Final answer:
The volume occupied by 24.3 g of argon gas at the given conditions is 235.8 L.
Step-by-step explanation:
To solve this problem, 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, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15: T = 414 K.
From the equation, we can rearrange it to solve for volume: V = (nRT) / P.
Substituting the given values: V = (24.3 g / 39.95 g/mol) * (0.0821 L*atm/mol*K) * 414 K / 1.38 atm = 235.8 L.
Therefore, the volume occupied by 24.3 g of argon gas at a pressure of 1.38 atm and a temperature of 414 K is 235.8 L.