Calculate the ratio of the lifting powers of helium (He) gas and hydrogen (H \(_{2}\) ) gas under identical circumstances. Assume that the molar mass of air is \(29.5 \mathrm{~g} / \mathrm{mol}\).

To find the lifting power of each gas, first, we need the molar masses of helium (He) and hydrogen (H\(_{2}\)). Using the periodic table, we can find that the molar mass of He is 4 g/mol, while the molar mass of H\(_{2}\) is 2 g/mol (since there are two hydrogen atoms in a H\(_{2}\) molecule).

The lifting force of a gas is determined by the difference in weight between the gas and the air it displaces. We can use the ideal gas law, \(PV = nRT\), to find the mass of the gas in a specific volume, where P, V, n, R, and T represent pressure, volume, moles, gas constant, and temperature, respectively. Assuming identical conditions for both gases, we can say that \(m_{He} = n_{He} \cdot M_{He}\) and \(m_{H_2} = n_{H_2} \cdot M_{H_2}\), where \(m_{He}\) and \(m_{H_2}\) are the masses of helium and hydrogen in the specific volume, respectively, and \(M_{He}\) and \(M_{H_2}\) are their molar masses. Using the ideal gas law and rearranging, we have \(n_{He}= \frac{PV}{RT}\) and \(n_{H_2} = \frac{PV}{RT}\). Since both gases are under identical conditions, we can find the ratio of moles of each gas without knowing the specific values of P, V, R, and T.

To determine the ratio of lifting forces, we need to find the difference in mass between each gas and the air it displaces and then find the ratio of these differences. We will use \(M_{air}\), which is given as \(29.5 \mathrm{~g} / \mathrm{mol}\). We have \(m_{air} = n_{air} \cdot M_{air}\), where \(n_{air}\) is the number of moles of air. Since identical volumes of He, H\(_{2}\), and air are being compared under the same conditions, we have \(n_{air} = n_{He} = n_{H_2}\). Now, we can find the differences in mass and the ratio of the lifting forces: \(\frac{L_{He}}{L_{H_2}} = \frac{m_{air} - m_{He}}{m_{air} - m_{H_2}}\) Substituting the expressions for \(m_{air}\), \(m_{He}\), and \(m_{H_2}\): \(\frac{L_{He}}{L_{H_2}} = \frac{n_{air}M_{air} - n_{He}M_{He}}{n_{air}M_{air} - n_{H_2}M_{H_2}}\) Since \(n_{air} = n_{He} = n_{H_2}\), we have: \(\frac{L_{He}}{L_{H_2}} = \frac{M_{air} - M_{He}}{M_{air} - M_{H_2}}\)

Now, we can plug in the values of the molar masses and find the ratio of lifting powers of helium and hydrogen: \(\frac{L_{He}}{L_{H_2}} = \frac{29.5 - 4}{29.5 - 2}\) \(\frac{L_{He}}{L_{H_2}} = \frac{25.5}{27.5}\) \(\frac{L_{He}}{L_{H_2}} = 0.926\) The ratio of the lifting powers of helium and hydrogen under identical conditions is approximately 0.926.