Final answer:
Using the Ideal Gas Law PV = nRT, the volume occupied by 0.248 mol of helium gas at 1.29 atm and 303 K is approximately 4.78 liters.
Step-by-step explanation:
To calculate the volume occupied by 0.248 mol of helium gas at 1.29 atm and 303 K, we can use the Ideal Gas Law which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we rearrange the Ideal Gas Law to solve for V: V = nRT/P.
We can then substitute the values into the equation: V = (0.248 mol)(0.0821 L·atm/K·mol)(303 K) / (1.29 atm).
After calculating, the volume V is approximately 4.78 liters.