Question

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) is pressurized into liquid and stored in cylinders to be used as a fuel. The normal boiling point of propane is listed as \(-42^{\circ} \mathrm{C}\). (a) When converting propane into liquid at room temperature of \(25^{\circ} \mathrm{C}\), would you expect the pressure in the tank to be greater or less than atmospheric pressure? How does the pressure within the tank depend on how much liquid propane is in it? (b) Suppose the fuel tank leaks and a few liters of propane escape rapidly. What do you expect would happen to the temperature of the remaining liquid propane in the tank? Explain. (c) How much heat must be added to vaporize \(20 \mathrm{~g}\) of propane if its heat of vaporization is \(18.8 \mathrm{~kJ} / \mathrm{mol} ?\) What volume does this amount of propane occupy at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C} ?\)

Short answer

(a) Higher than atmospheric pressure; pressure depends on vapor pressure at temperature. (b) Temperature decreases as vaporization absorbs heat. (c) 8.55 kJ needed; volume is 11.1 L.

Step-by-step solution

Understanding Boiling Point and Pressure

The normal boiling point of propane is -42°C. This means at this temperature, propane boils at 1 atm pressure. Given that room temperature is 25°C, which is higher than the boiling point, the propane needs to be pressurized to remain liquid. Therefore, the pressure inside the tank must be greater than atmospheric pressure, which is typically 1 atm.

Pressure Dependence on Liquid Content

The pressure in the tank depends on the temperature and the vapor pressure of propane above the liquid. As long as there is liquid present, the pressure will remain constant and equal to the vapor pressure at that temperature, regardless of the amount of liquid. Once all the liquid is gone, pressure might decrease as gas escapes.

Effect of Leakage on Temperature

If propane escapes rapidly from the tank, the pressure reduces, allowing the liquid to vaporize. This vaporization is an endothermic process, meaning it absorbs heat from the surroundings, causing the temperature of the remaining liquid to decrease.

Heat Required for Vaporization

To find the heat required to vaporize 20 g of propane, we'll use the formula: \[ Q = m \cdot \Delta H_{vap} \] where \( m \) is the number of moles and \( \Delta H_{vap} \) is the heat of vaporization. First, find the moles of propane: \[ \text{Molar mass of } \text{C}_3\text{H}_8 = 44 \text{ g/mol} \] \[ n = \frac{20 \text{ g}}{44 \text{ g/mol}} = 0.4545 \text{ mol} \] Now calculate the heat: \[ Q = 0.4545 \text{ mol} \times 18.8 \text{ kJ/mol} = 8.55 \text{ kJ} \]

Calculating Propane Gas Volume

Use the ideal gas law to calculate the volume occupied by gas: \[ PV = nRT \] where \( P = 100 \text{ kPa} = 1 \text{ atm} \), \( n = 0.4545 \text{ mol} \), and \( T = 25 + 273.15 = 298.15 \text{ K} \), \( R = 0.0821 \text{ L atm/mol K} \). Solve for \( V \): \[ V = \frac{nRT}{P} = \frac{0.4545 \times 0.0821 \times 298.15}{1} = 11.1 \text{ L} \]

Key concepts

Boiling Point

The boiling point of a substance is the temperature at which its liquid form turns into a gas. For propane, the normal boiling point is -42°C at 1 atm pressure. This is the temperature at which the vapor pressure of propane equals atmospheric pressure, allowing it to enter the gas phase. When the surrounding temperature is higher than the boiling point, such as at room temperature (25°C), the substance needs to be pressurized to remain liquid. For propane in a tank, the pressure must be greater than the atmospheric pressure to keep it in liquid form.
It's important to note that in closed containers, the pressure is determined by the temperature and vapor pressure rather than the liquid content itself.
The pressure remains stable as long as some liquid propane is present.

Vaporization

Vaporization is the process of changing from liquid to gas. This occurs in two main ways: evaporation and boiling. In our context with propane, vaporization happens when liquid propane absorbs enough heat to turn into a gas. If a propane tank leaks, the liquid rapidly vaporizes to fill the new space.
This process is endothermic, requiring heat absorption. Consequently, the temperature of the remaining liquid decreases as heat is drawn from it during vaporization.
Ensuring proper handling and storage of propane prevents unwanted vaporization and maintains safety.

Ideal Gas Law

The ideal gas law provides a simple relation between pressure, volume, temperature, and number of moles of a gas. It's expressed by the formula: \[ PV = nRT \] Where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume.
• \( n \) is the number of moles.
• \( R \) is the ideal gas constant (0.0821 L atm/mol K).
• \( T \) is the temperature in Kelvin.

In the case of propane, once vaporized, the ideal gas law helps us calculate the volume of the gas at specific pressures and temperatures. For example, the volume occupied by propane vapor at 100 kPa and 25°C can be calculated using this law, illustrating its practical applications in predicting the behavior of gases under non-standard conditions.

Enthalpy Change

Enthalpy change, specifically the heat of vaporization (\( \Delta H_{vap} \)), is crucial in understanding the energy required for a phase change.
It's the amount of energy needed to convert a liquid into a gas at its boiling point, without changing temperature. For propane, this value is given as 18.8 kJ/mol. To calculate the heat required to vaporize a specific amount of propane, use the formula:\[ Q = n \times \Delta H_{vap} \]Where \( Q \) is the heat absorbed, \( n \) is the number of moles vaporized, and \( \Delta H_{vap} \) is the enthalpy change.
For example, vaporizing 20 g of propane needs 8.55 kJ of energy. Understanding enthalpy changes helps in efficient energy management during industrial and domestic applications.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form. It depends on temperature, increasing with higher temperatures. In propane tanks, the vapor pressure dictates the pressure required to keep propane liquid inside. At temperatures above the boiling point, vapor pressure exceeds atmospheric pressure, necessitating containment. If the propane escapes from a damaged tank, the surrounding pressure decreases, causing rapid vaporization, which may induce cooling. Maintaining proper vapor pressure prevents accidental vaporization, keeping propane safely stored and ready for use. Understanding vapor pressure is essential for handling liquefied gases effectively, ensuring their seamless application in various fields.