Question

7. At what temperature would 343.01 g of P2O3 gas occupy 67.5 L if the pressure is 115 kPa?

Answer 1

Answer:

26.1°C

Explanation:

T = PV / nR = (115 kPa) x (67.5 L) / ((343.01 g P2O3 / (109.94572 g P2O3/mol)) x (8.3144626 L kPa/K mol)) = 299.3 K = 26.1°C

Answer 2

The temperature at which 343.01 g of P₂O₃ gas will occupy 67.5 L at 115 KPa is 299.4 K

How to determine the mole of P₂O₃

• Mass of P₂O₃ = 343.01 g
• Molar mass of P₂O₃ = (2×31) + (3×16) = 110 g/mol
• Mole of P₂O₃ =?

Mole = mass / molar mass

Mole of P₂O₃ = 343.01 / 110

Mole of P₂O₃ = 3.118 moles

How to determine the temperature

• Volume (V) = 67.5 L
• Number of mole (n) = 3.118 moles  
• Pressure (P) = 115 KPa
• Gas constant (R) = 8.314 LKPa/Kmol

• Temperature (T) =?

Using the ideal gas equation, the temperature of the gas can be obtained as follow:

PV = nRT

Divide both side by nR

T = PV / nR

T = (115 × 67.5) / (3.118 × 8.314)

T = 299.4 K

Thus, the temperature of the gas is 299.4 K

Learn more about ideal gas equation:

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