Why Does the Official Key State X=2.05 Instead of X=1 in This Adiabatic Process?

[bhupesh]

Homework Statement

A spherical bubble inside water has radius 𝑅. Take the pressure inside the bubble and the water pressure to be 𝑝0. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes (𝑅 − 𝑎). For 𝑎 ≪ 𝑅 the magnitude of the work done in the process is given by (4𝜋𝑝0𝑅𝑎2 )𝑋, where 𝑋 is a constant and 𝛾 = 𝐶𝑝⁄𝐶𝑉 = 41⁄30. The value of 𝑋 is ____

Relevant Equations

work done in a process=external pressure*change in volume
First law of thermodynamics, dX signifies change in X,
dQ=dU+dW
Q=heat energy
U=internal energy
W=work done
for any process, dU=n*Cv*dT
T=temperature of system,n=number of moles of gas undergoing the process
Cv=molar heat capacity at constant volume
Cp=molar heat capacity at constant pressure

as the process has been given as an adiabatic one, dQ=0, further attempt is given in the attached files, but the problem is I got X=1 while the official key states it to be X=2.05, could anyone explain why

 

[bhupesh]

I have also got a solution given on the internet that I am not able to comprehend, anyone wants it??

 

[kshitij]

I have also got a solution given on the internet that I am not able to comprehend, anyone wants it??

Bro this is literally from JEE adv 2020 (I even appeared in that paper), you can find thousands of explanations online!

for example, here's one

 

[bhupesh]

Bro this is literally from JEE adv 2020 (I even appeared in that paper), you can find thousands of explanations online!

for example, here's one

I couldn't understand thatt solution and why my solution is wrong, i have written that in the question
clearly

 

[kshitij]

I couldn't understand thatt solution and why my solution is wrong, i have written that in the question
clearly

What you couldn't understand in that solution? Also I didn't go through your solution well enough to comment on that.

 

[bhupesh]

Bro this is literally from JEE adv 2020 (I even appeared in that paper), you can find thousands of explanations online!

for example, here's one

could you also share how was your experience of that particular exam

 

[kshitij]

Also I didn't go through your solution well enough to comment on that.

I see that you didn't write $dV=4\pi r^2a$, use this formula just as we used to do when we took the change in volume for of an elementary shell here, $dr$ can be written as $a$

 

[Delta2]

I am a scrub in thermodynamics, maybe @Chestermiller who is expert can have a look at this.

However the problem statement gives you the value of $\gamma$ (I believe that's the $\phi$ in your work) but you nowhere use the value of $\phi$ in your work, because according to your work it gets simplified. Maybe that thing you do about keeping only the first term in the binomial expansion is wrong, maybe you got to look for another approximation.

 

[Chestermiller]

My result for the work to compress the bubble adiabatically and reversibly is $4\pi R^2 a p_0$, independent of $\gamma$, exactly the same as yours. It seems to me your analysis is flawless.

 

[haruspex]

My result for the work to compress the bubble adiabatically and reversibly is $4\pi R^2 a p_0$, independent of $\gamma$, exactly the same as yours. It seems to me your analysis is flawless.

I note that in post #1 the power of 2 is on a, not R. It also seems obvious that the adiabatic aspect would not affect the first order term.
This suggests to me that the original question says, or intended to say
"The work done is $4\pi R^2 a p_0+4\pi R a ^2p_0X$."
I believe $\gamma$ would feature in the expression for X.

 

[Chestermiller]

I note that in post #1 the power of 2 is on a, not R. It also seems obvious that the adiabatic aspect would not affect the first order term.
This suggests to me that the original question says, or intended to say
"The work done is $4\pi R^2 a p_0+4\pi R a ^2p_0X$."
I believe $\gamma$ would feature in the expression for X.

What do you get for X when you do this? I get X = 1.05

 

[haruspex]

What do you get for X when you do this? I get X = 1.05

Hmm.. that suggests the author confused $R^2a$ with $Ra^2$ and added the terms together.

 

[Delta2]

It also seems obvious that the adiabatic aspect would not affect the first order term.

I would be very interesting to know how you come to this conclusion. Such is the case but how do you say that it is obvious.

 

[haruspex]

I would be very interesting to know how you come to this conclusion. Such is the case but how do you say that it is obvious.

Because for a small displacement $\delta r$ the change in pressure will be of order o(1), e.g. proportional to $\delta r$, so the work done will be like $p_0(\delta r+o(\delta r))$, e.g. $p_0(\delta r+O(\delta r^2))$.

 

[Delta2]

Because for a small displacement $\delta r$ the change in pressure will be of order o(1), e.g. proportional to $\delta r$, so the work done will be like $p_0(\delta r+o(\delta r))$, e.g. $p_0(\delta r+O(\delta r^2))$.

Sorry I don't understand what you trying to say, I am abit confused about the big o small o notation and also pressure times displacement doesn't give us work, pressure times volume does.

