# Finding z @ STP

1. Sep 20, 2007

### jbowers9

1. The problem statement, all variables and given/known data
Beginning Calculations in Physical Chemistry
Barry R. Johnson & Stephen K. Scott

Calculate the average number of collisions per second between $$N_{2}$$ and $$O_{2}$$ molecules in air @ 298K and @ a pressure of 1 X $$10^5$$ in a room of volume V = 100$$m^3$$. (Assume air has the composition 0.8$$N_{2}$$ & 0.2$$O_{2}$$)

2. Relevant equations

I follow the solution in the text, however, do you need the given volume to calculate z?
Can't it be calculated from z = $$\sigma$$ $$\overline{c}_{rel}$$ $$\eta$$ using p/kT?

3. The attempt at a solution

Last edited: Sep 20, 2007
2. Sep 21, 2007

### Staff: Mentor

To get the collision density, n (collsions/ cm3-s), i.e. collisions per unit volume per second, one does not need the volume.

To get the total number of collisions/sec, N, in the room as called for in the problem, one needs N = nV, where V is the total volume.

3. Sep 21, 2007

### jbowers9

zzzzzz @ STP

Alrightee then. When I calculate z, I get an answer on the order of magnitude of $$10^9$$, and the book/authors get something on the order of $$10^38$$. What is the descrepancy here?

Last edited: Sep 21, 2007
4. Sep 21, 2007

### Staff: Mentor

5. Sep 22, 2007

### jbowers9

...will the real z @STP please stand up

If
z= $$\sigma_{{N_2}{O_2}}$$ $$\overline{c}_{rel}$$ $$\eta$$

I can't figure out how to make the above read sigma c rel eta. I keep getting the error above even though the syntax is the same as below

and $$\sigma_{{N_2}{O_2}}$$ = 0.415$$nm^2$$

$$\mu_{{N_2}{O_2}}$$ = 14.93gs/mol

gives $$\overline{c}_{rel}$$ = 650m/s

$$\eta_{{N_2}{O_2}}$$ = N/V =n$$N_a$$/V = n$$N_a$$p/nRT

and after canceling out Avogadro's number gives
$$\eta_{{N_2}{O_2}}$$ = p/kT

So z = (0.415 x $$10^{-18}$$)(650)(2.43 x $$10^{25}$$)
z= 6.55 x $$10^9$$ sec-1

The authors get z = 2.55 x $$10^{36}$$ sec-1

They calculate $$Z__{{N_2}{O_2}}$$ = $$\sigma_{{N_2}{O_2}}$$ $$\overline{c}_{rel}$$ $$N_a{^2}$$ [$$N_2$$][$$O_2$$] = 2.55 x $$10^{34}$$ sec-1 m-3
and multiply this value by the volume 100$$m^3$$ to get 2.55 x $$10^{36}$$ sec-1.

So where is my error?

Last edited: Sep 22, 2007
6. Sep 24, 2007

### jbowers9

z @ STP

I still can't figure out why there is such a large difference in the book's calculated value for z and my value. They do emphasize "total" collisions but this is because it is a mixture of gases.