Respuesta :
Answer: The value of work for the system is -935.23 J
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of sodium azide = 16.5 g
Molar mass of sodium azide = 65 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of sodium azide}=\frac{16.5g}{65g/mol}=0.254mol[/tex]
The given chemical equation follows:
[tex]2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)[/tex]
By Stoichiometry of the reaction:
2 moles of sodium azide produces 3 moles of nitrogen gas
So, 0.254 moles of sodium azide will produce = [tex]\frac{3}{2}\times 0.254=0.381mol[/tex] of nitrogen gas
To calculate volume of the gas given, we use the equation given by ideal gas, which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 1.00 atm
V = Volume of the gas = ?
T = Temperature of the gas = [tex]22^oC=[22+273]K=295K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of nitrogen gas = 0.381 moles
Putting values in above equation, we get:
[tex]1.00atm\times V=0.381mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 295K\\\\V=\frac{0.381\times 0.0821\times 295}{1.00}=9.23L[/tex]
To calculate the work done for expansion, we use the equation:
[tex]W=-P\Delta V[/tex]
We are given:
P = pressure of the system = [tex]1atm=1.01325\times 10^5Pa[/tex] (Conversion factor: 1 atm = 101325 Pa)
[tex]\Delta V[/tex] = change in volume = [tex]9.23L=9.23\times 10^{-3}m^3[/tex] (Conversion factor: [tex]1m^3=1000L[/tex] )
Putting values in above equation, we get:
[tex]W=-1.01325\times 10^5Pa\times 9.23\times 10^{-3}m^3\\\\W=-935.23J[/tex]
Hence, the value of work for the system is -935.23 J
The work done is -919 J.
The equation of the reaction is;
2 NaN 3 ( s ) ⟶ 2 Na ( s ) + 3 N 2 ( g )
Number of moles of NaN3 = 16.5 g/65 g/mol = 0.25 moles
If 2 moles of NaNa3 yields 3 moles of N2
0.25 moles of NaN3 yields 0.25 moles × 3 moles/2 moles
= 0.375 moles of N2
We need to find the volume change using;
PV = nRT
P = 1.00 atm
V = ?
n = 0.375 moles of N2
R = 0.082 atmLK-1mol-1
T = 22 ∘ C + 273 = 295 K
V = nRT/P
V = 0.375 × 0.082 × 295/ 1.00
V = 9.07 L
Recall that during expansion the gas does work. Work done by the gas is;
W = -PΔV
W =-( 1 atm × 9.07 L)
W = -9 atmL
Again;
1 L atm = 101.325 J
So,
-9 atmL = -9 atmL × 101.325 J/1 L atm
= -919 J
The work done is -919 J.
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