# 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?

Jul 7, 2017

Concept of partial pressure is to be applied here.

Let the partial pressure of $A r$ in the final mixture be ${p}_{\text{Ar}}$

Initially before mixing $A r$ had

Pressure ${P}_{\text{Ar}} = 1.29$ atm

Temperature${T}_{\text{Ar}} = 223 + 273 = 500$ K

Volume ${V}_{\text{Ar}} = 0.600$ L

In the mixture it acquires

Pressure p_"Ar"=? atm

Temperature ${T}_{\text{mAr}} = 23 + 227 = 250$ K

Volume ${V}_{\text{mAr}} = 0.400$ 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"

$\implies {p}_{\text{Ar"=(P_"Ar"xxV_"Ar")/T_"Ar"xxT_ "mAr"/V_"mAr}}$

$\implies {p}_{\text{Ar}} = \frac{1.29 \times 0.600}{500} \times \frac{250}{0.400} = 0.9675$ 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 $\approx 0.535$ 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 ${T}_{m {O}_{2}} = 23 + 227 = 250$ K

Volume ${V}_{m {O}_{2}} = 0.400$ L

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

$\frac{{p}_{{O}_{2}} \times {V}_{m {O}_{2}}}{T} _ \left(m {O}_{2}\right) = \frac{{P}_{{O}_{2}} \times {V}_{{O}_{2}}}{T} _ \left({O}_{2}\right)$

$\implies {p}_{{O}_{2}} = \frac{{P}_{{O}_{2}} \times {V}_{{O}_{2}}}{T} _ \left({O}_{2}\right) \times {T}_{m {O}_{2}} / {V}_{m {O}_{2}}$

$\implies {p}_{{O}_{2}} = \frac{0.535 \times 0.200}{379} \times \frac{250}{0.400} = 0.1765$ atm

So total pressure in the flask

$P = {p}_{\text{Ar}} + {p}_{{O}_{2}} = 0.9675 + 0.1765 = 1.144$ atm