What is the volume occupied by 0.103 mol of helium gas at a pressure of 0.95 atm and a temperature of 303 K?

The pressure of He gas, P = 0.95 atm

Temperature, T = 303 K

Gas constant, R = 0.0821 L. atm. K⁻¹mol⁻¹

Number of moles of He gas, n = 0.103 mol

According to an ideal gas equation, PV = nRT

V = nRT/P

= (0.103 mol) (0.0821 L. atm. K⁻¹mol⁻¹) (303 K) / 0.95 atm

= 2.7 L

Thus, the volume of the He gas is 2.7 L.