Final answer:
Using the ideal gas law PV = nRT, the temperature of 2.52 moles of an ideal gas occupying 3.30 L at 3.50 atm is calculated to be 55.85 K.
Step-by-step explanation:
To calculate the temperature of 2.52 moles of an ideal gas occupying 3.30 L at 3.50 atm, we need to use the ideal gas law, which is 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.
First, we ensure that our pressure is in the correct units for R = 0.08206 L atm mol⁻¹ K⁻¹ since our pressure is already given in atm. Now, we can plug the values into the ideal gas law:
PV = nRT
(3.50 atm) × (3.30 L) = (2.52 moles) × (0.08206 L atm mol⁻¹ K⁻¹) × T
To isolate T, we'll divide both sides by (2.52 moles × 0.08206 L atm mol⁻¹ K⁻¹):
T = ⅔[{(3.50 atm) × (3.30 L)} ÷ (2.52 moles × 0.08206 L atm mol⁻¹ K⁻¹)]
T = (11.55 atm·L)/(0.2068 atm·L mol⁻¹ K⁻¹)
T = 55.85 K
Thus, the temperature of the gas is 55.85 K.