Question

A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01x10⁵ Pa. Assume that the density of beer is the same as that of fresh water. If the temperature and number of moles of CO₂ in the bubble remain constant as the bubble rises, find the ratio of the bubble's volume at the top to its volume at the bottom.

Answer 1

Answer:

$\frac{1.019}{1}$

Explanation:

To solve this equation we will have to consider that the bubble is filled with an Ideal Gas and as such we can use the Ideal Gas Law

$PV = nRT$

Where

$P$ = Pressure

$V$ = Volume

$n$ = Moles

$R$ = Ideal Gas Constant

$T$  = Temperature

Now since we know that the value for the temperature and moles is constant we can simply use Boyles Law for the two states

$P_{1} V_{1} =P_{2} V_{2}$

Let us look at the two states

State 1 (at top)

Pressure = $1.01*10^5$

Volume = $V_{1}$

State 2 (at bottom)

Pressure = $1.01*10^5 + dgh$

Where

$d$ = Density of liquid (1000 kg/m³)

$d$ = Acceleration due to gravity (9.8 m/s²)

$d$ = Height of liquid (0.200 m)

Pressure = $102,962$

Volume = $V_{2}$

Inputting these values into the Boyles Law

$P_{1} V_{1} =P_{2} V_{2}\\ (101000)V_{1} = (102962)V_{2}\\ \frac{V_{1}}{V_{2}} = \frac{102962}{101000} \\ \frac{V_{1}}{V_{2}} = \frac{1.019}{1}$