Answer :
Answer : The value of work done for the system is 581.462 J
Explanation :
First we have to calculate the moles of [tex]NaN_3[/tex]
[tex]\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}[/tex]
Molar mass of [tex]NaN_3[/tex] = 65.01 g/mole
[tex]\text{Moles of }NaN_3=\frac{10.3g}{65.01g/mole}=0.158mole[/tex]
Now we have to calculate the moles of nitrogen gas.
The balanced chemical reaction is,
[tex]2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]NaN_3[/tex] react to give 3 mole of [tex]N_2[/tex]
So, 0.158 moles of [tex]NaN_3[/tex] react to give [tex]\frac{0.158}{2}\times 3=0.237[/tex] moles of [tex]N_2[/tex]
Now we have to calculate the volume of nitrogen gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = Pressure of [tex]N_2[/tex] gas = 1.00 atm
V = Volume of [tex]N_2[/tex] gas = ?
n = number of moles [tex]N_2[/tex] = 0.237 mole
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of [tex]N_2[/tex] gas = [tex]22^oC=273+22=295K[/tex]
Putting values in above equation, we get:
[tex]1.00atm\times V=0.237mole\times (0.0821L.atm/mol.K)\times 295K[/tex]
[tex]V=5.74L[/tex]
As initially no nitrogen was present. So,
Volume expanded = Volume of nitrogen evolved
Thus,
Expansion work = Pressure × Volume
Expansion work = 1.00 atm × 5.74 L
Expansion work = 5.74 L.atm
Conversion used : (1 L.atm = 101.3 J)
Expansion work = 5.74 × 101.3 = 581.462 J
Therefore, the value of work done for the system is 581.462 J