24.131 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Using -century technology, hydrogen fusion requires temperatures around 10⁸K . But, lower initial temperatures are used if the hydrogen is compressed. In the late 24^th century, the starship Leinad uses such methods to fuse hydrogen at 10⁶K  . 

(a) What is the kinetic energy of an  atom at $1.00\times {10}^{6}K$? 

(b) How many H atoms are heated to $1.00\times {10}^{6}K$ from the energy of one H and one anti-H atom annihilating each other? 

(c) If these H atoms fuse into ⁴He atoms (with the loss of two positrons per ⁴He  formed), how much energy (in Jr) is generated?

(d) How much more energy is generated by the fusion in than by the hydrogen-antihydrogen collision in ? 

(e) Should the captain of the Leinad change the technology and produce (mass = 3.01603amu)  instead of ⁴He ?

Step-by-Step Solution

Answer

The kinetic energy is $E_{kave}=2.0709\times {10}^{-17}\frac{J}{atom}H$.

(b) The number of H atoms is  = 14525864.33 atoms.

(c) The energy is $1.4960\times {10}^{-5} J$.

(d) The generated energy is $1.4960\times {10}^{-5} J$ by the fusion.

(e) The captain of the Leinad should not change the technology.

 

Step 1: Definition of kinetic energy

The “energy generated” by an object obtained from its motion is known as kinetic energy. When an item is configured to accelerate, that means forces must be applied. 

 

The use of force necessitates the completion of work, after which the energy is supplied to the object.

Step 2: Find the kinetic energy

(a) Kinetic energy is the energy generated by an object's motion. The mass (m) and speed of an item (v) are inversely related for a given E_k , which means that a heavier object going slower can have the same kinetic energy as a lighter one moving faster.

 

The formula is,

$E_k=\frac{1}{2}m{v}^2$

 

The average kinetic energy of the particles (E_k) is equal to the absolute temperature multiplied by a constant, $E_k=c\times T$. We can alternatively express this relationship using definitions of velocity, momentum, force, and pressure as follows:

$E_{kave}=\frac{3}{2}\left(\frac{R}{N_A}\right)\left(T\right)$

Where,

• R denotes the gas constant, and 
• N_A denotes Avogadro's number. 

 

This equation establishes the crucial fact that temperature is a measure of the particles' average kinetic energy: when T rises, E_K rises, and vice versa. This formula will be used to calculate kinetic energy.

 $$\begin{gathered} E_k=\frac{3}{2}\left(\frac{R}{N_A}\right)\left(T\right) \\ {E}_k=\frac{3}{2}\left(\frac{\left(8.314\frac{J}{mol.K}\right)\left(1x{10}^{6}K\right)}{6.022x{10}^{23}\frac{atom}{mol}}\right) \\ {E}_k=2.0709\times {10}^{-17}\frac{J}{atom}H \end{gathered}$$

 

Therefore, the kinetic energy is $E_{kave}=2.0709\times {10}^{-17}\frac{J}{atom}H$ .

Step 3: Find the H atoms

b. The energy of one and one anti-H atom annihilating each other was used to calculate the number of atoms heated to 1.003106 K . We'll start by utilising Einstein's equation to get the energy. 

 

The number of atoms is then calculated using Avogadro's number.

$$\begin{gathered} E=m{c}^2 \\ E=2kg{\left(2.99792x{10}^8\frac{m}{s}\right)}^2\left(\frac{J}{m^2{s}^2}\right) \\ E=1.7975x{10}^{17}JHatoms \\ =\frac{1.7975x{10}^{17}J}{1kgH}\left(\frac{1kg}{1000g}\right)\left(\frac{1.0078gH}{1molH}\right)\left(\frac{1molH}{6.022x{10}^{23}atoms}\right)\left(\frac{1Hatom}{2.0709066\times {10}^{-17} J}\right) \end{gathered}$$

 

Therefore, the H atoms is  = 14525864.53 atoms.

Step 4: Find how much energy is generated

c. Balanced reaction is,

$4_1^{1}H{\to }_2^{4}He+2_1^0\beta$

 

The mass defect is calculated as:

 $$\begin{gathered} 4m=\left[4\right(1.007825amu\left)\right]-[4.0026amu+2(0.000549amu) \\ 4m=1.6645\times {10}^{-22} kg\backslash \end{gathered}$$

 $$\begin{gathered} \text{Energy }=4m{c}^2 \\ =1.6645x{10}^{-22} kg{\left(2.99792x{10}^8\frac{m}{s}\right)}^2\left(\frac{J}{kg{m}^2 s^2}\right) \\ =1.4960\times {10}^{-5} J \end{gathered}$$

 

 

Step 5: How much more energy is generated by the fusion

 

d. To compute for the energy generated for part b.

 $$\begin{gathered} Energy=\frac{1.7975x{10}^{17} J}{1kgH}\left(\frac{1 kg}{1000g}\right)\left(\frac{1.0078gH}{1molH}\right)\left(\frac{1molH}{6.022\times {10}^{23}atoms}\right) \\ =3.0082x{10}^{-10} J \end{gathered}$$

 $$\begin{gathered} \text{Energy generated }=1.4960\times {10}^{-5} J-3.0082\times {10}^{-10} J \\ =1.4960\times {10}^{-5} J \end{gathered}$$

 

Therefore, the generated energy is $1.4960\times {10}^{-5} J$by the fusion.

 

Step 6: Change the technology and produce 3 He

e. The equation is, 

$3_1^{1}H{\to }_2^{3}He{+}_1^0\beta$

 

The mass defect is calculated as:

$$\begin{gathered} 4m=\left[\left[3\right(1.007825amu\left)\right]-[3.01603amu+0.000549amu]=0.0069\frac{g}{molHe}\right. \\ 4m=0.0069\frac{g}{molHe}\left(\frac{1kg}{1000g}\right)\left(\frac{1molHe}{3molH}\right)\left(\frac{1molH}{6.022\times {10}^{23}atoms}\right)\left(1.4525903\times {10}^{7}Hatoms\right) \\ 4m=5.5479\times {10}^{-23} kg \end{gathered}$$

$$\begin{gathered} Energy=4m{c}^2 \\ =5.5479x{10}^{-23} kg{\left(2.99792x{10}^8\frac{m}{s}\right)}^2\left(\frac{J}{kg{m}^2 s^2}\right) \\ =4.9862\times {10}^{-6} \end{gathered}$$

 

Therefore, the captain of the Leinad ‘should not’ change the technology.