Question 5 of 20: Partial Pressures of Digester Gas Components

Question 5: Partial Pressures of Digester Gas Components (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Digester gas, 1,000 kg total, composition by mass 68% $\text{CH}_4$ / 30% $\text{CO}_2$ / 2% $\text{H}_2\text{S}$; total tank pressure $P=300\ \text{kPa}$.

Find. Partial pressure of each gas component.

Approach. Dalton’s Law gives each partial pressure as its mole fraction times the total pressure, so the mass percentages must first be converted to moles via each gas’s molecular weight, then to mole fractions, before applying $P_i=y_iP_{\text{total}}$.

1. Mass of each component in the 1,000 kg charge. $$m_{\text{CH}_4}=680\ \text{kg},\quad m_{\text{CO}_2}=300\ \text{kg},\quad m_{\text{H}_2\text{S}}=20\ \text{kg}$$
2. Molecular weights. $$M_{\text{CH}_4}=12+4(1)=16,\quad M_{\text{CO}_2}=12+2(16)=44,\quad M_{\text{H}_2\text{S}}=2(1)+32=34$$
3. Moles of each component (kmol, using tonnes-scale kg→kmol). $$n_{\text{CH}_4}=\dfrac{680}{16}=42.50,\quad n_{\text{CO}_2}=\dfrac{300}{44}=6.818,\quad n_{\text{H}_2\text{S}}=\dfrac{20}{34}=0.588\ \text{kmol}$$ Total moles $n_T=42.50+6.818+0.588=49.91\ \text{kmol}$.
4. Mole fractions. $$y_{\text{CH}_4}=\dfrac{42.50}{49.91}=0.852,\quad y_{\text{CO}_2}=\dfrac{6.818}{49.91}=0.137,\quad y_{\text{H}_2\text{S}}=\dfrac{0.588}{49.91}=0.0118$$
5. Partial pressures (Dalton’s Law, $P_i=y_iP$). $$P_{\text{CH}_4}=0.852(300)=\boxed{255.5\ \text{kPa}}\qquad P_{\text{CO}_2}=0.137(300)=\boxed{41.0\ \text{kPa}}\qquad P_{\text{H}_2\text{S}}=0.0118(300)=\boxed{3.5\ \text{kPa}}$$ Check: $255.5+41.0+3.5=300.0\ \text{kPa}$, matching the given total exactly.