The equation \(d / P=M / R T,\) which can be derived from equation \((6.14),\) suggests that the ratio of the density \((d)\) to pressure (P) of a gas at constant temperature should be a constant. The gas density data at the end of this question were obtained for \(\mathrm{O}_{2}(\mathrm{g})\) at various pressures at \(273.15 \mathrm{K}\) (a) Calculate values of \(d / P,\) and with a graph or by other means determine the ideal value of the term \(d / P\) for \(\mathrm{O}_{2}(\mathrm{g})\) at \(273.15 \mathrm{K}\) [Hint: The ideal value is that associated with a perfect (ideal) gas.] (b) Use the value of \(d / P\) from part (a) to calculate a precise value for the atomic mass of oxygen, and compare this value with that listed on the inside front cover. $$\begin{array}{lllll} P, \mathrm{mmHg}: & 760.00 & 570.00 & 380.00 & 190.00 \\ d, \mathrm{g} / \mathrm{L}: & 1.428962 & 1.071485 & 0.714154 & 0.356985 \end{array}$$

Step by step solution

01

Calculate the Ratio \(d/P\)

To calculate the ratio \(d/P\), we divide each measurement of density by its corresponding pressure. Given that pressure is in units of mmHg, for use in further calculations it is advantageous to convert this to atm using the conversion factor 1 atm = 760 mmHg. The following calculations are carried out:\n\nFor P = 760.00 mmHg and d = 1.428962 g / L:\n\(d/P = 1.428962 / (760.00/760) = 1.428962 g/L atm\)\nFor P = 570.00 mmHg and d = 1.071485 g / L:\n\(d/P = 1.071485 / (570.00/760) = 1.424521 g/L atm\)\nFor P = 380.00 mmHg and d = 0.714154 g / L:\n\(d/P = 0.714154 / (380.00/760) = 1.426585 g/L atm\)\nFor P = 190.00 mmHg and d = 0.356985 g / L:\n\(d/P = 0.356985 / (190.00/760) = 1.425194 g/L atm\)

02

Determine the Ideal Value of \(d/P\)

The ideal value of \(d/P\) is that associated with a perfect (ideal) gas. It is the mass per mole per unit temperature. We can take the average of the calculated \(d/P\) values as the ideal value. This gives:\nIdeal \(d/P = (1.428962 + 1.424521 + 1.426585 + 1.425194) / 4 = 1.426315 g/L atm

03

Calculate the Atomic Mass of Oxygen

From the ideal gas law, we know that \(d/P = M/RT\), where M is the molar mass of the gas, R is the universal gas constant equal to 0.0821 L atm / mol K, and T is the temperature in Kelvin. Rearranging the equation and substituting the given temperature (T = 273.15 K) and the calculated ideal \(d/P\) value, we can calculate M as follows:\nM = ideal \(d/P\) * R * T\n= 1.426315 g/L atm * 0.0821 L atm / mol K * 273.15 K\n= 31.998 g/mol, which is approximately the accepted value for the molar mass of oxygen (32.00 g/mol)

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