Answer :
Answer : The work done is, [tex]1.98\times 10^4J[/tex]
Explanation :
The given balanced chemical reaction is:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
When 4 moles of [tex]N_2[/tex] react with 12 moles of [tex]H_2[/tex] then it gives 8 moles of [tex]NH_3[/tex]
First we have to calculate the change in moles of gas.
Moles on reactant side = Moles of [tex]N_2[/tex] + Moles of [tex]H_2[/tex]
Moles on reactant side = 4 + 12 = 16 moles
Moles on product side = Moles of [tex]NH_3[/tex]
Moles on reactant side = 8 moles
Change in moles of gas = 16 - 8 = 8 moles
Now we have to calculate the change in volume of gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = Pressure of gas = 1.0 atm
V = Volume of gas = ?
n = number of moles of gas = 8 mole
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of gas = [tex]25^oC=273+25=298K[/tex]
Putting values in above equation, we get:
[tex]1.0atm\times V=8mole\times (0.0821L.atm/mol.K)\times 298K[/tex]
[tex]V=195.7L[/tex]
As the number of moles of gas decreased. So, the volume also deceased. Thus, the volume of gas will be, -195.7 L
Now we have to calculate the work done.
Formula used :
[tex]w=-p\Delta V[/tex]
where,
w = work done
p = pressure of the gas = 1.0 atm
[tex]\Delta V[/tex] = change in volume = -195.7 L
Now put all the given values in the above formula, we get:
[tex]w=-p\Delta V[/tex]
[tex]w=-(1.0atm)\times (-195.7L)[/tex]
[tex]w=195.7L.artm=195.7\times 101.3J=19824.41J=1.98\times 10^4J[/tex]
conversion used : (1 L.atm = 101.3 J)
Thus, the work done is, [tex]1.98\times 10^4J[/tex]