At what Kelvin temperature will 25.7 grams of oxygen occupy a volume of 13.5 L at a pressure of 650. torr?
To find the temperature in Kelvin, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, we need to convert the pressure from torr to atm:
1 atm = 760 torr
So, the pressure in atm is:
650 torr / 760 torr/atm = 0.8553 atm
Next, we need to find the number of moles (n) of oxygen. We can use the following formula:
moles = mass / molar mass
The molar mass of oxygen (O2) is 32.00 g/mol.
moles = 25.7 g / 32.00 g/mol
moles ≈ 0.8031 mol
Now we can rearrange the Ideal Gas Law equation to solve for temperature (T):
T = PV / nR
T = (0.8553 atm) x (13.5 L) / (0.8031 mol) x (0.0821 L·atm/mol·K)
T = 11.535 L·atm / (0.0659 L·atm/mol·K)
T ≈ 175.4 K
Therefore, at approximately 175.4 Kelvin temperature, 25.7 grams of oxygen will occupy a volume of 13.5 L at a pressure of 650 torr.
To determine the Kelvin temperature, we can use the ideal gas law equation:
PV = nRT
where:
P is the pressure (650. torr),
V is the volume (13.5 L),
n is the number of moles of gas,
R is the ideal gas constant (0.0821 L·atm/(K·mol)), and
T is the temperature in Kelvin (what we want to find).
First, let's calculate the number of moles (n) of oxygen using the given mass (25.7 grams) and the molar mass of oxygen (32.00 g/mol).
n = mass / molar mass
n = 25.7 g / 32.00 g/mol
n ≈ 0.80313 mol
Now, rearrange the ideal gas law equation to solve for temperature (T):
T = PV / (nR)
Substitute the given values:
T = (650. torr) * (13.5 L) / (0.80313 mol * 0.0821 L·atm/(K·mol))
Now, let's convert torr to atmospheres (atm):
1 atm = 760 torr
T ≈ (650 / 760) * 13.5 / (0.80313 * 0.0821)
T ≈ 0.85526 * 13.5 / 0.0659728533
T ≈ 11.547840225 K
Therefore, at approximately 11.55 Kelvin temperature, 25.7 grams of oxygen will occupy a volume of 13.5 L at a pressure of 650. torr.