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 1.01 105 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:

Given values

$P_{2} =1.01*10^{5} Pa$

h=0.2m

using ideal gas law keeping temperature and number of moles as constants

$P_{1} V_{1} =nRT-----(1)$

$P_{2} V_{2} =nRT------(2)$

comparing the equation first and second we get

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

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

The pressure at bottom of beer is sum of pressure at top and beer pressure given by

$P_{1} =P_{2} +pgh$

               =$1.01*10^{5} +1000(9.8)(0.2)$

               =102960

Thus equation third can be written as

$\frac{V_{2} }{V_{1} } =\frac{102960}{1.01*10^{5} }$

                                      =1.02

The ratio of bubbles volume is 1.02