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A sample of air has 3.55 moles of N₂ and 2.50 moles of O₂. What is the total pressure for the sample of air described, if the volume is 29.51 L and the temperature is 300 K?

a. 1.01 atm
b. 5.06 atm
c. 4.25 atm
d. 3.13 atm

1 Answer

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Final answer:

The total pressure of a sample of air containing 3.55 moles of N₂ and 2.50 moles of O₂ in a 29.51 L container at 300 K is calculated using the Ideal Gas Law. The total pressure is found to be approximately 5.06 atm.

Step-by-step explanation:

The student has asked how to calculate the total pressure of a sample of air containing 3.55 moles of N₂ and 2.50 moles of O₂ at a volume of 29.51 L and temperature of 300 K. To find the total pressure of the gas mixture, we can use the Ideal Gas Law which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, calculate the total number of moles (ntotal) present in the gas mixture:

  • ntotal = moles of N₂ + moles of O₂
  • ntotal = 3.55 + 2.50
  • ntotal = 6.05 moles

Now use the Ideal Gas Law to find the pressure:

  • PV = nRT
  • P = (nRT) / V
  • P = (6.05 moles * 0.0821 L·atm/(mol·K) * 300 K) / 29.51 L
  • P ≈ 5.06 atm

Therefore, the total pressure of the gas mixture is approximately 5.06 atm, which corresponds to option (b).

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