At same temperature n2o4 is dissociated to 40 nd 50 t total pressure p1 and p2 atm respectively in no2. then the ratio of p1 and p2 is:______
The dissociation of N2O4 into NO2 can be represented by the following equation:
N2O4(g) ⇌ 2NO2(g)
Let the initial moles of N2O4 be x. Since N2O4 dissociates into 2 moles of NO2, the moles of NO2 formed will be 2x.
According to the problem, at the same temperature, the total pressure of the system when N2O4 is dissociated to 40 atm is p1, and when it is dissociated to 50 atm is p2.
Using the ideal gas law, we can relate the total pressure to the moles of the gases:
P = (nRT)/V
Since the volume remains constant, we can write the following equation:
Ptotal = P(N2O4) + P(NO2)
= 40 + 2x
= p1 (initial dissociation)
Similarly, for the second dissociation:
Ptotal = P(N2O4) + P(NO2)
= 50 + 2x
= p2 (final dissociation)
Now, we can find the ratio of p1 and p2:
p1/p2 = (40 + 2x)/(50 + 2x)
Since the moles of N2O4 dissociated remains the same for both dissociations:
40 = 50x/(50 + 2x)
Solving this equation, we get x = 10
Substitute x = 10 into the ratio of p1/p2 formula:
p1/p2 = (40 + 2(10))/(50 + 2(10))
p1/p2 = 60/70
p1/p2 = 6/7
Therefore, the ratio of p1 to p2 is 6:7.