Question

In a period of $1.00\, s ,\, 5.00 \times 10^{23}$ nitrogen molecules strike a wall with an area of $8.00\, cm ^{2}$. Assume the molecules move with a speed of $300\, m / s$ 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 $N =5.00 \times 10^{23}, m_{0}=4.65 \times 10^{-26} kg$
$\Delta v =[300-(-300)]=2 \times 300\, m / s$
$=\frac{\left(5.00 \times 10^{23}\right)\left[\left(4.65 \times 10^{-26} kg \right) 2(300 \,m / s )\right]}{1.00 \,.s }$
$=14.0\, N$
The pressure is then
$P=\frac{\bar{F}}{A}=\frac{14.0 N }{8.00 \times 10^{-4} m ^{2}}$
$=17.4\, kPa$