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?

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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?

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Answer 1 · 2017-07-07T04:09:14

Concept of partial pressure is to be applied here.

Let the partial pressure of $A r$ $Ar$ in the final mixture be $p_{Ar}$ $p_"Ar"$

Initially before mixing $A r$ $Ar$ had

Pressure $P_{Ar} = 1.29$ $P_"Ar"=1.29$ atm

Temperature $T_{Ar} = 223 + 273 = 500$ $T_"Ar"=223+273=500$ K

Volume $V_{Ar} = 0.600$ $V_"Ar"=0.600$ L

In the mixture it acquires

Pressure $p_{Ar} = ?$ $p_"Ar"=?$ atm

Temperature $T_{mAr} = 23 + 227 = 250$ $T_"mAr"=23+227=250$ K

Volume $V_{mAr} = 0.400$ $V_"mAr"=0.400$ L

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

$\frac{p_{Ar} \times V_{mAr}}{T_{mAr}} = \frac{P_{Ar} \times V_{Ar}}{T_{Ar}}$ $(p_"Ar"xxV_ "mAr")/T_"mAr"=(P_"Ar"xxV_"Ar")/T_"Ar"$

$\Rightarrow p_{Ar} = \frac{P_{Ar} \times V_{Ar}}{T_{Ar}} \times \frac{T_{mAr}}{V_{mAr}}$ $=>p_"Ar"=(P_"Ar"xxV_"Ar")/T_"Ar"xxT_ "mAr"/V_"mAr"$

$\Rightarrow p_{Ar} = \frac{1.29 \times 0.600}{500} \times \frac{250}{0.400} = 0.9675$ $=>p_"Ar"=(1.29xx0.600)/500xx250/0.400=0.9675$ atm

Again let the partial pressure of $O_{2}$ $O_2$ in the final mixture be $p_{O_{2}}$ $p_(O_2)$

Initially before mixing $O_{2}$ $O_2$ had

Pressure $P_{O_{2}} = 535$ $P_(O_2)=535$ torr $\approx 0.535$ $~~0.535$ atm

Temperature $T_{O_{2}} = 106 + 273 = 379$ $T_(O_2)=106+273=379$ K

Volume $V_{O_{2}} = 0.200$ $V_(O_2)=0.200$ L

In the mixture it acquires

Pressure $p_{O_{2}} = ?$ $p_(O_2)=?$ atm

Temperature $T_{m O_{2}} = 23 + 227 = 250$ $T_(mO_2)=23+227=250$ K

Volume $V_{m O_{2}} = 0.400$ $V_(mO_2)=0.400$ L

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

$\frac{p_{O_{2}} \times V_{m O_{2}}}{T_{m O_{2}}} = \frac{P_{O_{2}} \times V_{O_{2}}}{T_{O_{2}}}$ $(p_(O_2)xxV_ (mO_2))/T_(mO_2)=(P_(O_2)xxV_(O_2))/T_(O_2)$

$\Rightarrow p_{O_{2}} = \frac{P_{O_{2}} \times V_{O_{2}}}{T_{O_{2}}} \times \frac{T_{m O_{2}}}{V_{m O_{2}}}$ $=>p_(O_2)=(P_(O_2)xxV_(O_2))/T_(O_2)xxT_ (mO_2)/V_(mO_2)$

$\Rightarrow p_{O_{2}} = \frac{0.535 \times 0.200}{379} \times \frac{250}{0.400} = 0.1765$ $=>p_(O_2)=(0.535xx0.200)/379xx250/0.400=0.1765$ atm

So total pressure in the flask

$P = p_{Ar} + p_{O_{2}} = 0.9675 + 0.1765 = 1.144$ $P=p_"Ar"+p_(O_2)=0.9675+0.1765=1.144$ atm

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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?

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