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Back to the Ocean! d) went scuba diving again, and started wondering about the pressures that were affecting me. I was not paying attention and ended up going too deep. I had to make three decompression stops upon resurfacing and started thinking chemistry! At decompression stop 1, I noticed the bubbles were quite small as l exhaled them. My external pressure gauge read 1400 mmHg. My external temperature gauge read 50 F. What was the volume of the bubble at this depth I wondered? What about the volume at the surface, where the external pressure was 1 atm and the external temperature was 65 F? I remembered the ideal gas equation, but I didn't know how much air I was exhaling. Calculate the volume of the air bubble exhaled at decompression stop 1 and the surfacefor right below to be a bubbleassuming I exhaled 1 mole of gases. Show work. Surface e But is that right to assume 1 mole??? Should we ever assume? My tank was filled with a mixture of oxygen and nitrogen at a ratio of 21% 02 to 78 % N2. The total pressure on the tank gauge reads 15 atm on my 25 L tank at the surface. a) What is the partial pressure of oxygen at the filling of the tank? Show work. b] Now, calculate the number of moles of oxygen in my tankfat the surface) from this information? Show work.

Explanation / Answer

Pressure of the air at depth p = 1400 mmHg = 1400/760 = 1.8421 atm

Temperature at that depth = 50 0F

convert degree farhenheit to celcius then T in 0C = (T 0F -32) *5/9 = (50-32)*5/9 = 10 0C = 10+273.15 = 283.15 K

number of moles of the gas = 1 mole

gas constant R = 0.0821 l atm /mol K

according to ideal gas equation PV = nRT

1.8421 * V = 1 * 0.0821 * 283.15

V = volume of the air in depth of ocean = 12.62 L

Pressure of the air at surface p = 1 atm

Temperature T = 65 0F

convert degree farhenheit to celcius then T in 0C = (T 0F -32) *5/9 = (65-32)*5/9 = 18.33 0C = 18.33+273.15 = 291.48 K

number of moles of the gas = 1 mole

gas constant R = 0.0821 l atm /mol K

according to ideal gas equation PV = nRT

1 * V = 1 * 0.0821 * 291.48

V = volume of air at surface = 23.931 L

e)

partial pressure of gas = total pressure * mole fraction of the gas

PO2 = Pt * XO2

PO2 = 15 *(21/(21+78))

PO2 = 3.182 atm

number of moles of O2 = PV/RT from ideal gas equation

n = 3.182 * 25/(0.0821 * 291.48) (since temperature is obtained from the previous problem)

n = 3.3242 moles of O2 is present in that mixture

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