Question

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 \(3.00 \mathrm{~atm}\) 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 kineticmolecular theory of gases.

Short answer

In a 110-L container, \(13.5\) moles of propane gas can be stored at \(3.00~atm\) and \(27^\circ C\). However, if stored as a liquid, approximately \(1472\) moles of propane can be stored in the same volume, with a density of \(0.590~g/mL\). The ratio of moles of liquid propane to moles of gas propane is around \(109\), demonstrating the difference in particle arrangements and kinetic energies between the two states, as predicted by the kinetic molecular theory of gases and liquids.

Step-by-step solution

(Part a: Calculate the moles of propane gas using Ideal Gas Law)

The ideal gas law is given by the equation: \[PV = nRT\] Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in kelvin. First, we have to convert the given temperature to Kelvin: \( T(K) = T(^{\circ}C) + 273.15 = 27+273.15 = 300K\) Now we can rearrange the ideal gas law equation to find the number of moles (n): \[ n = \frac{PV}{RT}\] Substitute the given values: \[n = \frac{(3.00 atm)(110 L)}{(0.0821 L atm / (mol K))(300 K )}\] Solve for n: \[n \approx 13.5 moles\]

(Part b: Calculate the moles of liquid propane using density and volume conversion)

We are given the density of liquid propane (0.590 g/mL) and the volume of the container (110 L). First, we need to convert the volume to milliliters (mL) and then use the density to find the mass of liquid propane that can be stored in the container. Then, we can convert the mass to moles. Volume conversion: \[110 L = 110,000 mL\] Mass of liquid propane: \[mass = volume \times density = 110,000 mL \times 0.590 g/mL = 64,900 g\] Finally, we can find the number of moles using the molar mass of propane: \[n = \frac{mass}{molar~mass} = \frac{64,900 g}{44.09g/mol} = 1472 moles\]

(Part c: Calculate the ratio of moles of liquid propane to moles of gas propane)

Now that we have calculated the number of moles of gaseous and liquid propane that can be stored in the container, we can find the ratio: \[ ratio = \frac{moles~of~liquid~propane}{moles~of~gas~propane} =\frac{1472}{13.5}\] \(ratio \approx 109\) This ratio of approximately 109:1 tells us that for every 1 mole of propane gas stored in the container, about 109 moles of propane can be stored as a liquid. This is because, in the liquid state, the particles are closely packed and can occupy a smaller volume compared to their gaseous state where particles are farther apart, occupying a larger volume. This demonstrates the difference in the nature of the kinetic molecular theory of gases and liquids. The gas particles have much more kinetic energy and are in constant motion while the liquid particles have less kinetic energy and are in a more condensed state.

Key concepts

Kinetic Molecular Theory

The Kinetic Molecular Theory (KMT) provides insight into how gas particles behave under different conditions. According to KMT, gas particles are in constant random motion and exhibit elastic collisions. This means they don’t lose energy when they bump into each other or the walls of their container.
Gas particles are far apart, which not only gives them the freedom to move but also ensures that they occupy a larger volume. This theory explains why gases can be compressed more easily compared to liquids.

• Higher temperatures increase the kinetic energy of gas particles, making them move faster.
• Pressure is created by gas particles hitting the walls, and this increases when the number of collisions increases.

In the exercise, the liquefaction of propane reduces the volume occupied by the gas significantly. This is because liquid particles, according to KMT, are much closer together compared to gas particles, following a transition from high kinetic energy states to lower ones as they cool or are pressurized into the liquid state.

Density Calculations

Density is a crucial concept when dealing with phases of matter, particularly when converting between liquids and gases. It is defined as mass per unit volume. For propane, knowing the density allows us to compute mass from volume, which is essential when the state of the substance changes.
In the exercise, we use the density of liquid propane (0.590 g/mL) to determine how much mass can fit into a certain volume. Here's how you can approach density:

• Convert all your measurements to compatible units. Liquid propane's density is given in grams per milliliter, while the container's volume is initially in liters. Thus, conversion is necessary.
• Use the formula \( \text{mass} = \text{density} \times \text{volume} \) to find out how much propane fits in the container.

Converting the liquid's density into useful numbers allows us to calculate the number of moles when compared against the molar mass.

Mole Calculation

Calculating moles is a basic yet essential task in chemistry. It allows us to convert between various units of chemical measurement. The mole concept offers a bridge between the microscopic particles and macroscopic amounts we can measure.
In the exercise, knowing the moles of propane helps us determine how much of the substance is available in both its gaseous and liquid states.

• For gases, the Ideal Gas Law, \( PV = nRT \), is used. Rearrange to find \( n \) (moles), using known values of pressure (\( P \)), volume (\( V \)), the ideal gas constant (\( R \)), and temperature (\( T \)).
• For liquids, convert from mass to moles using the formula \( n = \frac{\text{mass}}{\text{molar mass}} \), where the mass is derived from density calculations.

These calculations enable us to compare substances in different states smartly.

Volume Conversion

Volume conversion plays a significant role in chemistry, especially when measurements need to convert between units like liters and milliliters for accurate calculations.
In the exercise, propane held as a gas or a liquid requires different volumes due to its density. Understanding how to perform these conversions is crucial.

• Always be aware of the unit differences and convert volumes appropriately. For example, 1 L = 1000 mL.
• Conversion is used for recalibrating measurements – in the example, the 110 L container translates to 110,000 mL.

This methodology simplifies calculations and allows us to use uniform units across different properties. Accurate volume conversions help ensure all subsequent calculations, such as those related to moles and density, use the correct numbers.