Question

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?

Answer 1

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