# Homework Help: What equation to use

1. Sep 7, 2007

### zeshkani

1. The problem statement, all variables and given/known data

If 1.0 x 10^23 carbon dioxide molecules strike 1.0 cm2 of wall per second at a 90o angle to the wall when moving toward it with a speed of 45,000 cm/s, what pressure (in atm) do they exert on the wall?

does anybody know what equation to use for this problem ??

2. Sep 7, 2007

### robphy

Start at the beginning....

what is the definition of pressure?

3. Sep 7, 2007

### zeshkani

well iit can't be P=nRT/V

4. Sep 7, 2007

### robphy

That's correct... that is not the definition of pressure.. but merely an important equation that is applicable in certain situations.

5. Sep 7, 2007

### zeshkani

well cant i just convert the molecules into grams, then into kilograms then use F=MA
to get force then convert it into atm, but what do i do with the 1.0cm2 wall, those that even matter
oh and the definition of pressure is P=F/A

6. Sep 7, 2007

### Dick

F/A is right. I think robphy would now ask you if you remember that force is also the time rate of change of momentum. Since momentum=mv and F=ma and a=dv/dt. What's the time rate of change of momentum for your collection of molecules?

7. Sep 11, 2007

### zeshkani

ok well i cant use F=ma since the problem has a velocity and not acceleration, but momentum can be used in this way, i hope iam right :)

so F = P/t
So for P =(CO2 in Kg * 450m/s) then F= (CO2Kg*450m/s)/(1sec) then when i get Force put it bake into P=F/A
so then

P= (CO2Kg*450m/s)/(1sec)/(1.00cm^2) those this seem right ??

8. Sep 11, 2007

### Dick

Does the CO2 bounce off the wall or stick to it? There's a factor of two in the momentum change between the two cases since after the bounce they are now going 450m/sec AWAY from the wall. I'm assuming you mean them to bounce.

9. Sep 11, 2007

### zeshkani

well since its a gas it would bounce of the wall and this is the question again

"If 1.0 x 10^23 carbon dioxide molecules strike 1.0 cm2 of wall per second at a 90o angle to the wall when moving toward it with a speed of 45,000 cm/s, what pressure (in atm) do they exert on the wall?"

and this is what i got when i sloved it

P=(.007308Kg of CO2)(450m/s)
P= 3.28869
so then

F= 3.288869/1sec
F= 3.28869
and fianlly

P= 3.28869/.0001m^2
i converted the area of 1.00cm^2 to .0001m^2 and used this as area instead does this seem right
and my finaly answer in Pressure in Pascal = 32886.91

10. Sep 11, 2007

### Dick

Your numbers seem right, except you weren't listening. F is not equal to P/t. It's equal to delta(P)/delta(t). P is +mv before it hits the wall and -mv after it hits the wall. What's the delta P?

Last edited: Sep 12, 2007
11. Sep 12, 2007

### Dick

And before someone stomps me here, yes, the real expression is F=dP/dt. It's a rate.

12. Sep 12, 2007

### zeshkani

well then iam just confued now

13. Sep 12, 2007

### Dick

The force exerted by the gas on a patch of the wall is equal to the rate at which that patch of wall is changing the momentum of the gas molecules. Your answer is off by a factor of two, because to change P to -P I need a delta(P) of 2*P. I seem to be just repeating myself...

14. Sep 12, 2007

### zeshkani

so would be something like this if iam understanding right

and this is what i got when i sloved it

P=(.007308Kg of CO2)(450m/s)*2
P= 6.57738
so then

F= 6.57738/1sec
F= 6.57738
and fianlly

P= 6.57738/.0001m^2
i converted the area of 1.00cm^2 to .0001m^2 and used this as area instead does this seem right
and my finaly answer in Pressure in Pascal = 65773.8

15. Sep 13, 2007

### Dick

Yes. That looks better.