Suppose that Daniel has a 2.00 L bottle that contains a mixture of O2, N2, and CO2 under a total pressure of 5.00 atm. He knows that the mixture contains 0.29 mol N2 and that the partial pressure of CO2 is 0.350 atm. If the temperature is 273 K, what is the partial pressure of O2? answer in atm
given
volume = 2.00 L
total pressure = 5.00 atm
mixture of O2, N2, and CO2
moles of N2 = 0.29 moles
partial pressure of CO2 = 0.350 atm
temperature = 273 K
we know that
PV = nRT
P is the pressure
V is the volume
n is the no of moles
R is the universal gas constant = 0.08206 L atm/mol K
T is the temperature
so
lets find total no of moles
PV = nRT
n = PV/RT
= (5.00 atm * 2.00 L)/( 0.08206 L atm/mol K * 273 K)
= 0.446 moles
Now
no of moles of CO2
n = PV/RT
= (0.350 atm * 2.00 L)/( 0.08206 L atm/mol K * 273 K)
= 0.0312 moles CO2
total no of moles = moles of O2 + moles of N2 + moles of CO2
0.446 moles = moles of O2 + 0.29 moles + 0.0312 moles
moles of O2 = 0.125 moles
now
partial pressure of O2 = (no of moles of O2)*R*T/V
= (0.125 moles*0.08206 L atm/mol K * 273 K)/ 2.00 L
= 1.40 atm
partial pressure of O2 = 1.40 atm
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