Question

A steel rigid vessel of volume 1 L is filled with a mixture of methane and oxygen at a total pressure of one atm at 27°C. The gas mixture is exploded. What would be the final pressure of the products at 127°C?

• A. 3.13 atm
• B. 1.29 atm
• C. 1.33 atm
• D. 1.37 atm

Answer 1

Answer: (C)

1.33 atm

Answer 2

The reaction is $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$. The change in moles of gas is $(1+2) - (1+2) = 0$. Since the volume is constant and the change in moles of gas is zero, the pressure remains constant at the initial temperature. Using Gay-Lussac's law ($P_1/T_1 = P_2/T_2$): $P_{final} = P_{initial} \times \frac{T_{final}}{T_{initial}} = 1 \text{ atm} \times \frac{127^\circ C + 273.15}{27^\circ C + 273.15} = 1 \text{ atm} \times \frac{400.15 \text{ K}}{300.15 \text{ K}} \approx 1.33 \text{ atm}$. The aqueous tension is irrelevant as water is a product and not present initially.