Problem 96 - Chemistry and Chemical Reactivity

Question

To store \(2.88 \mathrm{kg}\) of gasoline with an energy equivalence of \(1.43 \times 10^{8} \mathrm{J}\) requires a volume of \(4.1 \mathrm{L}\) In comparison, 1.0 kg of \(\mathrm{H}_{2}\) has the same energy equivalence. What volume is required if this quantity of \(\mathrm{H}_{2}\) is to be stored at \(25^{\circ} \mathrm{C}\) and 1.0 atm of pressure?

Step-by-Step Solution

Answer

The volume required is approximately 12,239 L.

Step 1: Determine the Energy Content per kg of Gasoline

Given that 2.88 kg of gasoline has an energy equivalence of \(1.43 \times 10^{8} \text{ J}\), the energy content per kg of gasoline is calculated by dividing the total energy by the mass: \[\text{Energy per kg of gasoline} = \frac{1.43 \times 10^{8} \text{ J}}{2.88 \text{ kg}}\] This results in an approximate energy equivalence per kilogram of gasoline.

Step 2: Calculate Volume Needed for 1 kg \\( \text{H}_{2}\\) Using Ideal Gas Law

We know the ideal gas law is \( PV = nRT \), where:- \( P = 1.0 \, \text{atm} \)- \( V \) is the volume we need to find- \( n \) is the number of moles- \( R = 0.0821 \, \text{L atm} / \text{mol K} \), the ideal gas constant- \( T = 25^\circ \text{C} = 298 \text{K} \)First, convert 1 kg of \( \text{H}_{2} \) to moles using the molar mass (2 g/mol):\[1 \, \text{kg} \times \frac{1000 \, \text{g}}{1 \, \text{kg}} \times \frac{1 \, \text{mol}}{2 \, \text{g}} = 500 \, \text{mol}\]Next, solve for \( V \):\[V = \frac{nRT}{P} = \frac{500 \, \text{mol} \times 0.0821 \, \text{L atm/mol K} \times 298 \, \text{K}}{1.0 \, \text{atm}} \approx 12238.9 \, \text{L}\]

Step 3: Verify Energy Equivalence of 1 kg \\( \text{H}_{2}\\) with Gasoline

Since 1 kg of \( \mathrm{H}_{2} \) has the same energy equivalence as \(1.43 \times 10^{8} \text{ J} \), it confirms that using the calculated volume following the ideal gas law provides the required energy equivalence.

Key Concepts

Energy Content

Understanding energy content is crucial when comparing fuels. Energy content refers to the amount of energy stored in a given quantity of a substance. In the context of gasoline, we know that 2.88 kg contains an energy equivalent of \(1.43 \times 10^{8} \text{ J}\). To find how much energy there is per kilogram, simply divide the total energy by the mass.

Here's how it's done:

• Calculate: \( \frac{1.43 \times 10^{8} \text{ J}}{2.88 \text{ kg}} \)
• This division tells us the energy content per kilogram.

Understanding the energy content lets us compare different fuels, like gasoline and hydrogen, on an equal footing.

Moles Calculation

Converting mass into moles is a key step in using the ideal gas law, \( PV = nRT \). Here, we want to convert 1 kg of \(\text{H}_{2}\) into moles. First, recognize the molar mass of hydrogen is 2 g/mol.

For the conversion:

• Start with 1 kg of \(\text{H}_{2}\).
• Convert kilograms to grams: \( 1 \, \text{kg} \times \frac{1000 \, \text{g}}{1 \, \text{kg}} = 1000 \, \text{g} \).
• Use the molar mass to convert grams to moles: \( 1000 \, \text{g} \times \frac{1 \, \text{mol}}{2 \, \text{g}} = 500 \, \text{mol}\).

This calculation is essential for determining the volume of hydrogen gas using the ideal gas law.

Gasoline Energy Equivalence

When comparing the energy equivalence of fuels, we need to ensure we're looking at the same amount of energy. Both 2.88 kg of gasoline and 1 kg of \( \text{H}_{2} \) are set to have an energy equivalence of \(1.43 \times 10^{8} \text{ J}\).

The challenge is to find out how much volume 1 kg of \(\text{H}_{2}\) will require under specific conditions.
Using the ideal gas law, solve for volume \( V \) with:

• Pressure \( P = 1.0 \, \text{atm} \)
• Temperature \( T = 298 \, \text{K} \)
• The number of moles \( n = 500 \, \text{mol} \)
• Gas constant \( R = 0.0821 \, \text{L atm/mol K} \)

The formula becomes:
\[ V = \frac{nRT}{P} = \frac{500 \, \text{mol} \times 0.0821 \, \text{L atm/mol K} \times 298 \, \text{K}}{1.0 \, \text{atm}} \approx 12238.9 \, \text{L} \]
This gives us the volume needed for hydrogen to match the energy content of gasoline.