Conceptual misunderstanding in applying formula for work done in adiabatic compression

[physicsissohard]

Homework Statement

Q.13 A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water pressure to
be $p_0$. 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 $(4\pi p_0Ra^2
)X$, where $X$ is a constant
and $\gamma = C_p/C_v = 41/30$. The value of $X$ is________ .

Relevant Equations

This is how I solved it: $W_g$ is work done by gas and $W_l$ is work done by liquid.

I was trying to solve a JEE ADVANCED 2020 paper 1 question 13 on thermodynamics. I attempted the problem and got a different answer than they did. I did it in a different way, but I can't find a mistake in both methods. So now I am stuck. I think the problem is in the first step probably using the direct formula but I can't pinpoint it.I also get their method which is more simple basically look at the area under the graph, and you see that whay you need to find is half of (dP)(dP) if you approximate it to a triangel. But I don't understand where I went wrong with mine. I used work done in adibatic conditions.

 

[Chestermiller]

Your work is unreadable. Please enter your work using LaTex. There is a guide at the bottom left of the Reply page.

Was what you stated the exact statement (word for word) of the problem?

 

[kuruman]

Your first equation is

Total work done = W_g + W_l

I am suspicious of questions that ask for the "work done" without specifying who is doing the work and on what. Here we are looking for a magnitude so the sign of the work does not matter. However, I would think that we are looking for the work done on the gas by the liquid. The sum of the works on the right hand side of your equation is zero.

I would not assume that the work is the area of a triangle in the pV-diagram. I would use pV^γ = const. and the ideal gas law to find the final temperature and then use the first law of thermodynamics to find the work. Since you did not post the solution that "they" got, I cannot tell you if it is right or wrong.

 

[Chestermiller]

if $pV^{\gamma}=const$, then $$p=p_0\left(\frac{R}{r}\right)^{3\gamma}$$In addition, $$dW=pdV=p(4\pi r^2)dr$$So, $$dW=p_0\left(\frac{R}{r}\right)^{3\gamma}(4\pi r^2)dr$$
Are you able to integrate that from R to R-a ?

 

[kuruman]

Are you able to integrate that from R to R-a ?

Yes, that's a more direct way to do it.

 

[Chestermiller]

With all due respect to whomever obtained the answer in the OP, I get $$W=-4\pi R^2 p_0 a$$ irrespective of $\gamma$.

 

[kuruman]

With all due respect to whomever obtained the answer in the OP, I get $$W=-4\pi R^2 p_0 a$$ irrespective of $\gamma$.

I obtained your result by integrating first and then finding an approximate expression when $\frac{R-a}{R}<<1.$

Here is how Sanjeew Sir does it to whom "respect" might be due. The spoken language sounds like a melange of about 20% English and 80% something else but the math is transparent. First he defines parameter $x$ which is the amount of radial compression from the initial radius, then finds an approximate expression for $pdV$ when $x<<R$ and finally does the integral. I think that one should save approximations until the very end of other mathematical manipulations such as integrals.

 

[haruspex]

I think that one should save approximations until the very end of other mathematical manipulations such as integrals.

That's certainly the safer way, but in the present case I would argue that:

1. ignoring the resulting increase in pressure, we get $4\pi R^2ap_0$ trivially
2. the extra work from the increase in pressure must be a higher order term

 

[kuruman]

That's certainly the safer way, but in the present case I would argue that:

1. ignoring the resulting increase in pressure, we get $4\pi R^2ap_0$ trivially
2. the extra work from the increase in pressure must be a higher order term

I have questions about this. Does the integral suggested by @Chestermiller in post #4, $$W=4\pi p_0\int_R^{R-a}\left(\frac{R}{r}\right)^{3\gamma}(r^2)dr$$not give the exact expression for the work? If it does, the leading term in the series expansion of $W$ when $\frac{R-a}{R}<<1$ is the trivial expression $4\pi R^2ap_0$. Are you saying that one should go beyond the leading term, which is linear in $a$, and add to it a quadratic term in $a$? There is nothing in the statement of the problem suggesting that and the conventional interpretation of $a<<R$ is "keep the leading term only."

 

[haruspex]

Are you saying that one should go beyond the leading term

No. The form of the question makes it clear that only the leading term is wanted. In post #8 I am saying we can find that leading term very easily.
It would have been clearer if I had ended the second bullet with "… so can be ignored".

 

[kuruman]

No. The form of the question makes it clear that only the leading term is wanted. In post #8 I am saying we can find that leading term very easily.
It would have been clearer if I had ended the second bullet with "… so can be ignored".

All is clear now. Thanks.