Solve "What Equation to Use" for Homework Problem

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To calculate the pressure exerted by carbon dioxide molecules on a wall, the relevant equation involves the change in momentum of the gas molecules upon collision. The pressure is defined as P = F/A, where F is the force and A is the area. The force can be derived from the momentum change, considering that the molecules bounce off the wall, which doubles the momentum change. After performing the calculations, the final pressure in Pascals was determined to be 65773.8, indicating a correct understanding of the principles involved. The discussion emphasizes the importance of correctly applying momentum concepts to derive the pressure from molecular impacts.
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Homework Statement



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 ??
 
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Start at the beginning...

what is the definition of pressure?
 
well iit can't be P=nRT/V
 
zeshkani said:
well iit can't be P=nRT/V

That's correct... that is not the definition of pressure.. but merely an important equation that is applicable in certain situations.
 
well can't 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
 
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?
 
ok well i can't 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 ??
 
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.
 
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
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?
 
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  • #11
And before someone stomps me here, yes, the real expression is F=dP/dt. It's a rate.
 
  • #12
well then iam just confued now
 
  • #13
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
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
Yes. That looks better.
 
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