After 0.600 L of Ar at 1.29 atm and 223 degrees Celsius is mixed with 0.200 L of O2 at 535 torr and 106 degrees Celsius in a 400.-mL flask at 23 degrees Celsius, what is the pressure in the flask?

1 Answer
Jul 7, 2017

Concept of partial pressure is to be applied here.

Let the partial pressure of #Ar# in the final mixture be #p_"Ar"#

Initially before mixing # Ar# had

Pressure #P_"Ar"=1.29# atm

Temperature#T_"Ar"=223+273=500# K

Volume #V_"Ar"=0.600# L

In the mixture it acquires

Pressure #p_"Ar"=?# atm

Temperature #T_"mAr"=23+227=250# K

Volume #V_"mAr"=0.400# L

So by combined Boyle's and Charles's equation for ideal gas we can write

#(p_"Ar"xxV_ "mAr")/T_"mAr"=(P_"Ar"xxV_"Ar")/T_"Ar"#

#=>p_"Ar"=(P_"Ar"xxV_"Ar")/T_"Ar"xxT_ "mAr"/V_"mAr"#

#=>p_"Ar"=(1.29xx0.600)/500xx250/0.400=0.9675# atm

Again let the partial pressure of #O_2# in the final mixture be #p_(O_2)#

Initially before mixing # O_2# had

Pressure #P_(O_2)=535# torr #~~0.535# atm

Temperature#T_(O_2)=106+273=379# K

Volume #V_(O_2)=0.200# L

In the mixture it acquires

Pressure #p_(O_2)=?# atm

Temperature #T_(mO_2)=23+227=250# K

Volume #V_(mO_2)=0.400# L

So by combined Boyle's and Charles's equation for ideal gas we can write

#(p_(O_2)xxV_ (mO_2))/T_(mO_2)=(P_(O_2)xxV_(O_2))/T_(O_2)#

#=>p_(O_2)=(P_(O_2)xxV_(O_2))/T_(O_2)xxT_ (mO_2)/V_(mO_2)#

#=>p_(O_2)=(0.535xx0.200)/379xx250/0.400=0.1765# atm

So total pressure in the flask

#P=p_"Ar"+p_(O_2)=0.9675+0.1765=1.144# atm