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Chapter 10: Problem 102

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.

Short Answer

Expert verified
The number of moles of propane gas in a 110-L container at 3 atm and 27°C is approximately 14.6 moles. In its liquid state, occupying the same volume with a density of 0.590 g/mL, there are approximately 1470 moles of liquid propane. The ratio of moles between liquid and gas propane is approximately 100, which means there are 100 times more moles of liquid propane than gas propane in the same volume. This supports the kinetic-molecular theory of gases, as it demonstrates the large volume occupied by gas propane due to the empty space between its particles compared to the closely packed particles of liquid propane.

Step by step solution

01

Calculate the number of moles of propane gas

To calculate the number of moles of propane gas in gaseous state, we will use the Ideal Gas Law: \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant and \(T\) is the temperature in Kelvin. Given: \(P = 3.00\,\text{atm}\) \(V = 110 \,\text{L}\) \(T = 27^\circ \text{C} = 300 \text{K}\) (Adding 273 to the given temperature in Celsius) Ideal gas constant, \(R = 0.0821\,\frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}}\) We need to find the value of \(n\). Rearrange the ideal gas equation for \(n\): \(n = \frac{PV}{RT}\)
02

Calculate the value of n for propane gas

Plug the given values into the equation: \(n = \frac{(3.00\,\text{atm})(110\,\text{L})}{(0.0821\,\frac{\text{atm} \cdot \text{L}}{\text{mol} \cdot \text{K}})(300\,\text{K})}\) \(n \approx 14.6\,\text{moles}\) So, there are approximately 14.6 moles of propane gas in the container.
03

Calculate the mass of liquid propane in the container

Given the density of liquid propane is: \(\rho = 0.590\,\frac{\text{g}}{\text{mL}}\) We also know the volume of the container is 110 L. Convert the volume from liters to milliliters: \(V = 110\,\text{L} \times \frac{1000\,\text{mL}}{\text{L}} = 110,000\,\text{mL}\) Now, we can calculate the mass of liquid propane using the formula: mass = density × volume mass = \((0.590\,\frac{\text{g}}{\text{mL}})(110,000\,\text{mL}) \approx 64900\,\text{g}\)
04

Calculate the number of moles of liquid propane

To find the number of moles of liquid propane, use the formula: \(n = \frac{\text{mass}}{\text{molar mass}}\) The molar mass of propane (C3H8) is 3(12.01) + 8(1.01) \(= 44.11\,\text{g/mol}\). Now to find the number of moles: \(n = \frac{64900\,\text{g}}{44.11\,\text{g/mol}} \approx 1470\,\text{moles}\) So, there are approximately 1470 moles of liquid propane in the container.
05

Calculate the ratio of moles between liquid and gas propane

To find the ratio of moles of liquid propane to gas propane, simply divide the number of moles of liquid propane by the number of moles of gas propane: Ratio = \(\frac{\text{moles of liquid propane}}{\text{moles of gas propane}}\) Ratio = \(\frac{1470\,\text{moles}}{14.6\,\text{moles}} \approx 100\)
06

Discuss the ratio in light of the kinetic-molecular theory of gases

According to the kinetic-molecular theory of gases, the particles in a gas are assumed to be far apart, and there is a lot of empty space between these particles. This assumption does not hold true for liquids, where the particles are much closer together. The ratio of 100 moles of liquid propane to 1 mole of gas propane supports this theory as it shows that gas propane takes up such a large volume compared to liquid propane because of the large amount of empty space between gas particles. Therefore, converting the propane from gas to the liquid state increases the number of available moles that can be stored in the same volume by a factor of 100.

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Most popular questions from this chapter

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? \begin{equation}\begin{array}{l}{\text { (a) Because Xe is a noble gas, there is less tendency for the Xe }} \\ {\text { atoms to repel one another, so they pack more densely in }} \\ {\text { the gaseous state. }} \\ {\text { (b) Xe atoms have a higher mass than } \mathrm{N}_{2} \text { molecules. Because }} \\ {\text { both gases at STP have the same number of molecules per }} \\ {\text { unit volume, the Xe gas must be denser. }}\\\\{\text { (c) The Xe atoms are larger than } \mathrm{N}_{2} \text { molecules and thus take }} \\ {\text { up a larger fraction of the space occupied by the gas. }} \\\ {\text { (d) Because the Xe atoms are much more massive than the }} \\\ {\mathrm{N}_{2} \text { molecules, they move more slowly and thus exert }} \\\ {\text { less upward force on the gas container and make the gas }} \\ {\text { appear denser. }}\end{array}\end{equation}

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