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?

Md Ariful Islam

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Md Ariful Islam

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Concept of 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

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