Final answer:
The volume of 3.20 mol of gas at 1.50 atm and 20.0 °C is calculated to be 49.424 L using the ideal gas law equation and converting the temperature to Kelvin.
Step-by-step explanation:
To calculate the volume of the gas in the rigid container, we need to use the ideal gas law equation, which is PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvin. Given the values, P = 1.50 atm, n = 3.20 mol, and T = 20.0 °C, we first need to convert the temperature to Kelvin (T(K) = T(°C) + 273.15). So T = 20.0 + 273.15 = 293.15 K. The ideal gas constant R when dealing with atmospheres for pressure is 0.0821 L atm/(mol K).
Now, substituting the known values into the ideal gas law equation, we can solve for the volume (V).
V = \(\frac{nRT}{P}\) = \(\frac{3.20 \text{ mol} \times 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1} \times 293.15 \text{ K}}{1.50 \text{ atm}}\)
After performing the calculation, V = 49.424 L, which is the volume of the gas.