Ideal Mixed Gas Law problems
1. What is the pressure generated by a mixture of 25.6 g SO2 and 12.6 g Ne in a container of 72.45 L and temperature of 131 K?
2. What is the pressure of a mixture of 5.80 moles Kr and 2.71 moles Xe in a container with a volume of 205.6 Liters and temperature of 288 K?
3. What is the pressure of a mixture of 528.5 g Br2 and 52.8 g NO in a container of 101.3 L and temperature of 214.2 oC?
4. What is the pressure of a sample of gasses in a 491.8 Liter container at a temperature of -193 oC if it consists of a mixture of 5.9 moles H2S, 8.2 moles CH4, 4.8 moles CO2, and 6.3 moles H2?
5. What is the pressure of a gas mixture made when mixing 0.828 moles of Ne and 0.491 moles Ar in a container of 37.2 L at a temperature of 50oC?
6. What is the gas pressure of a mixture of 0.593 moles O2, 0.824 moles N2, 1.03 moles He, and 0.596 moles Xe in a container of 41.7 L and 1,049 K?
7. What is the pressure of a mixture of 194.3 g F2 and 215.8 g Ne if they occupy 12.3 L at 300 K?
8. An unknown pressure of a gas mixture is resting at 271 K in a volume of 121.9 Liters. What is the pressure if there is 14.14 moles of H2, 12.12 moles of O2, and 10.10 moles of Kr?
To solve these Ideal Gas Law problems, we can use the equation:
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.
Let's go through each problem step by step:
1. For this problem, we are given the mass of SO2 (25.6 g) and Ne (12.6 g), volume (72.45 L), and temperature (131 K).
To find the number of moles (n), we need to divide the mass of each gas by their respective molar masses. The molar mass of SO2 is 64.06 g/mol, and the molar mass of Ne is 20.18 g/mol.
n(SO2) = 25.6 g / 64.06 g/mol
n(Ne) = 12.6 g / 20.18 g/mol
Now that we have the number of moles, we can substitute the values into the Ideal Gas Law equation:
P * 72.45 L = (n(SO2) + n(Ne)) * 0.0821 L*atm/(K*mol) * 131 K
Solve for P:
P = [(n(SO2) + n(Ne)) * 0.0821 * 131 K] / 72.45 L
Plug in the values:
P = [(25.6 g / 64.06 g/mol + 12.6 g / 20.18 g/mol) * 0.0821 * 131 K] / 72.45 L
P ≈ 0.867 atm
2. In this problem, we are given the number of moles of Kr (5.80 moles) and Xe (2.71 moles), volume (205.6 Liters), and temperature (288 K).
We can use the same steps as in problem 1:
P * 205.6 L = (5.80 moles + 2.71 moles) * 0.0821 L*atm/(K*mol) * 288 K
Solve for P:
P = [(5.80 moles + 2.71 moles) * 0.0821 * 288 K] / 205.6 L
Plug in the values:
P = [(5.80 moles + 2.71 moles) * 0.0821 * 288 K] / 205.6 L
P ≈ 1.648 atm
3. For this problem, we are given the mass of Br2 (528.5 g) and NO (52.8 g), volume (101.3 L), and temperature (214.2 °C).
First, we need to convert the temperature from Celsius to Kelvin:
T = 214.2 °C + 273.15
Next, find the number of moles:
n(Br2) = 528.5 g / molar mass of Br2
n(NO) = 52.8 g / molar mass of NO
Now, substitute the values into the Ideal Gas Law equation:
P * 101.3 L = (n(Br2) + n(NO)) * 0.0821 L*atm/(K*mol) * T
Solve for P:
P = [(n(Br2) + n(NO)) * 0.0821 * T] / 101.3 L
Plug in the values:
P = [(528.5 g / molar mass of Br2 + 52.8 g / molar mass of NO) * 0.0821 * T] / 101.3 L
P ≈ 1.472 atm
Continue below for the remaining problems.