Respuesta :
The change in pressure in a sealed 10.0L vessel is 5.28 atm
calculation
The pressure is calculated using the ideal gas equation
That is P=n RT
where;
P (pressure)= ?
v( volume) = 10.0 L
n( number of moles) which is calculated as below
write the equation for decomposition of NH₄NO₂
NH₄NO₂ → N₂ +2H₂O
Find the moles of NH₄NO₂
moles = molarity x volume in liters
= 2.40 l x 0.900 M =2.16 moles
Use the mole ratio to determine the moles of N₂
that is from equation above NH₄NO₂:N₂ is 1:1 therefore the moles of N₂ is also =2.16 moles
R(gas constant) =0.0821 l.atm/mol.K
T(temperature) = 25° c into kelvin = 25 +273 =298 K
make p the subject of the formula by diving both side by V
P = nRT/V
p ={ (2.16 moles x 0.0821 L.atm/mol.K x 298 K) /10.0 L} = 5.28 atm.
Answer : The pressure of the gas will be, 5.285 atm
Solution : Given,
Volume of [tex]N_2[/tex] gas = 10 L
Temperature of gas = [tex]25^oC=273+25=298K[/tex] [tex](0^oC=273K)[/tex]
Volume of [tex]NH_4NO_2[/tex] = 2.40 L
Molarity of the ammonium nitrate solution = 0.9 M
First we have to calculate the moles of ammonium nitrate.
[tex]\text{ Moles of }NH_4NO_2=\text{ Molarity of }NH_4NO_2\times \text{ Volume of }NH_4NO_2[/tex]
[tex]\text{ Moles of }NH_4NO_2=(0.9mole/L)\times (2.40L)=2.16moles[/tex]
The balanced chemical reaction is,
[tex]NH_4NO_2\rightarrow N_2+2H_2O[/tex]
From the balanced reaction, we conclude that
1 mole of [tex]NH_4NO_2[/tex] decomposes to give 1 mole of [tex]N_2[/tex] gas
2.16 moles of [tex]NH_4NO_2[/tex] decomposes to give 2.16 moles of [tex]N_2[/tex] gas
Now we have to calculate the pressure of the gas.
using ideal gas equation,
[tex]PV=nRT[/tex]
where,
P = pressure of the gas
V = volume of the gas
T= temperature of the gas
n = number of moles of gas
R = gas constant = 0.0821 Latm/moleK
Now put all the given values in ideal gas equation, we get pressure of the gas.
[tex]P=\frac{nRT}{V}=\frac{(2.16mole)\times (0.0821Latm/moleK)\times (298K)}{10L}=5.285atm[/tex]
Therefore, the pressure of the gas will be, 5.285 atm
