(a) 2.06 (4.20) A methane bubble 1.00 cm in diameter is released from the bottom

Question

(a) 2.06

(4.20) A methane bubble 1.00 cm in diameter is released from the bottom of a lake 5.00 m deep. The temperature at the bottom of the lake is 4 degreeC. (a) How many methane molecules are in the bubble? (b) What is the mass of the methane in the bubble? (If you don't know the chemical formula for methane, look it up.) (c) The bubble rises to the surface of the lake, where the temperature is 15 degreeC. (Assume that no gas diffuses in or out of the bubble as it rises.) What is the diameter of the bubble just before it breaks the surface? (a) 2.06 ?? 10 (b) 5.49 ?? 10 kg (c) 1.16 cm

Explanation / Answer

Part A)

Apply PV = nRT

P = 1.013 X 105 + pgh

P = (1.013 X 105) + (1000)(9.8)(5) = 150300

(150300)(4/3)(pi)(.005)3 = n(8.31)(277)

n = 3.42 X 10-5 moles

Multiply by Avogadros number (6.022 X 1023) and get 2.06 X 1019 molecules

Part B)

The molar mass of methane is 16.04 g/mole

So the mass is 16.04(3.42 X 10-5) = 5.49 X 10-4 g which is 5.49 X 10-7 kg

Part C)

PV/T = PV/T

(150300)(4/3)(pi)(.005)3/277 = (1.013 X 105)(4/3)(pi)(r3)/(288)

r = .00578 m

d = 2r = .0116 m which is 1.16 cm

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