Answer : The value of work done for the system is, 922.8 J
Explanation :
First we have to calculate the moles of ![NaN_3](https://tex.z-dn.net/?f=NaN_3)
![\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DNaN_3%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DNaN_3%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DNaN_3%7D)
Molar mass of
= 65.01 g/mole
![\text{Moles of }NaN_3=\frac{16.3g}{65.01g/mole}=0.251mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DNaN_3%3D%5Cfrac%7B16.3g%7D%7B65.01g%2Fmole%7D%3D0.251mole)
Now we have to calculate the moles of nitrogen gas.
The balanced chemical reaction is,
![2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)](https://tex.z-dn.net/?f=2NaN_3%28s%29%5Crightarrow%202Na%28s%29%2B3N_2%28g%29)
From the balanced reaction we conclude that
As, 2 mole of
react to give 3 mole of ![N_2](https://tex.z-dn.net/?f=N_2)
So, 0.251 moles of
react to give
moles of ![N_2](https://tex.z-dn.net/?f=N_2)
Now we have to calculate the volume of nitrogen gas.
Using ideal gas equation:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
where,
P = Pressure of
gas = 1.00 atm
V = Volume of
gas = ?
n = number of moles
= 0.376 mole
R = Gas constant = ![0.0821L.atm/mol.K](https://tex.z-dn.net/?f=0.0821L.atm%2Fmol.K)
T = Temperature of
gas = ![22^oC=273+22=295K](https://tex.z-dn.net/?f=22%5EoC%3D273%2B22%3D295K)
Putting values in above equation, we get:
![1.00atm\times V=0.376mole\times (0.0821L.atm/mol.K)\times 295K](https://tex.z-dn.net/?f=1.00atm%5Ctimes%20V%3D0.376mole%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20295K)
![V=9.11L](https://tex.z-dn.net/?f=V%3D9.11L)
As initially no nitrogen was present. So,
Volume expanded = Volume of nitrogen evolved
Thus,
Expansion work = Pressure × Volume
Expansion work = 1.00 atm × 9.11 L
Expansion work = 9.11 L.atm
Conversion used : (1 L.atm = 101.3 J)
Expansion work = 9.11 × 101.3 = 922.8 J
Therefore, the value of work done for the system is, 922.8 J