Question

Partial Pressure:

Part A:
Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 ∘C, the total pressure in the container is 4.10 atm . Calculate the partial pressure of each gas in the container.
Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.

Part B:
A gaseous mixture of O2 and N2 contains 36.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 805 mmHg ?
Express you answer numerically in millimeters of mercury.

Answer 1

Answer on Question #66236, Chemistry / General Chemistry

Partial Pressure:

Part A:

Three gases (8.00 g of methane, CH₄, 18.0 g of ethane, C₂H₆, and an unknown amount of propane, C₃H₈) were added to the same 10.0-L container. At 23.0 °C, the total pressure in the container is 4.10 atm. Calculate the partial pressure of each gas in the container. Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.

Solution:

We use the Mendeleev-Clapeyron equation

$$PV = nRT$$

Find the pressure

$$P = nRT / V$$ $$((8.00 \text{ g } CH_4) / (16.04 \text{ g } CH_4/mol)) \times (0.082 \text{ L atm/K mol}) \times (296 \text{ K}) / (10.0 \text{ L}) = 1.21 \text{ atm } CH_4$$ $$((18.00 \text{ g } C_2H_6) / (30.07 \text{ g } C_2H_6/mol)) \times (0.082 \text{ L atm/K mol}) \times (296 \text{ K}) / (10.0 \text{ L}) = 1.45 \text{ atm } C_2H_6$$ $$(4.10 \text{ atm}) - (1.21 \text{ atm}) - (1.45 \text{ atm}) = 1.44 \text{ atm } C_3H_8$$

**Answer:** 1.21 atm, 1.45 atm, 1.44 atm

Part B:

A gaseous mixture of O₂ and N₂ contains 36.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 805 mmHg? Express you answer numerically in millimeters of mercury.

Solution:

Then partial pressure O₂ = mole fraction O₂ × total pressure

Mass N₂ in 100 g = 36.8 % × 100 g = 36.8 g

Moles N₂ = mass / molar mass = 36.8 g / 28.02 g/mol = 1.31335 mol

Mass O₂ = 100 % - 36.8 % = 63.2 %

Mass O₂ in 100 g = 63.2 g

Moles O₂ = 63.2 g / 32.00 g/mol = 1.975 mol

Total moles gas = 1.31335 mol + 1.975 mol = 3.28835 mol

Mole fraction O₂ = 1.975 mol / 3.28835 mol

= 0.6006

Partial pressure O₂ = 0.6006 × 805 mmHg = 483 mmHg

**Answer:** 483 mmHg

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