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What is the total pressure exerted by a mixture of 1.50 g [tex]$H_2$[/tex] and 5.00 g [tex]$N_2$[/tex] in a 5.00-L vessel at 25 degrees Celsius?

Answer :

Final answer:

To find the total pressure exerted by a mixture of 1.50 g H2 and 5.00 g N2 in a 5.00-L vessel at 25 degrees Celsius, we convert the mass of each gas to moles, and use the ideal gas law to calculate the total pressure, resulting in 4.57 atm.

Explanation:

To determine the total pressure exerted by a mixture of 1.50 g H2 and 5.00 g N2 in a 5.00-L vessel at 25 degrees Celsius, we can use the ideal gas law, which states that the pressure of a gas is directly proportional to the number of moles of the gas and the temperature, and inversely proportional to the volume. First, we need to convert the mass of each gas to moles using their molar masses (H2 = 2 g/mol, N2 = 28 g/mol), and then use the ideal gas law formula: P(total) = (n(total)RT)/V.

Calculating the moles: n(H2) = 1.50 g / 2 g/mol = 0.75 mol, n(N2) = 5.00 g / 28 g/mol = 0.179 mol. Thus, n(total) = 0.75 mol + 0.179 mol = 0.929 mol.

Using the ideal gas law, R = 0.0821 L·atm/K·mol, T = 25 °C + 273.15 = 298.15 K, and V = 5.00 L, we can find the total pressure: P(total) = (0.929 mol × 0.0821 L·atm/K·mol × 298.15 K) / 5.00 L = 4.57 atm.

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