91Ó°ÊÓ

First, we need to convert the given temperatures \((a)\) \(10^{\circ}\mathrm{C}\) and \((b)\) \(30^{\circ}\mathrm{C}\) into Kelvin. To do this, add 273.15 to the Celsius temperature. \((a)\) \(T_1 = 10^{\circ}\mathrm{C} + 273.15 = 283.15\,\text{K}\) \((b)\) \(T_2 = 30^{\circ}\mathrm{C} + 273.15 = 303.15\,\text{K}\)

Then, we need to determine the ideal gas constant, \(R\), in units of \(\dfrac{\text{L}\cdot\text{atm}}{\text{K}\cdot\text{mol}}\). The value for \(R\) is approximately \(0.0821\dfrac{\text{L}\cdot\text{atm}}{\text{K}\cdot\text{mol}}\).

In terms of mass \(m\), density \(\rho\), and molar mass \(M\), the ideal gas law can be rewritten as \(PV=\frac{m}{M} RT\). We can solve for the mass density, \(\rho = \frac{m}{V}\): \(\rho = \frac{m}{V} = \frac{M}{RT}P\) Now we can use this formula to find the mass density for the given conditions.

Using the derived formula for mass density, we can now plug in the known values and the average molecular mass of air (\(M = 29\,\text{u} = 29\dfrac{\text{g}}{\text{mol}}\)): \((a)\) \(\rho_1 = \frac{29\dfrac{\text{g}}{\text{mol}}}{0.0821\dfrac{\text{L}\cdot\text{atm}}{\text{K}\cdot\text{mol}}\cdot 283.15\,\text{K}}\cdot 1.0\,\text{atm} = 1.205\,\dfrac{\text{g}}{\text{L}}\) \((b)\) \(\rho_2 = \frac{29\dfrac{\text{g}}{\text{mol}}}{0.0821\dfrac{\text{L}\cdot\text{atm}}{\text{K}\cdot\text{mol}}\cdot 303.15\,\text{K}}\cdot 1.0\,\text{atm} = 1.132\,\dfrac{\text{g}}{\text{L}}\) Therefore, the mass density of air at \(10^{\circ}\mathrm{C}\) and \(1.0\,\text{atm}\) is approximately \(1.205\,\dfrac{\text{g}}{\text{L}}\) and at \(30^{\circ}\mathrm{C}\) and \(1.0\,\text{atm}\) is approximately \(1.132\,\dfrac{\text{g}}{\text{L}}\).