
#1
Apr2708, 07:51 PM

P: 13

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 X 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 remain constant as the bubble rises, find the ratio of its volume at the top to that at the bottom.
What I did was use the equation [Pressure at bottom = Pressure at top + P("rho")*g*h]. Since n and T are constant, I know: (PV)top = (PV)bottom. I need one of the volume values to solve for the other (V at top or V at bottom). Is the volume at the top 0.200m^3? Thanks. 



#2
Apr2708, 09:03 PM

P: 112

You just need the ratio, not the true value for the volumes. So just solve Boyle's law for the ratio of the pressures. . .
[tex] P_1 V_1 = P_2 V_2 [/tex] 


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