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} ?$

Step by step solution

01

Compare Boiling Points

Propane's normal boiling point is -42°C, while room temperature is 25°C. Since the boiling point is lower than room temperature, propane must be pressurized to become a liquid at room temperature.

02

Assess Pressure in the Tank

The pressure in the propane tank must be greater than the atmospheric pressure to keep propane in a liquid state at room temperature, as we want to maintain a higher pressure to prevent propane from vaporizing into gas. The pressure inside the tank does not depend on the amount of liquid propane in it, but rather on the temperature because, according to the ideal gas law, pressure and temperature are directly proportional. #b: Leaked Propane and Change in Temperature#

03

The effect of Leakage

When propane leaks from the tank, the pressure inside the tank decreases. As the pressure decreases, the remaining liquid propane's boiling point is lowered. This allows the propane to start vaporizing at a lower temperature.

04

Explain Change in Temperature

As the remaining liquid propane vaporizes, it absorbs heat from the surroundings and induces evaporative cooling. Consequently, the temperature of the remaining liquid propane in the tank decreases. #c: Heat Required to Vaporize Propane and Volume Occupied#

05

Calculate Heat Required to Vaporize Propane

Given the mass of propane (20g), and the heat of vaporization (18.8 kJ/mol), we will first convert the mass to moles and then calculate the heat required. 20 g Propane * (1 mol/44.097 g) = 0.453 moles of Propane Heat required = (0.453 moles) * (18.8 kJ/mol) = 8.52 kJ So, 8.52 kJ of heat must be added to vaporize 20 g of propane.

06

Calculate the Volume of Vaporized Propane

Now, we need to find the volume of the vaporized propane at 100 kPa and 25°C. We will use the ideal gas law: PV = nRT. Pressure (P) = 100 kPa Moles (n) = 0.453 R (gas constant) = 8.314 J/(mol·K) (converted to kPa·L/(mol·K)): 8.314 * 1/1000 = 0.008314 kPa·L/(mol·K) Temperature (T) = 25 + 273.15 = 298.15 K Volume (V) = (nRT)/P = (0.453 * 0.008314 * 298.15) / 100 = 0.0112 m³ Therefore, the volume of vaporized propane at 100 kPa and 25°C is 0.0112 m³.

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