somebigguy

05-29-2005, 08:59 PM

The whole article is here: http://www.hawaii.indymedia.org/news/2003/12/3961.php

Here is one of the finer points:

The Unexplored Option -- Explosives.

We will consider the explosive amatol which is a mixture of trinitrotoluene (TNT) and ammonium nitrate (AN).

Trinitrotoluene (left) has the chemical composition C7H5(NO2)3 and ammonium nitrate (right) has the chemical composition NH4NO3.

We choose the TNT to AN molar ratio in the amatol mixture to be 4 : 42.

This translates to a 4 x 227 : 42 x 80 = 908 : 3360 = 21 : 79 ratio by weight.

In this case the explosion proceeds according to the equation:

4 C7H5(NO2)3 + 42 NH4NO3 => 28 CO2 + 94 H2O + 48 N2

From the equation we see that 4 moles of TNT and 42 moles of AN produces 28 + 94 + 48 = 170 moles of gaseous product.

At standard temperature and pressure, 170 moles of gas occupies 170 x 22.4 = 3,808 liters. This is the volume that the explosion products would occupy if, after the explosion, they were cooled to 25 C (with the assumption that the water remains a vapor). In order to calculate the volume of the hot gaseous products generated by the explosion, we initially need to know the amount of heat released by the explosion of 4 moles of TNT and 42 moles of AN. We also need to know the amount of heat required to raise each of the gases, by one degree K. That is, we need to know the enthalpy of explosion ΔexplosionH and the heat capacities Cp (also known as specific heats) of each of the gases.

ΔexplosionH = - [4 Δf H°solid(TNT) + 21 Δf H°solid(AN)] + [28 Δf H°gas(CO2) + 94 Δf H°gas(H2O) + 48 Δf H°gas(N2)]

= - [4(-60) + 21(-365.5)] + [28(-393.5) + 94(-242) + 48(0.00)]

= 240 + 7,675.5 - 11,018 - 22,748

= - 25,850 kJ per 4 moles of TNT and 42 moles of AN.

Here we have used the following facts:

Δf H°gas(H2O) = -242 kJ/mol

Δf H°gas(CO) = -110.5 kJ/mol

Δf H°gas(CO2) = -393.5 kJ/mol

Δf H°solid(TNT) = -60 kJ/mol

Δf H°solid(AN) = -365.5 kJ/mol

The heat capacities Cp for the gases involved are:

Cpgas (N2) = 28.87 J/mol*K

Cpgas (O2) = 28.91 J/mol*K

Cpgas (H2O) = 30.43 J/mol*K

Cpgas (CO2) = 37.12 J/mol*K

Since the heat capacities are all approximately 30 J/mol*K, we will assume this value for all the gases, i.e., we assume:

Cpgas (all relevant gases) = 30 J/mol*K

So, by assumption, 30 joules of energy will raise the temperature of one mole of the gaseous product (of the explosion) by 1 degree K.

Summarizing from earlier, we have that 4 moles of TNT and 42 moles of AN,

1. produces 170 moles of gaseous product and

2. liberates 25,850 kJ of energy.

We will assume that the original 170 moles of explosion products mix with N moles of air.

This 170 + N mole mixture of gases is initially assumed to be at the temperature T0 = 298 degrees K (25 C).

This 170 + N mole mixture of gases has an initial volume of V0 = 22.4 (170 + N) liters.

We calculate the increase in temperature ΔT of this 170 + N moles of gases after being heated by the 25,850 kJ of energy released by the explosion of the 4 moles of TNT and 42 moles of AN.

Now 30 joules raises the temperature of one mole of the gases by 1 degree K.

Hence, 25,850,000 joules raises the temperature of the 170 + N moles by

ΔT = 25,850,000/(30 x (170 + N)).

We now calculate the volume V1 of this 170 + N moles of gas after being heated by the explosion to the temperature

T1 = T0 + ΔT

From the ideal gas law we have that V1 / V0 = T1 / T0. Rearranging we obtain

V1 = V0 (T1 / T0) = V0 (T0 + ΔT) / T0 = V0 + V0 ΔT / T0. On substituting we obtain

V1 = V0 + 22.4 x (170 + N) x 25,850,000 / (30 x (170 + N) x 298) = V0 + 22.4 x 25,850,000 / (30 x 298) = V0 + 64,770 liters.

Hence, the increase in volume of the mixture of gases ΔV = V1 - V0 = 64,770 liters.

So, summing up, the explosion of 4 moles of TNT and 42 moles of AN produce 64,770 + 22.4 x 170 = 64,770 + 3,808 = 68,578 liters of hot gases.

That is, the explosion of 4,268 grams of amatol produces 68,578 liters of hot gases.

That is, the explosion of one kilogram of amatol produces 68,578 x 1,000 / 4,268 = 16,068 liters of hot gaseous product.

Hence the 200,000,000 liter expansion calculated by Hoffman can be explained by the detonation of

200,000,000/16,068 = 12,447 kg = 12.5 tonnes (14 tons) of the high explosive amatol.

Summary

The 200,000,000 liter expansion calculated by Hoffman can be explained by the detonation of 12.5 tonnes (14 tons) of the high explosive amatol.

