Pressure of Gas Exerted on Wall

[pkpiotr517]

In a period of 1.1 s, 5.0 x 10^23 nitrogen molecules strike a wall of area 9.0 cm^2 . If the molecules move at 260 m/s and strike the wall head on in a perfectly elastic collision, find the pressure exerted on the wall. (The mass of one N2 molecule is 4.68 10-26 kg.)




Molar mass=mass of gas/mols of gas
n=N/N0
PV=nRT
Vrms^2=(3RT)/M




I used PV=nRT, solved for P, and found temperature by using the formula Vrms^2=3RT/M, solved for T.

So to clarify:

Vrms^2=3RT/M
T=Vrms^2(M) / 3R
T=( (260m/s )^2*(5.63e-26 kg/mol) ) / ( 3(8.31 J/(mol*K)) )
T=1.53e-22 K

(I found M by dividing the mass of the gas, by the number of moles ) (and I found the number of moles by dividing the number of molecules by Avogadro's constant)

To clarify this...
n=N/N0
n=5e23 molecules/6.02e23 moles

n=8.305e-1 moles

then I placed this in ...\/... to find the Molar Mass (M)

Molar Mass = Kg/Moles
M=4.68e-26 kg/8.305e-1 moles

M=5.634e-26
......... ...

Once I found the Temperature, I tried to find the pressure.

I used the formula: PV=nRT
Solved for P
P=nRT/V

Plugged everything in

P=( (8.305e-1 moles) (8.31 J/(mol*K)) (1.5279e-22 K) ) / (V)

I don't Know what the Volume should be or is... so I cannot solve for P

What am I doing wrong?

 

[ehild]

Read the problem again. You are given the number of molecules colliding with a wall in 1.1 s. It is not the number of molecules in the vessel. The mass of one nitrogen molecule is given, not the molar mass of the nitrogen. And it is said that the collision of the molecules is perfectly elastic with the wall. Treat the molecules as balls hitting a wall. What do you know about elastic collision?

ehild

 

[pkpiotr517]

So elastic collisions end up being it not sticking together...

oooooohohhhh okay...
J=Ft= change p

F=changep/t

f=m(vf-vi)/t

p=m(vf-vi)/t / a
I think I got it...

Is it 12,290.9 Pascals?

 

[ehild]

Great! You got it!

ehild

 

[pkpiotr517]

Thanks so much!