About 1 of every 3300 water molecules contains one deuterium atom. If all the de

Question

About 1 of every 3300 water molecules contains one deuterium atom. If all the deuterium nuclei in 28 L of water are fused in pairs according to the D-D fusion reaction (below), how much energy is liberated?^2H +^2H rightarrow^3He + n + 3.27 Mev ________J Burning liquid ethane produces about 2.70 times 10^7 J/L. Calculate the energy liberated from the burning of 28 L of liquid ethane. __________ J Compare this answer with the energy obtainable from the fusion of the deuterium in 28 L of water. E_liquid ethane/E_fusion = ___________

Explanation / Answer

First we need to calculate the number of water molecules in 1 L of water. We know that one mole of a substance contains Avogadro number of molecules.

The molecular weight of water is 18.015 g/mole.

And 1L of water = 1000 g

So, number of moles in 1000 g of water = 1000/18.015 = 55.5093 moles

Number of molecules in 1 L of water = 55.5093* 6.02214*1023 = 3.3428*1025

Number of molecules in 28 L of water = 3.3428*1025 * 28 = 9.3599*1026

Now, 1 of every 3300 water molecules is deuterium. So,

Number of deuterium molecules in 28 L of water = 9.3599*1026/3300 = 2.836*1023

(a) Fusing 2 deuterium nuclei liberates 3.27 MeV energy.

     So, 2.836*1023 nuclei will give 1/2*(2.836*1023 * 3.27) = 4.637*1023 MeV = 4.637*1023 * 1.602*10-19* 106 = 7.429*1010 J

      (1eV = 1.602*10-19 J)

(b) 1L of ethane produces 2.7*107 J

   28L of ethane will produce   2.7*107 *28 = 7.560*108J

(c) Eliquid ethane/Efusion = (7.560*108)/(7.429*1010) = 1.017*10-2

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