# Solutions 2

## Q1.1

Calculate the mean free path and the binary number of collisions per liter per second between Ar atoms at 298 K and 1.00 atm. Use 3.62 Å as the collision diameter of the Ar molecules. Assume ideal gas behavior.

## S1.1

Calculate the mean free path

Step 1: Calculate $$\frac{N}{V}$$

$PV=NRT=\frac{N}{N_A}RT$

$$\dfrac{N}{V}=\dfrac{PN_A}{RT}=\dfrac{(1atm)(6.022\times 10^{23}mol^{-1})}{(0.08206L\,atm\,K^{-1}mol^{-1})(298K)}=2.46\times 10^{22}L^{-1}=2.46\times10^{25}m^{-3}$$

Step 2: Calculate the mean free path

$$\lambda=\dfrac{1}{\sqrt{2}\pi d^{2}(\dfrac{N}{V})}=\dfrac{1}{\sqrt{2}\pi (3.62\times 10^{-10}m)^{2}(2.46\times 10^{25}m^{-3})}=6.982\times 10^{-8}m$$

Calculate the binary number of collisions

Step 3: Calculate the average molecular speed

$$\bar{C}=\sqrt{\dfrac{8RT}{\pi M}}=\sqrt{\dfrac{8(8.3145JK^{-1}mol^{-1}(298K)}{\pi (39.948\times 10^{-3}\dfrac{kg}{mol})}}=397.419 m/s$$