1.2 bar
0.5 bar
To solve this problem, we use Boyle's Law, which states:
\( P_1 V_1 = P_2 V_2 \)
Given:
We substitute the values into the formula:
\( 1 \times 0.5 = P_2 \times 1 \)
Solving for \( P_2 \):
\( P_2 = \frac{0.5}{1} = 0.5 \) bar
0.5 bar
The reaction:
\[ \text{H}_2\text{O}(g) + \text{Cl}_2\text{O}(g) \rightleftharpoons 2 \text{HOCl}(g) \]
is allowed to attain equilibrium at 400K. At equilibrium, the partial pressures are given as:
The value of \( K_p \) for the reaction at 400K is:
\[ K_p = \frac{P_{\text{HOCl}}^2}{P_{\text{H}_2\text{O}} \cdot P_{\text{Cl}_2\text{O}}} \]
Kepler's second law (law of areas) of planetary motion leads to law of conservation of