# Pressure of Gas Exerted on Wall

• pkpiotr517
In summary, the problem involves finding the pressure exerted on a wall by a certain number of nitrogen molecules colliding with it in a given time period. Using the formulas PV=nRT and Vrms^2=3RT/M, we can calculate the temperature and molar mass of the nitrogen gas. Then, using the formula P=nRT/V and the concept of elastic collisions, we can find the pressure exerted on the wall. The final answer is 12,290.9 Pascals.
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?

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

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?

Great! You got it!

ehild

Thanks so much!

## What is the definition of "Pressure of Gas Exerted on Wall"?

The pressure of gas exerted on a wall is the force per unit area that the gas molecules exert on the surface of the wall.

## How is the pressure of gas exerted on a wall calculated?

The pressure of gas exerted on a wall can be calculated using the formula P = F/A, where P is the pressure, F is the force exerted by the gas molecules, and A is the area of the wall.

## What factors affect the pressure of gas exerted on a wall?

The pressure of gas exerted on a wall is affected by the number of gas molecules present, the temperature of the gas, and the volume of the container holding the gas.

## Why does the pressure of gas exerted on a wall increase with an increase in temperature?

According to the ideal gas law, an increase in temperature leads to an increase in the average kinetic energy of gas molecules. This results in an increase in the number of collisions between gas molecules and the walls of the container, leading to an increase in pressure.

## How does changing the volume of the container affect the pressure of gas exerted on a wall?

As the volume of the container decreases, the gas molecules have less space to move around. This leads to an increase in the number of collisions with the wall, resulting in an increase in pressure. Conversely, as the volume of the container increases, the pressure decreases.

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