A sample containing 4.8g of O2 gas has a volume of 15.0L. Pressure and temperature remain constant.
What is the new volume if 0.500 mole of O2 gas is added?
Convert 4.8 g O2 to mols, add 0.5 mol to that and use (n1/V1) = (n2/V2)
11. A sample containing 4.80 g of O2 gas has an initial volume of 15.0 L. What is the final volume, in liters, when each of the following changes occur in the quantity of the gas at constant pressure and temperature?
To solve this problem, you can use the ideal gas law 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.
Since the pressure and temperature remain constant, you can rearrange the ideal gas law equation as V1/n1 = V2/n2, where V1 is the initial volume, n1 is the initial number of moles, V2 is the final volume, and n2 is the final number of moles.
First, let's calculate the initial number of moles using the given mass and the molar mass of O2 gas (32 g/mol).
n1 = (m1 / M)
= (4.8 g / 32 g/mol)
= 0.15 mol
Now, let's calculate the final volume by substituting the known values into the rearranged ideal gas law equation.
V1/n1 = V2/n2
15.0 L / 0.15 mol = V2 / (0.15 mol + 0.500 mol)
100 L/mol = V2 / 0.650 mol
Cross-multiplying, we get:
V2 = (100 L/mol) * (0.650 mol)
V2 = 65.0 L
Therefore, the new volume when 0.500 mole of O2 gas is added is 65.0 L.