Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), liquefies under modest pressure, allowing a large amount to be stored in a container. (a) Calculate the number of moles of propane gas in a 110 -L container at 3.00 atm and \(27^{\circ} \mathrm{C} .\) (b) Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is \(0.590 \mathrm{~g} / \mathrm{mL} .\) (c) Calculate the ratio of the number of moles of liquid to moles of gas. Discuss this ratio in light of the kinetic-molecular theory of gases.

In order to use the ideal gas law, we must first convert the Celsius temperature to Kelvin. The formula is: \[ T (K) = T ( ^\circ C) + 273.15 \] For a temperature of 27°C: \[T (K) = 27 + 273.15 = 300.15 K\]

We can use the ideal gas law equation \(PV = nRT\) to find the number of moles of propane gas (\(n\)). Plugging in the values, we have: \[3.00\,\text{atm} \times 110\,\text{L} = n \times 0.0821\, (\text{L}\cdot\text{atm/mol}\cdot\text{K}) \times 300.15\, \text{K}\] Now, solve for \(n\): \[n = \frac{3.00 \times 110}{0.0821 \times 300.15} = 13.42 \, \text{moles}\]

To calculate the mass of liquid propane in the container, we will use the density formula: \[mass = density \times volume\] Given the density of liquid propane as 0.590 g/mL, we first convert the container volume from liters to milliliters: \[110\, \text{L} \times \frac{1000\, \text{mL}}{1\, \text{L}} = 110,000\, \text{mL}\] Now, using the density formula, we have: \[mass = 0.590\, \text{g/mL} \times 110,000\, \text{mL} = 64,900\, \text{g}\]

Next, we will convert the mass of liquid propane to the number of moles using the molecular weight of propane (C3H8), which is: \[(3 \times 12.01\, \text{g/mol}) + (8 \times 1.01\, \text{g/mol}) = 44.1\, \text{g/mol}\] Now, divide the mass by the molecular weight to get the number of moles: \[n = \frac{64,900\, \text{g}}{44.1\, \text{g/mol}} = 1471.43\, \text{moles}\]

Finally, we will determine the ratio of the number of moles of liquid propane to the number of moles of propane gas: \[\text{Ratio} = \frac{n_{liquid}}{n_{gas}} = \frac{1471.43}{13.42} = 109.54\] #Answer# (a) 13.42 moles of propane gas (b) 1471.43 moles of liquid propane (c) The ratio of the number of moles of liquid propane to propane gas is 109.54. This large ratio indicates that, under modest pressure, the gas can be condensed into a liquid, which can be stored much more compactly and efficiently than in gas form. This is consistent with the kinetic-molecular theory, as the gas molecules are forced closer together under pressure, resulting in an increase in intermolecular forces and ultimately allowing the substance to turn into a liquid.