An 0.727 mol sample of argon gas at a temperature of 22.0 °C is found to occupy a volume of 26.2 liters. The pressure of this gas sample is how many mm Hg?
Use PV = nRT and solve for P (in atmospheres) and convert to mm Hg.
1 atm = 760 mm Hg. Don't forget T must be in kelvin.
To find the pressure of the gas sample in mm Hg, you can use the Ideal Gas Law. The Ideal Gas Law equation is:
PV = nRT
Where:
P = Pressure (in units such as atm or mm Hg)
V = Volume (in units such as liters)
n = Number of moles of gas
R = Ideal Gas Constant (0.0821 L·atm/mol·K for units in atm)
T = Temperature (in units such as Kelvin)
First, let's convert the given temperature from Celsius to Kelvin. The conversion formula is:
K = °C + 273.15
So, the temperature in Kelvin is:
T = 22.0 °C + 273.15 = 295.15 K
Now, we can plug in the values into the Ideal Gas Law equation:
P * V = n * R * T
P * 26.2 L = 0.727 mol * (0.0821 L·atm/mol·K) * 295.15 K
Simplifying the equation:
P * 26.2 = 0.727 * 0.0821 * 295.15
P * 26.2 = 18.2956
Finally, divide both sides of the equation by 26.2 to isolate the pressure:
P = 18.2956 / 26.2
P ≈ 0.698 atm
To convert from atm to mm Hg, we can multiply by a conversion factor. The conversion factor is:
1 atm = 760 mm Hg
So, multiply the pressure by the conversion factor:
P_mmHg = P * (760 mm Hg / 1 atm)
P_mmHg = 0.698 atm * 760 mm Hg
P_mmHg ≈ 530.48 mm Hg
Therefore, the pressure of the argon gas sample is approximately 530.48 mm Hg.