 

[Chestermiller]

Hmm.. that suggests the author confused $R^2a$ with $Ra^2$ and added the terms together.

Or a typo

 

[Twigg]

@Delta2 I think what's going on is that since this is adiabatic, $pV^\gamma$ is a constant, so $\frac{dp}{p} = -\gamma \frac{dV}{V}$. Making a simple first order approximation, $\frac{\Delta p}{p} = -\gamma \frac{\Delta V}{V}$, where $\Delta$ means finite difference as opposed to differentials. $\Delta V$ scales linearly with radial displacement $\delta r$, so $\Delta p$ scales linearly with $\delta r$. If you work it out, you get a pressure $p$ vs displacement $\delta r$ given by $p\approx p_0 (1 + 3\gamma \frac{\delta r}{R})$. Since the term with $\gamma$ in it is small compared to 1, it vanishes at lowest order, and the work will just be $W \approx p_0 \Delta V$, which is first order in $a$, aka $O(a^1)$. The variation in pressure becomes a second order term in a.

 

[Twigg]

What do you get for X when you do this? I get X = 1.05

I ended up with 2.05 (from $\frac{3}{2} \gamma$), but I trust your work over mine

 

[bhupesh]

but the exam is such you can challenge a particular question for bonus if its a typo or its answer given is wrong, many challenged the answer as 2.05 but finally ans was not changed, it remained 2.05
I have a solution available on the internet that is doubtful to me and i have attached that, check question no. 13 in the pdf file

 

[Chestermiller]

I ended up with 2.05 (from $\frac{3}{2} \gamma$), but I trust your work over mine

I ended up with that minus 1

 

[bhupesh]

What do you get for X when you do this? I get X = 1.05

but they have clearly mentioned that the whole process is adiabatic??

 

[Twigg]

@bhupesh, I still agree with everyone else on the thread in saying that this problem is somewhere on the spectrum between poorly worded and dead wrong. In the solutions you posted, they define the average pressure as half the change in pressure. That makes no sense to me, unless there's a constraint force balancing p0, which is not at all stated explicitly in the problem. I got the same number as the solutions, but I would've challenged their answer too.

I'm sorry they weren't more open-minded in reviewing the challenges. That sucks

Edit: after re-reading the problem statement, they say explicitly that the pressure inside the bubble is p0, so there's no room for a constraint force balancing p0. They're just wrong.

 

[bhupesh]

hmm, thanks for helping

 

[Delta2]

@bhupesh if you consider only the second order term in the binomial expansion what do you get that its coefficient is?

 

[Chestermiller]

I ended up with 2.05 (from $\frac{3}{2} \gamma$), but I trust your work over mine

My solution for the work done by the gas on the surroundings was $$W=p_0v_0\left[\frac{(1-x)^{3-3\gamma}-1)}{(1-\gamma)}\right]$$where x = a/R. If I expand this in a Taylor series in x, I get $$W=-p_0v_0\left(3x+(3\gamma-2)\frac{x^2}{2}\right)$$which leads to X=1.05.

The reason you ended up with 2.05 is that you left out the third term in the expansion of $(R-a)^3$

 

[bhupesh]

@bhupesh if you consider only the second order term in the binomial expansion what do you get that its coefficient is?

I am getting 1.05 with only second term included, neglecting the first term

 

[Twigg]

@Chestermiller Oh! That makes sense, thanks! Looks like I should've expanded the volume change to 2nd order in a, and multiplied that by the 0th order in p.

 

[bhupesh]

My solution for the work done by the gas on the surroundings was $$W=p_0v_0\left[\frac{(1-x)^{3-3\gamma}-1)}{(1-\gamma)}\right]$$where x = a/R. If I expand this in a Taylor series in x, I get $$W=-p_0v_0\left(3x+(3\gamma-2)\frac{x^2}{2}\right)$$which leads to X=1.05.

The reason you ended up with 2.05 is that you left out the third term in the expansion of $(R-a)^3$

why are we adding these terms when a is a infinitesimal quantity

 

[Twigg]

why are we adding these terms when a is a infinitesimal quantity

We are treating a as a small but finite quantity, not an infinitesimal.

 

[bhupesh]

We are treating a as a small but finite quantity, not an infinitesimal.

then also we cannot add them and write X=1+1.05, it would be X=(1/a )+ 1.05

 

[Twigg]

Sorry, you lost me on this. I think we're talking about different things. X can't be 1/a + 1.05 because 1/a has units of inverse meters?

 

[Delta2]

The sum of the coefficients of the first and second term is 2.05, but yes we can't add the coefficients together, they don't have common factor.

 

[Chestermiller]

why are we adding these terms when a is a infinitesimal quantity

We are expanding the exact expression for the work in a Taylor series to 2nd order terms in a/R. So the result is not exact; we are retaining only the first 3 terms of the expansion.