What is the temperature in K of 0.0420 mole of gas at 16.3 psi and that occupies 981 mL?

To find the temperature in Kelvin (K) of the gas, we can use the Ideal Gas Law, ( PV = nRT ). Rearranging for temperature gives ( T = \frac{PV}{nR} ). First, convert the pressure from psi to atm (1 psi ≈ 0.068046 atm), which gives approximately 1.11 atm. The volume in liters is 0.981 L, and using the gas constant ( R = 0.0821 , \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} ), we calculate:

[ T = \frac{(1.11 , \text{atm})(0.981 , \text{L})}{(0.0420 , \text{mol})(0.0821 , \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol})} \approx 268.8 , \text{K} ]

So, the temperature is approximately 268.8 K.