Questions: After 0.600 L of Ar at 1.06 atm and 245.0°C is mixed with 0.200 L of O2 at 313 torr and 104.0°C in a 400 mL flask at 25.0°C, what is the pressure in the flask? Be sure your answer has the correct number of significant figures.

Solution Steps

• Convert the initial temperatures from Celsius to Kelvin using the formula \( T(K) = T(°C) + 273.15 \).
• For Ar: \( 245.0^{\circ} \mathrm{C} = 245.0 + 273.15 = 518.15 \, \mathrm{K} \).
• For \( \mathrm{O}_{2} \): \( 104.0^{\circ} \mathrm{C} = 104.0 + 273.15 = 377.15 \, \mathrm{K} \).
• Final temperature in the flask: \( 25.0^{\circ} \mathrm{C} = 25.0 + 273.15 = 298.15 \, \mathrm{K} \).

• Convert the pressure of \( \mathrm{O}_{2} \) from torr to atm using the conversion \( 1 \, \mathrm{atm} = 760 \, \mathrm{torr} \).
• \( 313 \, \mathrm{torr} = \frac{313}{760} \, \mathrm{atm} \approx 0.4118 \, \mathrm{atm} \).

• Use the ideal gas law \( PV = nRT \) to find the moles of each gas.
• For Ar: \( n = \frac{PV}{RT} = \frac{1.06 \, \mathrm{atm} \times 0.600 \, \mathrm{L}}{0.0821 \, \mathrm{L \cdot atm \cdot K^{-1} \cdot mol^{-1}} \times 518.15 \, \mathrm{K}} \).
• For \( \mathrm{O}_{2} \): \( n = \frac{PV}{RT} = \frac{0.4118 \, \mathrm{atm} \times 0.200 \, \mathrm{L}}{0.0821 \, \mathrm{L \cdot atm \cdot K^{-1} \cdot mol^{-1}} \times 377.15 \, \mathrm{K}} \).

• Add the moles of Ar and \( \mathrm{O}_{2} \) to find the total moles in the flask.

• Use the ideal gas law again to find the final pressure in the flask with the total moles.
• \( P = \frac{nRT}{V} \), where \( n \) is the total moles, \( R = 0.0821 \, \mathrm{L \cdot atm \cdot K^{-1} \cdot mol^{-1}} \), \( T = 298.15 \, \mathrm{K} \), and \( V = 0.400 \, \mathrm{L} \).

• Ensure the final pressure is reported with the correct number of significant figures based on the given data.

Final Answer