Question

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

Answer #1

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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