Question

In a period of $1.00s,5.00\times 10^{23}$ nitrogen molecules strike a wall with an area of $8.00c{m}^2$. Assume the molecules move with a speed of $300m/s$ and strike the wall head-on in elastic collisions. What is the pressure exerted on the wall? Note: The mass of one $N_2$ molecule is $4.65\times 10^{-26}kg$.

• A. 17.4 kPa
• B. 24.5 kPa
• C. 36.2 kPa
• D. 8.24 kPa

Answer 1

Answer: (A)

To find the pressure exerted by the nitrogen molecules, we first calculated the average force exerted by the molecules:

$\bar{F}=N{m}_0\frac{\Delta v}{\Delta t}$
Here $N=5.00\times 10^{23},m_0=4.65\times 10^{-26}kg$
$\Delta v=[300-(-300)]=2\times 300m/s$
$=\frac{\left(5.00\times 10^{23}\right)\left[\left(4.65\times 10^{-26}kg\right)2(300m/s)\right]}{1.00.s}$
$=14.0N$
The pressure is then
$P=\frac{\bar{F}}{A}=\frac{14.0N}{8.00\times 10^{-4}{m}^2}$
$=17.4kPa$