Final answer:
The volume of 10.3 moles of O2 gas at 25°C and 2 ATM pressure is 126.6 liters. This is calculated using the ideal gas law where the temperature is converted to Kelvin and the known values are substituted into the equation V = nRT/P.
Step-by-step explanation:
The volume of 10.3 moles of oxygen gas (O₂) at 25°C and 2 ATM can be calculated using the ideal gas law, which is PV = nRT. Following this law, we need to rearrange the formula to solve for the volume (V = nRT/P), with R being the ideal gas constant (0.0821 L·atm/(mol·K)), n being the number of moles, T the temperature in Kelvin, and P the pressure in atm.
First, convert the temperature from Celsius to Kelvin by adding 273.15 to the given temperature, which yields T = 25 + 273.15 = 298.15 K. Then substitute the values into the formula:
V = (10.3 mol × 0.0821 L·atm/(mol·K) × 298.15 K) / (2 atm)
After performing the multiplication and the division, we get the volume of oxygen gas:
V = (10.3 mol × 0.0821 L·atm/(mol·K) × 298.15 K) / (2 atm) = 126.6 L
The volume of the oxygen gas under these conditions is calculated to be 126.6 liters.