_{rev}}{T} = ∫\frac {dU + dw}{T} = C_v ln \frac {T_f}{T_i} + nR ln \frac{V_f}{V_i}$

Since it's an isothermal process
$$ ΔS = nR ln \frac{V_f}{V_i} = nR ln \frac {\frac {nRTA}{Mg}} {\frac{nRTA}{Mg + mg}} = nR ln \frac {M + m}{M} $$

 

[Chestermiller]

Since entropy is a state variable, we can find the ΔS for the reversible and it would still apply for the irreversible.
$ΔS = ∫\frac {dQ_{rev}}{T} = ∫\frac {dU + dw}{T} = C_v ln \frac {T_f}{T_i} + nR ln \frac{V_f}{V_i}$

Since it's an isothermal process
$$ ΔS = nR ln \frac{V_f}{V_i} = nR ln \frac {\frac {nRTA}{Mg}} {\frac{nRTA}{Mg + mg}} = nR ln \frac {M + m}{M} $$

Excellent! Now, please take a look at the comparison between this result for ΔS with the value for the integral of dQ/T_I in post #36. Please calculate each of these for a range of values for m/M so we can make the comparison. We want to see whether the Clausius inequality is properly satisfied by these results.

After this, we can tackle the case of irreversible adiabatic expansion, if you feel up to it.

Chet

 

[worryingchem]

I attached images of the graph for ΔS and Q/T. The x-value = M while the y-value = m.
So I noticed that the biggest difference between ΔS and Q/T is when m>>M so a sudden change in weight on the piston.
When I tried M>>m using M = 1000 kg while m = 0.1 kg and n=1 mol, I got ΔS = 8.313584*10^-4 J/K and Q/T = 8.313168*10^-4 J/K.
The difference is much smaller, but ΔS is still greater than Q/T. So Clausius inequality is true and the entropy change is greater than the heat transferred into the system and lost by the surrounding over T. So that's why ΔS_total was over 0.
Thank you so much for clearing this up for me.
We can continue to the irreversible adiabatic expansion then if you don't mind. Since Q will be 0, how will you compare the temperature and volume change to get a positive ΔS?

 

[Chestermiller]

I would have done this a little differently, but you certainly demonstrated what we were looking for. I would have plotted ΔS and Q/T as a function of m/M on the same graph. This would have shown all the results at once. But, "different strokes for different folks."

To do the adiabatic case, we do not know the final temperature from the outset. So we need to use the first law of thermodynamics to get the conditions in the final state. The equation you used to determine the work is still the same as before: $W=\frac{Mg}{A}(V_f-V_i)$ . Please write the first law equation that applies to this adiabatic irreversible expansion of an ideal gas, including the proper relationship for the change in internal energy ΔU as a function of the temperature change.

Chet

 

[worryingchem]

Since it's an adiabatic process, Q = 0 so ΔU = W.
For a perfect gas ΔU = C_vΔT
So the first law equation for this adiabatic irreversible expansion is $C_vΔT = \frac{Mg}{A} ΔV$

 

[Chestermiller]

Since it's an adiabatic process, Q = 0 so ΔU = W.
For a perfect gas ΔU = C_vΔT
So the first law equation for this adiabatic irreversible expansion is $C_vΔT = \frac{Mg}{A} ΔV$

You're missing a minus sign on the right hand sides of your equations (since we have been calling W the work done by the gas on its surroundings).

So now call the initial temperature T_i. Combine your first law equation with the ideal gas law to determine the final temperature and final volume. Don't forget that if you are using actual volumes (rather than specific volumes), then there should be an nCv on the left side of the first law equation.

Chet

 

[worryingchem]

For T_f:
$nC_vT_f - nRT_i = \frac {-Mg}{A}V_f + \frac{Mg}{A}V_i = \frac {-Mg}{A} \frac{nRT_f}{P_f} + \frac{Mg}{A}V_i = \frac {-Mg}{A} \frac{nRT_fA}{Mg} + \frac{Mg}{A}V_i$
so
$T_f (nC_v + nR) = nC_vT_i + \frac{Mg}{A}V_i$ and
$$ T_f = \frac{nC_vT_i + \frac{Mg}{A}V_i}{nC_v + nR} $$

