Final answer:
To determine the molar volume of hydrogen gas at 50.0kPa and 300K, we use the ideal gas law, resulting in a molar volume of approximately 22.71 L/mol, which matches Option 1.
Step-by-step explanation:
The question asks for the molar volume of hydrogen gas at 50.0kPa and 300K, which is not at standard temperature and pressure (STP). We know that at STP, which is defined as 1 atm of pressure and 273K (0°C) temperature, the molar volume of an ideal gas is 22.4 L/mol. To find the molar volume under the given conditions (50.0kPa and 300K), we use the ideal gas law equation PV=nRT. Transforming it to find V (molar volume), we get V=nRT/P. Assume n=1 mol of hydrogen gas for molar volume, R is the ideal gas constant 8.314 J/(mol·K) or 0.08206 L atm/(mol·K), T is the temperature 300K, and the pressure must be converted to atmospheres (1 atm = 101.325 kPa), so P = 50.0 kPa / 101.325 = 0.493 atm. Plugging these into the equation we get:
V = (1 mol)(0.08206 L atm/(mol·K))(300K) / (0.493 atm)
Calculating the value gives us an approximate molar volume of 22.71 L/mol, which corresponds to Option 1.