Answer : The the partial pressure of the nitrogen gas is 0.981 atm.
The total pressure in the tank is 2.94 atm.
Explanation :
The balanced chemical reaction will be:
![(CH_3)_2N_2H_2(l)+2N_4O_4(l)\rightarrow 3N_2(g)+4H_2O(g)+2CO_2(g)](https://tex.z-dn.net/?f=%28CH_3%29_2N_2H_2%28l%29%2B2N_4O_4%28l%29%5Crightarrow%203N_2%28g%29%2B4H_2O%28g%29%2B2CO_2%28g%29)
First we have to calculate the moles of dimethylhydrazine.
Mass of dimethylhydrazine = 150 g
Molar mass of dimethylhydrazine =60.104 g/mole
![\text{Moles of dimethylhydrazine}=\frac{\text{Mass of dimethylhydrazine}}{\text{Molar mass of dimethylhydrazine}}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20dimethylhydrazine%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20dimethylhydrazine%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20dimethylhydrazine%7D%7D)
![\text{Moles of dimethylhydrazine}=\frac{150g}{60.104g/mole}=2.49mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20dimethylhydrazine%7D%3D%5Cfrac%7B150g%7D%7B60.104g%2Fmole%7D%3D2.49mole)
Now we have to calculate the moles of
gas.
From the balanced chemical reaction we conclude that,
As, 1 mole of
react to give 3 moles of
gas
So, 2.49 mole of
react to give
moles of
gas
Now we have to calculate the partial pressure of nitrogen gas.
Using ideal gas equation :
![PV=nRT\\\\P_{N_2}=\frac{nRT}{V}](https://tex.z-dn.net/?f=PV%3DnRT%5C%5C%5C%5CP_%7BN_2%7D%3D%5Cfrac%7BnRT%7D%7BV%7D)
where,
P = Pressure of
gas = ?
V = Volume of
gas = 250 L
n = number of moles
gas = 7.47 mole
R = Gas constant = ![0.0821L.atm/mol.K](https://tex.z-dn.net/?f=0.0821L.atm%2Fmol.K)
T = Temperature of
gas = ![127^oC=273+127=400K](https://tex.z-dn.net/?f=127%5EoC%3D273%2B127%3D400K)
Putting values in above equation, we get:
![P_{N_2}=\frac{(7.47mole)\times (0.0821L.atm/mol.K)\times 400K}{250L}=0.981atm](https://tex.z-dn.net/?f=P_%7BN_2%7D%3D%5Cfrac%7B%287.47mole%29%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20400K%7D%7B250L%7D%3D0.981atm)
Thus, the partial pressure of the nitrogen gas is 0.981 atm.
Now we have to calculate the total pressure in the tank.
Formula used :
![P_{N_2}=X_{N_2}\times P_T](https://tex.z-dn.net/?f=P_%7BN_2%7D%3DX_%7BN_2%7D%5Ctimes%20P_T)
![P_T=\frac{1}{X_{N_2}}\times P_{N_2}](https://tex.z-dn.net/?f=P_T%3D%5Cfrac%7B1%7D%7BX_%7BN_2%7D%7D%5Ctimes%20P_%7BN_2%7D)
![P_T=\frac{1}{(\frac{n_{N_2}}{n_T})}\times P_{N_2}](https://tex.z-dn.net/?f=P_T%3D%5Cfrac%7B1%7D%7B%28%5Cfrac%7Bn_%7BN_2%7D%7D%7Bn_T%7D%29%7D%5Ctimes%20P_%7BN_2%7D)
![P_T=\frac{n_{T}}{n_{N_2}}\times P_{N_2}](https://tex.z-dn.net/?f=P_T%3D%5Cfrac%7Bn_%7BT%7D%7D%7Bn_%7BN_2%7D%7D%5Ctimes%20P_%7BN_2%7D)
where,
= total pressure = ?
= partial pressure of nitrogen gas = 0.981 atm
= moles of nitrogen gas = 3 mole (from the reaction)
= total moles of gas = (3+4+2) = 9 mole (from the reaction)
Now put all the given values in the above formula, we get:
![P_T=\frac{9mole}{3mole}\times 0.981atm=2.94atm](https://tex.z-dn.net/?f=P_T%3D%5Cfrac%7B9mole%7D%7B3mole%7D%5Ctimes%200.981atm%3D2.94atm)
Thus, the total pressure in the tank is 2.94 atm.