For V_f:
$\frac {-Mg}{A}V_f + \frac{Mg}{A}V_i = nC_vT_f - nRT_i = nC_v\frac{P_fV_f}{nR} - nRT_i = C_v\frac{MgV_f}{AR} - nRT_i$
so
$V_f (\frac{C_vMg}{AR} + \frac{Mg}{A}) = nC_vT_i + \frac{Mg}{A}V_i$
and $V_f = \frac{C_v(nRT_i)A + \frac{Mg}{A}V_iRA}{C_vMg + MgR} = \frac{C_v(P_iV_i)A + \frac{Mg}{A}V_iRA}{C_vMg + MgR} = \frac{C_v\frac{g(M+m)}{A}V_iA + \frac{Mg} {A}V_iRA}{C_vMg + MgR}$ and
$$ V_f = \frac{C_v(M+m)V_i + MV_iR}{C_vM + MR} $$

 

[Chestermiller]

You need to use the ideal gas law to get V_i on terms of T_i and the masses. Then substitute it into your equations.

Chet

 

[worryingchem]

$V_i = \frac{nRT_iA}{g(M+m)}$

For T_f:
$$ T_f = \frac{T_i(C_v(M + m) + R)}{(C_v + R)(M + m)} $$

For V_f:
$$ V_f = \frac{\frac{nRT_iA}{g(M +m)}(C_v(M + m) + MR)}{C_vM + MR} = \frac{nRT_iA(\frac{C_v}{g} + \frac{MR}{g(M + m)})}{C_vM + MR} $$

 

[Chestermiller]

Your equation for Tf has a typo. There should be an M multiplying the R in the numerator.
Your equation for Vf looks OK.

But, these equations can be simplified substantially if we make use of the following two relationships:
$$\frac{C_p}{C_v}=γ$$
$$C_p-C_v=R$$
With these substitutions, I obtain:
$$\frac{T_f}{T_i}=\frac{1}{γ}\left[1+(γ-1)\frac{M}{(M+m)}\right]$$
$$\frac{V_f}{V_i}=1+\frac{m}{γM}$$
Please check these results out and see if you can confirm them.

Now that we know the conditions at the initial and final thermodynamic equilibrium states of the system, do you know how to determine the change in entropy between these states?

Chet

 

[worryingchem]

K. Sorry about the long break. Just finished my finals.
Thank you for simplifying the equations, I was wondering if I did something wrong since they looked like a big mess.

So to find the entropy change in an adiabatic process, I'll find the entropy change in the reversible case.
$dS = \frac {dq_{rev}}{T} = \frac{dU + dw}{T} = \frac{C_vdT}{T} - \frac{PdV}{T} = \frac{C_vdT}{T} - \frac{nRdV}{V}$

$ΔS = C_vln(\frac{T_f}{T_i}) - nRln(\frac{V_f}{V_i})$

$$ ΔS = C_vln\left[ \frac{1}{γ} \left(1+\frac{(γ-1)M}{M+m} \right)\right] - nRln\left[ 1 + \frac{m}{γM} \right] $$

 

[Chestermiller]

K. Sorry about the long break. Just finished my finals.

Welcome back. Hope your finals turned out great!

Thank you for simplifying the equations, I was wondering if I did something wrong since they looked like a big mess.

I hope you had a chance to check my results. I'd like to get confirmation on the algebra.

So to find the entropy change in an adiabatic process, I'll find the entropy change in the reversible case.

I think you meant to use the word irreversible here, instead of adiabatic.

$dS = \frac {dq_{rev}}{T} = \frac{dU + dw}{T} = \frac{C_vdT}{T} - \frac{PdV}{T} = \frac{C_vdT}{T} - \frac{nRdV}{V}$

There are two minor errors in the above equation: You are missing an n in front of the C_v, and the work done by the system on the surroundings is pdV, not -pdV. So the equation should read:
$$ dS = \frac {dq_{rev}}{T} = \frac{dU + dw}{T} = n\frac{C_vdT}{T} + \frac{PdV}{T} = n\frac{C_vdT}{T} +\frac{nRdV}{V} $$

So,

$$ ΔS = nC_vln(\frac{T_f}{T_i}) + nRln(\frac{V_f}{V_i}) $$

$$ ΔS = nC_vln\left[ \frac{1}{γ} \left(1+\frac{(γ-1)M}{M+m} \right)\right] + nRln\left[ 1 + \frac{m}{γM} \right] $$

