GATE 2020 BT Q.22: solution and answer

GATE 2020 · BTQ.22 · 1 mark · NAT

To facilitate mass transfer from a gas to a liquid phase, a gas bubble of radius $r$ is introduced into the liquid. The gas bubble then breaks into 8 bubbles of equal radius. Upon this change, the ratio of the interfacial surface area to the gas phase volume for the system changes from $3/ r$ to $3 n / r$ . The value of $n$ is ________.

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Answer and solution

ANSWER

$n = 2$ . The interfacial area per unit gas volume doubles, from $3/ r$ to $6/ r$ .

Official answer: 2Source: Official answer key, GATE 2020 BT (IIT Delhi)

For one spherical bubble of radius $r$ , the interfacial area is $A = 4 π r^{2}$ and the gas volume is $V = \frac{4}{3} π r^{3}$ , so $\frac{A}{V} = \frac{4 π r ^{2}}{\frac{4}{3} π r ^{3}} = \frac{3}{r}$ . This matches the starting value given in the question.
When the bubble breaks up, the amount of gas stays the same, so the total gas volume is unchanged. If each of the 8 new bubbles has radius $r^{'}$ : $8 \cdot \frac{4}{3} π r^{'3} = \frac{4}{3} π r^{3} \Rightarrow r^{'3} = \frac{r ^{3}}{8} \Rightarrow r^{'} = \frac{r}{2}$ .
The total interfacial area of the 8 bubbles is $A^{'} = 8 \cdot 4 π (\frac{r}{2})^{2} = 8 π r^{2}$ . That is twice the original area of $4 π r^{2}$ .
The new ratio is $\frac{A ^{'}}{V} = \frac{8 π r ^{2}}{\frac{4}{3} π r ^{3}} = \frac{6}{r}$ . You get the same result from $\frac{3}{r ^{'}}$ with $r^{'} = \frac{r}{2}$ , because equal spheres have the same area-to-volume ratio taken together as each one has alone.
Setting $\frac{3 n}{r} = \frac{6}{r}$ gives $n = 2$ .

KEY IDEA

The gas volume stays the same when the bubble breaks up. Splitting one sphere into $N$ equal spheres makes each radius $r / N^{1/3}$ . Since the area-to-volume ratio of a sphere is $3/ radius$ , it rises by a factor of $N^{1/3}$ . Here $N = 8$ , so the factor is $8^{1/3} = 2$ . This is why smaller bubbles give a larger specific interfacial area, $a$ , and so a larger $k_{L} a$ in a bioreactor.

WATCH OUT FOR

One mistake is to assume the radius stays at $r$ and multiply the area by 8, which gives $n = 8$ . Another is to divide the radius by 8 instead of by $8^{1/3} = 2$ , which also gives $n = 8$ . The gas volume is fixed, so the new radius is $r /2$ and the ratio only doubles.

From the GATE 2020 BT paper. Every question of this paper · Download the full paper · All 24 Bioreactor operation questions, with notes

GATE 2020 BT Q.22: solution and answer

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