Calculating Work Done by Expanding Gas in Spherical Bubble

[AdityaDev]

Homework Statement

(1)The gas inside a spherical bubble expands slowly so that its radius is increases from R to 2R. Atmospheric pressure is $P_0$ and surface tension is S. The work done by the gas is ___________

Homework Equations

work done against surface tension $W_t = S.\Delta A$ where A is the surface area and S is the surface tension.

The Attempt at a Solution

initial area =$(4\pi R^2)$
final area = $(4\pi(4R^2-R^2))$
so W against S is $S.8\pi R^2$
but the answer given is $\frac{25\pi P_0R^3}{3} + 24\pi S R^2$
where did the first term containing $P_0$ come from and why is the second term $24\pi SR^2$

 

[Chestermiller]

Regarding the second term, your algebra is incorrect, your equation for the final area is incorrect (your second equation gives for the change in area), and the surface tension acts at both surfaces of the bubble (so you're missing a factor of 2). With regard to the first term, you omitted the PdV work to push back the atmosphere.

Chet

 

[AdityaDev]

With regard to the first term, you omitted the PdV work to push back the atmosphere.

$P_{int}$ is not a constant right?(as V becomes V+dV, P becomes P-dP)
so if I consider the process to be isothermal, I will get
$W_2=nRTlog(\frac{V_f}{V_i})$.
but there is no log term in the answer.[/QUOTE]

 

[Chestermiller]

$P_{int}$ is not a constant right?(as V becomes V+dV, P becomes P-dP)
so if I consider the process to be isothermal, I will get
$W_2=nRTlog(\frac{V_f}{V_i})$.
but there is no log term in the answer.

No this is not the correct equation to use. If assumes that the gas expansion occurs isothermally and reversibly. If the initial pressure inside the bubble is high enough to cause the bubble to spontaneously expand to the final equilibrium volume, then the expansion is irreversible. If the gas is at equilibrium initially, then to increase its volume you need to increase the temperature or the number of moles. In all these cases, the work done by the gas is P₀ΔV + 2SΔA.

Chet

 

[AdityaDev]

No this is not the correct equation to use. If assumes that the gas expansion occurs isothermally and reversibly. If the initial pressure inside the bubble is high enough to cause the bubble to spontaneously expand to the final equilibrium volume, then the expansion is irreversible. If the gas is at equilibrium initially, then to increase its volume you need to increase the temperature or the number of moles. In all these cases, the work done by the gas is P₀ΔV + 2SΔA.

Chet

understood. if an air bubble expands in a liquid then the expression will be $P_0\Delta V + S\Delta A$ right?

 

[Chestermiller]

understood. if an air bubble expands in a liquid then the expression will be $P_0\Delta V + S\Delta A$ right?

Not exactly. The problem statement and answer book solution implies that the bubble is not in a liquid. It is a bubble in a gas (think balloon). The gas pressure on the inside surface of the bubble is given by:
$$P=\frac{4S}{r}+P_0$$
The volume of the bubble is:
$$V=\frac{4}{3}\pi r^3$$
So that:
$$dV=4\pi r^2dr$$
So, the differential work is given by:
$$PdV=(16\pi S r+4\pi r^2)dr$$
(I guess the 25 in the answer book solution should really be a 28.)

So the work is $P_0\Delta V + 2S\Delta A$

The factor of 2 in front of the S is because a bubble in a gas has an inside surface and an outside surface.

Chet

 

[AdityaDev]

I

Not exactly. The problem statement and answer book solution implies that the bubble is not in a liquid. It is a bubble in a gas (think balloon). The gas pressure on the inside surface of the bubble is given by:
$$P=\frac{4S}{r}+P_0$$
The volume of the bubble is:
$$V=\frac{4}{3}\pi r^3$$
So that:
$$dV=4\pi r^2dr$$
So, the differential work is given by:
$$PdV=(16\pi S r+4\pi r^2)dr$$
(I guess the 25 in the answer book solution should really be a 28.)

So the work is $P_0\Delta V + 2S\Delta A$

The factor of 2 in front of the S is because a bubble in a gas has an inside surface and an outside surface.

Chet

Thank you sir. The answer given must be wrong.