Consider a chamber filled with gaseous N2 (molar mass 28.0 g/mol) and O2 (molar

Question

Consider a chamber filled with gaseous N2 (molar mass 28.0 g/mol) and O2 (molar mass 32.0 g/mol) which behave ideally (cv = (3/2)R). The chamber has a volume of 1.00 L and is initially at a temperature T = 25.0 OC ; the total initial pressure of all

gases is ptot,i = 760. torr. The initial partial pressure of oxygen is pO2,i = 593. torr. A vacuum pump attached to the chamber is turned on. The total pressure inside the chamber is reduced to 35 mtorr (i.e. milli-torr) and the vacuum pump is turned off. You may assume that the temperature does not change as the pressure is reduced. What is the final partial pressure of O2: pO2,f?

Explanation / Answer

This is the case of isothermal process

moles of gas initially, n =

PV/RT

P= 760 Torr =760/760 atm, V =1 L, R= 0.0821 L.atm/mole.K T= 25 deg.c =25+273.15= 298.15

n= PV/RT = 1*1/ (0.0821*298.15)= 0.041 moles

Partial pressure of oxygen= 593 Torr, partial pressure of nitrogen = 760-593= 167 mm Hg

moles of n2 = (593/760)*0.041 =0.032 moles, moles of oxygen =0.041-0.032 =0.009 moles

during the expansion, volume increases and number of moles rermain the same. Moles of gas will not change

Parital pressure of oxygen = (0.032/0.041)*35*10-3 Torr =27*10-3 Torr, partial pressure of nitrogen = 35*10-3-27*10-3= 8*10-3 Torr

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