At what temperature will 0.0100 mole of argon gas have a volume of 275 mL at 100.0 kPa?
In order to determine the temperature at which 0.0100 mole of argon gas will have a volume of 275 mL at a pressure of 100.0 kPa, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in Pa)
V = volume (in m^3)
n = number of moles
R = gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
First, let's convert the given values to appropriate units:
Volume: 275 mL = 0.275 L
Pressure: 100.0 kPa = 100,000 Pa
Number of moles: 0.0100 mole
Now, we can rearrange the ideal gas law equation to solve for temperature (T):
T = (PV) / (nR)
Substituting the values:
T = (100,000 Pa)(0.275 L) / (0.0100 mole)(8.314 J/(mol·K))
Calculating:
T ≈ 331 K
Therefore, at approximately 331 K, 0.0100 mole of argon gas will have a volume of 275 mL at a pressure of 100.0 kPa.
To determine the temperature at which 0.0100 mole of argon gas will have a volume of 275 mL at 100.0 kPa, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature
First, we need to rearrange the equation to solve for temperature (T):
T = (PV) / (nR)
Let's gather the required values:
P = 100.0 kPa (convert to atm)
V = 275 mL (convert to liters by dividing by 1000)
n = 0.0100 mole
R = 0.0821 atm L / (mol K) (ideal gas constant)
Converting units to match the units of the ideal gas constant:
P = 100.0 kPa * (1 atm / 101.3 kPa) ≈ 0.9869 atm
V = 275 mL / 1000 ≈ 0.275 L
Now, substitute the values into the rearranged equation for temperature:
T = (0.9869 atm * 0.275 L) / (0.0100 mol * 0.0821 atm L / (mol K))
Calculating:
T ≈ 33.86 K
Therefore, at approximately 33.86 Kelvin, 0.0100 mole of argon gas will occupy a volume of 275 mL at 100.0 kPa.
Use PV = nRT
Watch the units.