Question

At 27$^{\circ}$C, 1 L of H$_2$ with a pressure of 1 bar is mixed with 2 L of O$_2$ with a pressure of 2 bar in a 10 L flask. What is the pressure exerted by gaseous mixture in bar? (Assume H$_2$ and O$_2$ as ideal gases)

• A. 4
• B. 0.05
• C. 1
• D. 0.5

Hint:

Partial pressures: $P_{\text{total}} = \sum P_i$. At constant T: $P_1V_1 = P_2V_2$.

Answer 1

The Correct Option is D

We can use the ideal gas law ($PV = nRT$) and the concept of partial pressures. Since the temperature is constant, we can use the relation $P_1V_1 = P_2V_2$ for each gas. For H$_2$: $P_1 = 1 \, \text{bar}$, $V_1 = 1 \, \text{L}$, $V_2 = 10 \, \text{L}$. So, $P_2(\text{H}_2) = \frac{1 \times 1}{10} = 0.1 \, \text{bar}$. For O$_2$: $P_1 = 2 \, \text{bar}$, $V_1 = 2 \, \text{L}$, $V_2 = 10 \, \text{L}$. So, $P_2(\text{O}_2) = \frac{2 \times 2}{10} = 0.4 \, \text{bar}$. The total pressure of the mixture is the sum of the partial pressures: $$ P_{\text{total}} = P(\text{H}_2) + P(\text{O}_2) = 0.1 + 0.4 = 0.5 \, \text{bar} $$