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Mixtures of O_2 gas and He gas are often used as breathing mixtures for scuba di

ID: 537672 • Letter: M

Question

Mixtures of O_2 gas and He gas are often used as breathing mixtures for scuba diving in deep dives as a way to avoid the bends. For a particular dive, 46 L of O_2 at 25 degree C and 1.0 atm and 12 L of the at 25 degree C and 1.0 atm were both pumped into a 5.0 L scuba tank. Calculate the partial pressure of each gas and the total pressure at 25 degree C. A sample of solid KClO_3 is heated in the apparatus below and decomposed according to the reaction: 2 KClO_3 rightarrow KCl + 3 O_2 The oxygen thus produced is collected by displacement of water at 22 degree C. The resulting mixture ofo2 gas and water vapor has a pressure of 754 torr and a volume of 0.65 L. Calculate the partial pressure of and the number of moles of O_2 collected. The partial pressure of water vapor at 22 degree C is 21 torr.

Explanation / Answer

moles of gas can be calculated from PV=nRT

n = no of moles of gas

given for oxygen, V= 46L, T=25+273=298K, P= 1atm, R=0.0821 L.atm/mole.K, n= PV/RT= 46*1/(0.0821*298) =1.89 moles

for Helium, P= 1atm, V= 12L, T=298K, n= PV/RT= 12*1/(0.0821*298)= 0.49 moles

total moles after mixing= 1.89+0.49= 2.38 moles, V=5L, T=298K

P= total pressure= nRT/V= 2.38*0.0821*298/5 = 11.64 atm

partial pressure= mole fraction* total pressure

mole fraction = moles of the gas/total moles

moles of gas : O2=1.89, He= 0.49 , total moles= 2.38

Mole fractions O2: 1.89/2.38 = 0.79, He=1-0.79=0.21

partial pressure= mole fractions* total presure

Partial pressures : O2=0.79*11.64=9.2, He=11.64-9.20=2.44 atm

2.

Vapor pressure of water at 22deg.c= 21 Torr,

Partial pressure of Oxygen= total pressure- saturation vapor pressure= 754-21= 733 Torr

Since 760 torr= 1atm, 733 Torr= 733/760 atm =0.964 atm, V=0.45L, T=22+273=295K

Partial pressure of gas is the pressure exerted by the gas if the gas alone occupies the entire volume. Hence PO2=0.964 atm, V=0.65L, T=295K, R=0.0821 L.atm/mole.K

No of moles, n= PV/RT= 0.964*0.65/(0.0821*295)= 0.026 moles