What is the total pressure exerted by a mixture of 1.50 g of H₂ and 5.00 g of N₂ in a 5.00-L vessel at 25°C?

Question

11-03-2024
Chemistry

High School

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What is the total pressure exerted by a mixture of 1.50 g of H₂ and 5.00 g of N₂ in a 5.00-L vessel at 25°C?

Answer :

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TaylorSmith474 TaylorSmith474 · Answer 1 · 2024-10-06T04:13:46-04:00

Final answer:

To calculate the total pressure of a mixture of H₂ and N₂ gases in a container, determine the number of moles of each gas, convert temperature to Kelvin, apply the Ideal Gas Law individually for each gas, and then sum their pressures to find the total pressure.

Explanation:

The question is asking for the total pressure exerted by a mixture of gases in a given container using the Ideal Gas Law. First, we must determine the number of moles (n) of each gas using their respective molar masses (H₂ = 2.02 g/mol and N₂ = 28.02 g/mol). For 1.50 g of H₂, n = 1.50 g / 2.02 g/mol = 0.7426 mol. For 5.00 g of N₂, n = 5.00 g / 28.02 g/mol = 0.1783 mol. We then apply the Ideal Gas Law formula: P = (nRT) / V, where P is pressure, n is moles of gas, R is the ideal gas constant (0.0821 L·atm/mol·K), T is temperature in Kelvin, and V is volume.

Converting temperature to Kelvin (25°C + 273.15 = 298.15 K), and applying the Ideal Gas Law separately for H₂ and N₂:
PH₂ = (0.7426 mol × 0.0821 L·atm/mol·K × 298.15 K) / 5.00 L = 3.638 atm and
PN₂ = (0.1783 mol × 0.0821 L·atm/mol·K × 298.15 K) / 5.00 L = 0.876 atm.

Finally, we calculate the total pressure by adding the pressures of the individual gases: Total Pressure = PH₂ + PN₂ = 3.638 atm + 0.876 atm = 4.514 atm.