A sample containing 4.80g 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?
Am I converting the 4.80 O2 into moles first? How do I do that?
yes. Use PV = nRT and solve for n, then add 0.500 to that and use PV =nRT again and solve for V. Since P and T remain constant you can use any number you which as long as you stay consistent with the two calculations (I would use 1 for P and 1 for T. n won't mean anything then but the answer will come out right. The 0.500 mole of O2 added must be at the same T and P as the original sample.)
a sample containing 4.80 g O2 of gas has a volume of 15.0 L. Pressure and temperature remain constant. What is the new voume if 0.500 mole O2 gas is added?
Yes, you need to convert the given mass of O2 gas into moles first. To convert grams to moles, you can use the molar mass of O2. The molar mass of O2 is 2 * atomic mass of oxygen = 2 * 16.00 g/mol = 32.00 g/mol.
To calculate the number of moles of O2 gas present, you can use the formula:
moles = mass / molar mass
Substituting the given mass of O2 gas, we have:
moles = 4.80 g / 32.00 g/mol
moles = 0.15 mol
Therefore, the initial sample contains 0.15 moles of O2 gas.
Now, to find the new volume when 0.500 mole of O2 gas is added, since the pressure and temperature remain constant, we 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 (0.0821 L·atm/mol·K), and T is the temperature.
Substituting the initial conditions, we have:
(15.0 L) * (0.15 mol) = (n + 0.500 mol) * (0.0821 L·atm/mol·K)
Simplifying the equation:
2.25 L·mol = 0.0821 L·atm/mol·K * (n + 0.500 mol)
Dividing both sides of the equation by 0.0821 L·atm/mol·K:
2.25 L·mol / (0.0821 L·atm/mol·K) = n + 0.500 mol
27.42 mol = n + 0.500 mol
Subtracting 0.500 mol from both sides:
27.42 mol - 0.500 mol = n
26.92 mol = n
Therefore, the new volume when 0.500 mole of O2 gas is added is 26.92 L.
To convert grams of a substance to moles, you need to use the molar mass of the substance. In this case, you would need to convert the 4.80 grams of O2 gas to moles.
The molar mass of O2 gas is the sum of the molar masses of two oxygen atoms, which is 16.00 g/mol. Therefore, you can calculate the number of moles by dividing the mass by the molar mass:
moles = mass / molar mass
moles = 4.80 g / 32.00 g/mol (since O2 has a molar mass of 32.00 g/mol)
moles = 0.15 mol
Hence, the initial sample of O2 gas contains 0.15 moles.
Now, to find the new volume when 0.500 mole of O2 gas is added, you 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. Since the pressure and temperature remain constant, you can rewrite the 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.
Plugging in the values:
(15.0 L * 0.15 mol) = (V2 * 0.65 mol)
Simplifying:
2.25 L = 0.65 V2
Now, to find V2, you can divide both sides by 0.65:
V2 = 2.25 L / 0.65
V2 ≈ 3.46 L
Therefore, the new volume, when 0.500 mole of O2 gas is added to the initial sample, is approximately 3.46 liters.