Question

Two identical bubbles of gas form at the bottom of a lake, then rise to the surface. Because the pressure is much lower at the surface than at the bottom, both bubbles expand as they rise. However, bubble A rises very quickly, so that no heat is exchanged between it and the water. Meanwhile, bubble B rises slowly (impeded by a tangle of seaweed), so that it always remains in thermal equilibrium with the water (which has the same temperature everywhere). Which of the two bubbles is larger by the time they reach the surface? Explain your reasoning fully.

Short answer

Size of bubble B is larger than size of bubble A.

Step-by-step solution

Given information

Two bubbles of gas are identical in nature and pressure is much lower at the surface than at the bottom. Bubble Arises very quickly and bubble B rises slowly.

Explanation

The relationship between pressure P and Volume V can be given as

$P{V}^{\gamma }=\mathrm{const}............................\left(1\right)$

Where

P= pressure, V=volume and γ = adiabatic constant.

We can write Ideal gas equation in terms of Boltzmann constant as

P V=N k T ................................(2)

Where

P= pressure, V=volume , k= Boltzmann constant and T = temp

As given bubble A, there is no heat exchange by the system i.e. bubble experiences adiabatic expansion. So bubble A obeys equation (1)

But bubble B expands isothermally since it moves slowly which means the temperature difference is zero i.e.ΔT=0. So, bubble B, obeys equation (2).

P V = N k T= Constant .......................................(3)

Equation (1) and (2) can be written for bubble A and B respectively as below

$$\begin{gathered} P_f{V}_f^{\gamma }=P_i{V}_i^{\gamma }\left(\text{for}A\right) \\ {P}_f{V}_f=P_i{V}_i\left(\text{for}B\right) \end{gathered}$$

As initial volume V₀ and pressure P₁ is same for both the two bubbles so from above equation we can get final volume for both the bubbles as below

$$\begin{gathered} V_A^{\gamma }=\left(\frac{P_o}{P_1}\right)V_o^{\gamma }\left(\text{for}A\right) \\ {V}_B=\left(\frac{P_o}{P_1}\right)V_o\left(\text{for}B\right) \end{gathered}$$

Where V_A and V_B are the final volumes of bubbles.

$\left.\begin{array}{l}{V}_A={\left(\frac{P_o}{P_1}\right)}^{\left(\frac{1}{y}\right)}{V}_o\left(\text{for}A\right)\\ V_B=\left(\frac{P_o}{P_1}\right)V_o\left(\text{for}B\right)\end{array}\right\}...........................\left(4\right)$

As adiabatic constant $\gamma =\frac{f+2}{f}$is greater than 1 so the fraction $\frac{1}{\gamma }$is less than 1 , therefore from equation (6) can be written as

${\left(\frac{P_o}{P_1}\right)}^{\left(\frac{1}{\gamma }\right)}<\left(\frac{P_o}{P_1}\right)................................\left(5\right)$

In order to simplify and solve multiply both side of equation (5) with initial volume V₀, we get

${\left(\frac{P_o}{P_1}\right)}^{\left(\frac{1}{y}\right)}{V}_o<\left(\frac{P_o}{P_1}\right)V_o.............................\left(6\right)$

Substitute ${\left(\frac{P_o}{P_1}\right)}^{\left(\frac{1}{r}\right)}{V}_o=V_{A}and\left(\frac{P_o}{P_1}\right)V_o=V_B$

we get

V_A < V_B

So the final size of bubble b will be larger.

We can conclude that the bubble that rises very slowly(isothermally) will be larger than bubble rises quickly (adiabatic) when they reach the surface.

Bubble B expands isothermally this means that the temperature is constant throughout, so the received amount of heat is converted to work. It takes more time to reach the surface, more work will be done.

For bubble A, the heat exchange is zero so the thermal energy due to the temperature difference is equal to work done on the bubble and temperature of the lake is roughly same,

This work is less than the work in isothermal process.