According to the Clausius Inequality, this change in entropy must come out to be positive for all (positive) values of m/M (since, in adiabatic irreversible expansion, Q = 0). I have a suggestion that I think would be worthwhile for you to try. Rewrite the equation as:

$$ \frac{ΔS}{nC_v} = ln\left[ \frac{1}{γ} \left(1+\frac{(γ-1)}{1+(m/M)} \right)\right] + (γ-1)ln\left[ 1 + \frac{(m/M)}{γ} \right] $$

Incidentally, this result can be simplified further to the following (see if you can show it):

$$ \frac{ΔS}{nC_v} = ln\left[\frac{1}{1+(m/M)} \right] + γ\ln \left[ 1 + \frac{(m/M)}{γ} \right] $$

Then plot the left hand side of the equation (the dimensionless parameter $\frac{ΔS}{nC_v}$) as a function of the dimensionless parameter m/M for the case of γ=7/5 (a diatomic gas). Let's see whether it really does come out to be positive, and how the entropy changes as m/M gets larger.

Is there anything else you think we should be considering in this thread before we declare ourselves finished?

Chet

 

[worryingchem]

Thank you, I did great on my finals, the thermodynamic questions were the fastest for me to go through .

Starting at
$\frac{ΔS}{nC_v} = ln\left[ \frac{1}{γ} \left(1+\frac{(γ-1)}{1+(m/M)} \right)\right] + (γ-1)ln\left[ 1 + \frac{(m/M)}{γ} \right]$
$\frac{ΔS}{nC_v} = ln\left[ \frac{1}{γ} \left(1+\frac{(γ-1)}{1+(m/M)} \right)\right] - ln\left[ 1 + \frac{(m/M)}{γ} \right] + γln\left[ 1 + \frac{(m/M)}{γ} \right]$
$\frac{ΔS}{nC_v} = ln\left[ \frac{1+ \frac{γ-1}{1+(m/M)}} {γ +(m/M)} \right] + γln\left[ 1 + \frac{(m/M)}{γ} \right]$
$\frac{ΔS}{nC_v} = ln\left[ \frac{ \frac{1-1+γ+(m/M)}{1+(m/M)}} {γ +(m/M)} \right] + γln\left[ 1 + \frac{(m/M)}{γ} \right]$
$$ \frac{ΔS}{nC_v} = ln\left[ \frac{1}{1+(m/M)}\right] + γln\left[ 1 + \frac{(m/M)}{γ} \right] $$

For my graph, I used m/M as the x-axis and $\frac{\Delta S}{nC_v}$ as the y-axis like you said and it clearly shows that ΔS is positive.
Thank you so much for taking your time to show me with your examples.

 

[Chestermiller]

Thank you, I did great on my finals, the thermodynamic questions were the fastest for me to go through .

Starting at
$\frac{ΔS}{nC_v} = ln\left[ \frac{1}{γ} \left(1+\frac{(γ-1)}{1+(m/M)} \right)\right] + (γ-1)ln\left[ 1 + \frac{(m/M)}{γ} \right]$
$\frac{ΔS}{nC_v} = ln\left[ \frac{1}{γ} \left(1+\frac{(γ-1)}{1+(m/M)} \right)\right] - ln\left[ 1 + \frac{(m/M)}{γ} \right] + γln\left[ 1 + \frac{(m/M)}{γ} \right]$
$\frac{ΔS}{nC_v} = ln\left[ \frac{1+ \frac{γ-1}{1+(m/M)}} {γ +(m/M)} \right] + γln\left[ 1 + \frac{(m/M)}{γ} \right]$
$\frac{ΔS}{nC_v} = ln\left[ \frac{ \frac{1-1+γ+(m/M)}{1+(m/M)}} {γ +(m/M)} \right] + γln\left[ 1 + \frac{(m/M)}{γ} \right]$
$$ \frac{ΔS}{nC_v} = ln\left[ \frac{1}{1+(m/M)}\right] + γln\left[ 1 + \frac{(m/M)}{γ} \right] $$

For my graph, I used m/M as the x-axis and $\frac{\Delta S}{nC_v}$ as the y-axis like you said and it clearly shows that ΔS is positive.
Thank you so much for taking your time to show me with your examples.
View attachment 83673

Excellent job Worryingchem. I particularly like your graph which hints at a zero slope at m/M close to zero. This emphasizes that, at small changes in pressure/volume, the behavior is very close to reversible. Would it be possible to show a replotted version of the graph, with m/M only over the range from 0 to 10, so people can see this feature of the behavior more clearly?

Chet

 

[worryingchem]

K. Here it is.

 

[Chestermiller]

Thanks.

Chet