Answer : The work done is, ![1.98\times 10^4J](https://tex.z-dn.net/?f=1.98%5Ctimes%2010%5E4J)
Explanation :
The given balanced chemical reaction is:
![N_2(g)+3H_2(g)\rightarrow 2NH_3(g)](https://tex.z-dn.net/?f=N_2%28g%29%2B3H_2%28g%29%5Crightarrow%202NH_3%28g%29)
When 4 moles of
react with 12 moles of
then it gives 8 moles of ![NH_3](https://tex.z-dn.net/?f=NH_3)
First we have to calculate the change in moles of gas.
Moles on reactant side = Moles of
+ Moles of ![H_2](https://tex.z-dn.net/?f=H_2)
Moles on reactant side = 4 + 12 = 16 moles
Moles on product side = Moles of ![NH_3](https://tex.z-dn.net/?f=NH_3)
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:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
where,
P = Pressure of gas = 1.0 atm
V = Volume of gas = ?
n = number of moles of gas = 8 mole
R = Gas constant = ![0.0821L.atm/mol.K](https://tex.z-dn.net/?f=0.0821L.atm%2Fmol.K)
T = Temperature of gas = ![25^oC=273+25=298K](https://tex.z-dn.net/?f=25%5EoC%3D273%2B25%3D298K)
Putting values in above equation, we get:
![1.0atm\times V=8mole\times (0.0821L.atm/mol.K)\times 298K](https://tex.z-dn.net/?f=1.0atm%5Ctimes%20V%3D8mole%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20298K)
![V=195.7L](https://tex.z-dn.net/?f=V%3D195.7L)
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 :
![w=-p\Delta V](https://tex.z-dn.net/?f=w%3D-p%5CDelta%20V)
where,
w = work done
p = pressure of the gas = 1.0 atm
= change in volume = -195.7 L
Now put all the given values in the above formula, we get:
![w=-p\Delta V](https://tex.z-dn.net/?f=w%3D-p%5CDelta%20V)
![w=-(1.0atm)\times (-195.7L)](https://tex.z-dn.net/?f=w%3D-%281.0atm%29%5Ctimes%20%28-195.7L%29)
![w=195.7L.artm=195.7\times 101.3J=19824.41J=1.98\times 10^4J](https://tex.z-dn.net/?f=w%3D195.7L.artm%3D195.7%5Ctimes%20101.3J%3D19824.41J%3D1.98%5Ctimes%2010%5E4J)
conversion used : (1 L.atm = 101.3 J)
Thus, the work done is, ![1.98\times 10^4J](https://tex.z-dn.net/?f=1.98%5Ctimes%2010%5E4J)