Question: (a) The energy released in the explosion of 1.00 molof TNT is 340 mJ. The molar mass of TNT is. What weight of TNT is$\mathbf{0}\mathbf{.}\mathbf{227}\mathbf{}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$needed for an explosive release of $\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{}\mathit{J}$? (b) Can you carry that weight in a backpack, or is a truck or train required? (c) Suppose that in an explosion of a fission bomb, 0.080%of the fissionable mass is converted to released energy. What weight of fissionable material is needed for an explosive release of $\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{}\mathit{J}$? (d) Can you carry that weight in a backpack, or is a truck or train required?

One mol of TNT has a molar mass of $0.227kg$.

Energy released per unit mol$E_{released}=3.40MJ/mol$

Required energy to be released from the bomb is $1.8\times {10}^{14}J$.

TNT is still often utilised in the military and, less frequently, in industry. It is a component of many explosives and weapons, including grenades, bombs, and artillery shells. TNT is used in industry to make dyes, photographic chemicals, and for underwater blasting.

One mole of TNT releases 340 MJ of energy.

Required energy released in the bomb $1.8\times {10}^{14}J$.

No. of moles required to release this energy is$0.227kg$

$$\begin{gathered} n=\frac{1.8\times {10}^{14}J}{3.40\times {10}^{6}J/Mol} \\ =0.53\times {10}^{8}mol \end{gathered}$$

One mole weight 0.227 kg .

Therefore, $0.53\times {10}^{8}mol$weighs

$\left(0.53\times {10}^{8}mol\right)\left(0.227\frac{kg}{mol}\right)=1.2\times {10}^{7}kg$

Hence the weight of the TNT explosive is $1.2\times {10}^{7}kg$.

The mass of the TNT would be 12 million kilograms which would require a truck or train to carry.

The fission bomb converts 0.080% of mass into energy of$1.8\times {10}^{14}J$ . The mass-energy relation can be used to determine mass required.

$$\begin{gathered} m_o=\frac{E_{rel}}{0.0008c^2} \\ =\frac{1.8\times {10}^{14}J}{0.0008{\left(3\times {10}^8\right)}^2} \\ =2.5kg \end{gathered}$$

Hence the 2.5 kg of fission material is required to release $1.8\times {10}^{14}J$. This weight can be easily carried in a backpack.

The 2.5 kg of fission material is required to release$1.8\times {10}^{14}J$. This weight can be easily carried in a backpack