Question

In a hydrogen bomb, deuterium nuclei fuse to produce a helium nucleus. In the process, \(5 \mathrm{MeV}\) of energy is released per deuterium. If, \(1 / 7000\) of natural \(\mathrm{H}\) is \(\mathrm{D}\), how many liters of gasoline \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) are needed to be burned to produce the same amount of energy as derived from 1 liter of water. Assume the following: the heat of combustion of gasoline is \(5100 \mathrm{KJ} /\) mole the density of gasoline is \(.703 \mathrm{~g} / \mathrm{cm}^{3}\); \(\mathrm{leV}=1.60 \times 10^{-19} \mathrm{~J}\)

Short answer

Approximately \( 1.95 \times 10^{-4} \) liters of gasoline are needed to be burned to produce the same amount of energy as derived from 1 liter of water.

Step-by-step solution

Calculate the number of deuterium nuclei in 1 liter of water

To determine how many deuterium nuclei are present in 1 liter of water, we will find the number of hydrogen atoms in 1 liter of water, and then calculate how many are deuterium nuclei using the given ratio. First, we need to find the number of moles of water (H2O) in 1 liter. Since 1 mole of any substance has a volume of 22.4 liters at standard temperature and pressure (STP), we have: Moles of H2O in 1 liter = 1 L x (1 mole / 22.4 L) = \( \frac{1}{22.4} \) moles In one mole of water, there are two hydrogen atoms, so in 1/22.4 moles of water, we have: Number of hydrogen atoms in 1 liter of water = \( \frac{1}{22.4} \) moles x 2 atoms x \( 6.022 \times 10^{23} \) atoms/mole ≈ \( 5.36 \times 10^{22} \) atoms Given \$1 / 7000 \$ of natural H is D, we have: Number of deuterium nuclei in 1 liter of water = \( \frac{1}{7000} \) x \( 5.36 \times 10^{22} \) ≈ \( 7.66 \times 10^{18} \) nuclei

Calculate the energy released from the fusion of 1 liter of water

In this step, we will calculate the total energy released from the fusion of all deuterium nuclei present in 1 liter of water. Using the given amount of energy released per deuterium nucleus (5 MeV), we can calculate the total energy released: Energy released per deuterium nucleus = \( 5 \mathrm{MeV} \) × \(1.60 \times 10^{-19} \) J/MeV = \( 8.00 \times 10^{-19} \) J Total energy released from 1 liter of water = \( 7.66 \times 10^{18} \) nuclei × \( 8.00 \times 10^{-19} \) J/nucleus ≈ \( 6.13 \times 10^{3} \) J

Calculate the energy produced by 1 liter of gasoline

To determine how much energy is produced by 1 liter of gasoline, we will use the given heat of combustion and density of gasoline to find the total number of moles in 1 liter of gasoline, and from there, the total energy produced. First, we find the mass of 1 liter of gasoline: Mass of 1 liter of gasoline = 1 L × 1000 \( \mathrm{cm}^3 \) / L × .703 g/ \( \mathrm{cm}^3 \) = 703 g Now, we can calculate the number of moles of gasoline in 1 liter: Molar mass of gasoline (C8H18) = 12.01 x 8 + 1.008 x 18 ≈ 114 g/mole Moles of gasoline in 1 liter = mass of gasoline / molar mass = 703 g / 114 g/mole ≈ 6.16 moles Using the heat of combustion of gasoline, we can find the energy produced by 1 liter of gasoline: Energy produced = Moles of gasoline × heat of combustion = 6.16 moles × \( 5.1 \times 10^3 \) KJ/mole = \(3.14 \times 10^{4}\) KJ Convert the energy to Joules: \( 3.14 \times 10^{4} \) KJ x \( 10^3 \) J/KJ ≈ \(3.14 \times 10^7 \) J

Calculate the number of liters of gasoline needed to match the energy from 1 liter of water

Finally, we will determine the number of liters of gasoline needed to produce the same amount of energy as derived from 1 liter of water: Energy from 1 liter of water = \( 6.13 \times 10^{3} \) J Energy from 1 liter of gasoline = \( 3.14 \times 10^{7} \) J Liters of gasoline needed = Energy from 1 liter of water / Energy from 1 liter of gasoline = \( \frac{6.13 \times 10^{3}}{3.14 \times 10^{7}} \) ≈ \( 1.95 \times 10^{-4} \) liters Thus, approximately \( 1.95 \times 10^{-4} \) liters of gasoline are needed to be burned to produce the same amount of energy as derived from 1 liter of water.

Key concepts

Deuterium Nuclei

Deuterium nuclei are a type of hydrogen isotope, consisting of one proton and one neutron in their nucleus. Unlike the more common protium, which has no neutrons, deuterium plays a key role in nuclear fusion reactions such as those occurring in hydrogen bombs. In the fusion process, two deuterium nuclei come together under extreme pressure and temperature. This reaction produces a helium nucleus and releases a significant amount of energy, specifically measured in electron volts (eV) or mega-electron volts (MeV).

• Deuterium's rarity is highlighted by its natural prevalence—only about 1 in 7000 hydrogen atoms is deuterium.
• This rarity makes calculating its energy contribution essential.

Understanding deuterium nuclei is crucial for grasping how small amounts of hydrogen can yield massive energy outputs through fusion.

Energy Conversion

Energy conversion is the process of transforming energy from one form to another. In nuclear fusion, mass is converted to energy based on the principle of E=mc², Einstein's famous equation, demonstrating the relationship between mass (m) and energy (E). The energy release in a fusion reaction occurs when deuterium nuclei fuse, forming a helium nucleus. This release is significant because it involves the conversion of nuclear binding energy into usable energy in the form of heat and light.

• In our scenario, each deuterium nucleus fusion releases about 5 MeV, equivalent to \(8.00 \times 10^{-19}\) Joules.
• The conversion efficiency in fusion is far greater than chemical reactions.

Understanding how energy conversion in nuclear reactions compares to chemical processes helps us evaluate the potential of nuclear fusion as a powerful energy source.

Heat of Combustion

The heat of combustion refers to the amount of energy released as heat when a substance undergoes complete combustion with oxygen under standard conditions. Gasoline, with the chemical formula \(\text{C}_8\text{H}_{18}\), is a common fuel due to its high energy density. Its heat of combustion is about 5100 KJ per mole, making it a powerful energy source.In the exercise, we compare how the energy produced from burning gasoline equates to the energy from deuterium fusion:

• First, calculate the number of moles in 1 liter of gasoline using its mass and molar mass.
• Multiply the moles by the heat of combustion to find the total energy output.

This comparison shows that although chemical energy release from gasoline is significant, it pales in comparison to nuclear fusion outputs.

Gasoline Density

Gasoline density is a critical factor in calculating how much energy can be obtained from a given volume of gasoline. For gasoline, the density is approximately 0.703 g/cm³. This density helps us determine the mass of gasoline per unit volume, which is important for calculating how many moles are present in a liter of gasoline.

• By multiplying the density by the volume (1 liter = 1000 cm³), we find the mass of 1 liter of gasoline is 703 grams.
• Knowing the molar mass of gasoline allows us to convert this mass into moles, crucial for further energy calculations.

Understanding gasoline density aids in comprehending how much fuel is needed to achieve equivalent energy outputs, such as comparing gasoline combustion to nuclear fusion.