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 ???? 10 5 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:

the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019

Explanation:

given information

h = 0.2 m

$P_{0}$ = 1.01  x $10^{5}$ Pa

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

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

$P_{1}$  = $P_{0}$  + ρgh, ρ = 1000 kg/$m^{3}$

$P_{1}$  = 1.01 x $10^{5}$ Pa + (1000 x 9.8 x 0.2) = 1,0296 x $10^{5}$ Pa

$P_{2}$  = $P_{0}$  = $10^{5}$ Pa

thus,

$\frac{V_{2} }{V_{1}} = 1,0296 x [tex]10^{5}$/$10^{5}$ = 1.019