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What is the total pressure exerted by a mixture of 1.50 g of [tex]$H_2$[/tex] and 5.00 g of [tex]$N_2$[/tex] in a 5.00 L vessel at [tex]$25^\circ C$[/tex]?

Answer :

Final answer:

To find the total pressure of a mixture of H₂ and N₂ in a vessel, calculate the moles of each gas, convert temperature to Kelvin, and use the ideal gas law to find the partial pressures, then add them together.

Explanation:

To calculate the total pressure exerted by a mixture of H₂ and N₂ gases in a 5.00-L vessel at 25°C, we can use the ideal gas law, 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, we need to calculate the number of moles of each gas. The molar mass of H₂ is approximately 2.02 g/mol, and for N₂ it is approximately 28.02 g/mol.

  • Number of moles of H₂: n = 1.50 g / 2.02 g/mol = 0.7426 mol
  • Number of moles of N₂: n = 5.00 g / 28.02 g/mol = 0.1784 mol

Next, we convert the temperature to Kelvin: T = 25 + 273 = 298 K.

The ideal gas constant, R, is 0.0821 L·atm/K·mol.

We then calculate the partial pressures for each gas using the ideal gas law. Remember to use the same units for the volume.

  • Partial pressure of H₂: PH₂ = (0.7426 mol) x (0.0821 L·atm/K·mol) x (298 K) / (5.00 L)
  • Partial pressure of N₂: PN₂ = (0.1784 mol) x (0.0821 L·atm/K·mol) x (298 K) / (5.00 L)

Finally, the total pressure is the sum of partial pressures of H₂ and N₂.

P₂total = PH₂ + PN₂

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