Question

physical chemistry problem

Consider a sample of CO₂ a pressure of I atm. For mass is 0.044 kg/mol. gas the at a con temperature of 298 K and 0 = 5.21019 m² and the molar (a) calculate the average velocity of the molecules in the gas. 6) Calculate the mean free path of the coa molecules. (c) Calculate the total number of colisions Zil of the gas molectes.

Answer 1

ANSWER:

Average velocity of the molecules

• the average velocity (v) is defined as

$\bar{v}=\sqrt{\frac{8RT}{\pi M}}$

where

• R = gas constant = 8.314 J/mol.K,
• T = temperature of gas = 298 K
• M = molar mass of gas = 0.044 kg/mol

then,

$\bar{v}=\sqrt{\frac{8\times 8.314\, \frac{J}{mol.K}\times 298\: K}{3.142 \times 0.044\: kg/mol}}=\sqrt{143388.30\frac{J}{kg}}=\sqrt{143388.30\frac{\frac{kg.m^{2}}{s^{2}}}{kg}}$

$\bar{v}=\sqrt{143388.30 \frac{m^{2}}{s^{2}}}=\mathbf{378.67\: m/s}$

Mean free path of CO₂ molecules

• The mean free path is defined as

$\lambda =\frac{1}{\sqrt{2}\pi \sigma^{2} n_{v}}$

where

• σ² = square of collision diameter = 5.2x10^-19 m²,
• n_v = number of molecules per unit of volume
  • This is defined as

$n_{v} =\frac{nN_{A}}{V}$

• where n = moles of gas, N_A = Avogadro's number =6.02x10²³ molecules/mol and V = volume
• Replacing V by ideal gas law: V = nRT/P

$n_{v} =\frac{nN_{A}}{\frac{nRT}{P}}=\frac{N_{A}P}{RT}$

where R = gas constant = 0.082 atm.L/mol.K, T = temperature of gas = 298 K and P = Pressure of gas = 1 atm

• Then,

$n_{v} =\frac{6.02x10^{23}\: \frac{molecules}{mol}\times 1\: atm}{0.082\: \frac{atm.L}{mol.K}\times 298\: K}=2.46x10^{22}\frac{molecules}{L}$

transforming L in m³:

$n_{v} =2.46x10^{22}\frac{molecules}{L}\times \frac{1000\: L}{1\: m^{3}}=\mathbf{2.46x10^{25}\frac{molecules}{m^{3}}}$

then the mean free path is,

$\lambda =\frac{1}{\sqrt{2}\pi \sigma^{2} n_{v}}$

$\lambda =\frac{1}{\sqrt{2}\times 3.142 \times 5.2x10^{-19}\: m^{2} \times 2.46x10^{25}\: \frac{molecules}{m^{3}}}=\mathbf{1.76x10^{-8}\: \frac{m}{molecules}}$

Total number of collisions of CO₂ molecules

• The Total number of collisions is defined as

$Z=\frac{\pi \bar{v}\sigma ^{2}n_{v}^{2}}{\sqrt{2}}$

where

• v = average velocity of molecules = 378.68 m/s,
• σ² = square of collision diameter = 5.2x10^-19 m²,
• n_v = number of molecules per unit of volume = 2.46x10²⁵ molecules/m³

^{$Z=\frac{3.142\times 378.68\, \frac{m}{s}\times 5.2x10^{-19}\, m^{2}\times 2.46x10^{25}\, \frac{molecules}{m^{3}}}{\sqrt{2}}$}

^{$Z=\mathbf{1.08x10^{10}\, \frac{molecules}{s}}$}