a bubble having a diameter of 1.00 cm is released from the bottom of a swimming pool
Homework Answers
p = density *g *h + atmospheric pressure
p = 1x10^3 * 9.8 * 5 + 1 *10^5
p = 1.49*10^5
p at bottom diameter = 1 cm = .01 m = .005 m radius
p at top = 1 atm = 1*10^5 pascals
Volume = 4/3*pi*r^3
Volume =4/3*pi*(.005)^3
Volume = 5.24*10^-7
P1*V1/T1 = P2*V2/T2
T1 = 15 C = 288 K
T2 = 20 C = 293 K
V2 = P1*V1*T2/(T1*P2)
V2 = 1.49*10^5 * 5.24*10^-7 * 293 / (288 * 1*10^3)
V2 = 7.9*10-5
7.9*10^-5 = 4/3*pi*r^3
3*7.9*10^-5/(4*pi) = r^3
1.9*10^-5 = r^3
(1.9*10^5)^(1/3) = (r^3)^(1/3)
r = .0266 m = 2.66 cm
p = density *g *h + atmospheric pressure
p = 1x10^3 * 9.8 * 5 + 1 *10^5
p = 1.49*10^5
p at bottom diameter = 1 cm = .01 m = .005 m radius
p at top = 1 atm = 1*10^5 pascals
Volume = 4/3*pi*r^3
Volume =4/3*pi*(.005)^3
Volume = 5.24*10^-7
P1*V1/T1 = P2*V2/T2
T1 = 15 C = 288 K
T2 = 20 C = 293 K
V2 = P1*V1*T2/(T1*P2)
V2 = 1.49*10^5 * 5.24*10^-7 * 293 / (288 * 1*10^3)
V2 = 7.9*10-5
7.9*10^-5 = 4/3*pi*r^3
3*7.9*10^-5/(4*pi) = r^3
1.9*10^-5 = r^3
(1.9*10^5)^(1/3) = (r^3)^(1/3)
r = .0266 m = 2.66 cm
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