Here is one of the finer points:

The Unexplored Option -- Explosives.

We will consider the explosive amatol which is a mixture of trinitrotoluene (TNT) and ammonium nitrate (AN).

Trinitrotoluene (left) has the chemical composition C7H5(NO2)3 and ammonium nitrate (right) has the chemical composition NH4NO3.

We choose the TNT to AN molar ratio in the amatol mixture to be 4 : 42.

This translates to a 4 x 227 : 42 x 80 = 908 : 3360 = 21 : 79 ratio by weight.

In this case the explosion proceeds according to the equation:

4 C7H5(NO2)3 + 42 NH4NO3 => 28 CO2 + 94 H2O + 48 N2

From the equation we see that 4 moles of TNT and 42 moles of AN produces 28 + 94 + 48 = 170 moles of gaseous product.

At standard temperature and pressure, 170 moles of gas occupies 170 x 22.4 = 3,808 liters. This is the volume that the explosion products would occupy if, after the explosion, they were cooled to 25 C (with the assumption that the water remains a vapor). In order to calculate the volume of the hot gaseous products generated by the explosion, we initially need to know the amount of heat released by the explosion of 4 moles of TNT and 42 moles of AN. We also need to know the amount of heat required to raise each of the gases, by one degree K. That is, we need to know the enthalpy of explosion ΔexplosionH and the heat capacities Cp (also known as specific heats) of each of the gases.

ΔexplosionH = - [4 Δf H°solid(TNT) + 21 Δf H°solid(AN)] + [28 Δf H°gas(CO2) + 94 Δf H°gas(H2O) + 48 Δf H°gas(N2)]

= - [4(-60) + 21(-365.5)] + [28(-393.5) + 94(-242) + 48(0.00)]

= 240 + 7,675.5 - 11,018 - 22,748

= - 25,850 kJ per 4 moles of TNT and 42 moles of AN.

Here we have used the following facts:

Δf H°gas(H2O) = -242 kJ/mol

Δf H°gas(CO) = -110.5 kJ/mol

Δf H°gas(CO2) = -393.5 kJ/mol

Δf H°solid(TNT) = -60 kJ/mol

Δf H°solid(AN) = -365.5 kJ/mol

The heat capacities Cp for the gases involved are:

Cpgas (N2) = 28.87 J/mol*K

Cpgas (O2) = 28.91 J/mol*K

Cpgas (H2O) = 30.43 J/mol*K

Cpgas (CO2) = 37.12 J/mol*K

Since the heat capacities are all approximately 30 J/mol*K, we will assume this value for all the gases, i.e., we assume:

Cpgas (all relevant gases) = 30 J/mol*K

So, by assumption, 30 joules of energy will raise the temperature of one mole of the gaseous product (of the explosion) by 1 degree K.

Summarizing from earlier, we have that 4 moles of TNT and 42 moles of AN,

1. produces 170 moles of gaseous product and

2. liberates 25,850 kJ of energy.

We will assume that the original 170 moles of explosion products mix with N moles of air.

This 170 + N mole mixture of gases is initially assumed to be at the temperature T0 = 298 degrees K (25 C).

This 170 + N mole mixture of gases has an initial volume of V0 = 22.4 (170 + N) liters.

We calculate the increase in temperature ΔT of this 170 + N moles of gases after being heated by the 25,850 kJ of energy released by the explosion of the 4 moles of TNT and 42 moles of AN.

Now 30 joules raises the temperature of one mole of the gases by 1 degree K.

Hence, 25,850,000 joules raises the temperature of the 170 + N moles by

ΔT = 25,850,000/(30 x (170 + N)).

We now calculate the volume V1 of this 170 + N moles of gas after being heated by the explosion to the temperature

T1 = T0 + ΔT

From the ideal gas law we have that V1 / V0 = T1 / T0. Rearranging we obtain

V1 = V0 (T1 / T0) = V0 (T0 + ΔT) / T0 = V0 + V0 ΔT / T0. On substituting we obtain

V1 = V0 + 22.4 x (170 + N) x 25,850,000 / (30 x (170 + N) x 298) = V0 + 22.4 x 25,850,000 / (30 x 298) = V0 + 64,770 liters.

Hence, the increase in volume of the mixture of gases ΔV = V1 - V0 = 64,770 liters.

So, summing up, the explosion of 4 moles of TNT and 42 moles of AN produce 64,770 + 22.4 x 170 = 64,770 + 3,808 = 68,578 liters of hot gases.

That is, the explosion of 4,268 grams of amatol produces 68,578 liters of hot gases.

That is, the explosion of one kilogram of amatol produces 68,578 x 1,000 / 4,268 = 16,068 liters of hot gaseous product.

Hence the 200,000,000 liter expansion calculated by Hoffman can be explained by the detonation of

200,000,000/16,068 = 12,447 kg = 12.5 tonnes (14 tons) of the high explosive amatol.

Summary

The 200,000,000 liter expansion calculated by Hoffman can be explained by the detonation of 12.5 tonnes (14 tons) of the high explosive amatol.