Answer:
-245.5°C or 27.7 K
Step-by-step explanation:
To solve this problem, we can use the ideal gas law:
PV = nRT
∴ P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to calculate the number of moles of argon gas using its mass and molar mass:
n = m/M
∴ m is the mass of argon and M is its molar mass. The molar mass of argon is approximately 39.95 g/mol.
n = 120 g / 39.95 g/mol = 3.003 mol
Next, we can rearrange the ideal gas law to solve for the temperature:
T = PV/nR
Substituting the given values:
T = (200 kPa) (0.25 L) / (3.003 mol) (8.314 J/(mol·K))
T = 27.7 K
Therefore, the temperature at which 120 grams of argon gas will have a volume of 0.25 L at a pressure of 200 kPa is approximately -245.5°C (27.7 K - 273.15).