Answer :
Final answer:
The total pressure exerted by the mixture of 1.50 g of H2 and 5.00 g of N2 in a 5.00-L vessel at 25°C is 4.60 atm, calculated using the Ideal Gas Law.
Explanation:
The total pressure exerted by a mixture of gases can be calculated using the Ideal Gas Law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, 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. For hydrogen (H2), its molar mass is approximately 2 g/mol. For nitrogen (N2), the molar mass is approximately 28 g/mol.
Therefore, we have:
n(H2) = 1.50 g / 2 g/mol
= 0.75 mol of H2
n(N2) = 5.00 g / 28 g/mol
= 0.179 mol of N2
Next, we use the Ideal Gas Law to calculate the total pressure.
Assuming a standard temperature of 25℃ which is 298K (25 + 273), and using the ideal gas constant R = 0.0821 L·atm/mol·K, we get:
P = (n(H2) + n(N2)) × R × T / V
P = (0.75 mol + 0.179 mol) × 0.0821 L·atm/mol·K × 298 K / 5.00 L
P = (0.929 mol) × (0.0821 L·atm/mol·K) × (298 K) / (5.00 L)
P = 4.60 atm
