Final answer:
After adding 12.0 g of helium (He) to the existing gas sample containing 12.0 g of methane (CH4), the new volume of the gas mixture is calculated to be A. 70.0 L.
Step-by-step explanation:
The student's question is about determining the new volume of a gas mixture when 12.0 g of helium (He) is added to an existing volume of methane (CH4). Given that temperature and pressure do not change, the ideal gas law can be applied in an indirect manner, relying on Avogadro's law that equal volumes of all gases at the same temperature and pressure have the same number of molecules. First, calculate the number of moles of CH4 and He using their respective molar masses. For CH4, with a molar mass of 16.05 g/mol, we have (12.0 g) / (16.05 g/mol) = 0.747 moles. For He, with a molar mass of 4.00 g/mol, we have (12.0 g) / (4.00 g/mol) = 3.00 moles. Since the conditions of temperature and pressure remain constant and the number of moles of gas is directly proportional to volume, we can add the volume occupied by each gas to get the total volume.
Volume of CH4 = 14.0 L (given)
Volume of He = 3.00 moles / 0.747 moles * 14.0 L ≈ 56.0 L
Total volume = 14.0 L + 56.0 L = 70.0 L
Therefore, the new volume after adding 12.0 g of He is 70.0 L.