The vapour pressure of water at `80^(@)C` is `355` torr. A `100 ml` vessel contains water-saturated oxygen at `80^(@)C`, the total gas pressure being `760` torr. The contents of the vessel were pumped into a `50.0 ml` vessel at the same temperature. What were the partial pressures of oxygen and of water vapour, what was the total pressure in the final equilibrated state? Neglect the volume of any water which might condense.

Answer

To solve the problem step by step, we will analyze the situation in two vessels and apply Dalton's Law of Partial Pressures and Boyle's Law.

### Step 1: Identify the given data
- Vapour pressure of water at \(80^\circ C\) = \(355\) torr
- Total pressure in the \(100\) ml vessel = \(760\) torr
- Volume of the first vessel (V1) = \(100\) ml
- Volume of the second vessel (V2) = \(50\) ml

### Step 2: Calculate the partial pressure of oxygen in the first vessel
According to Dalton's Law of Partial Pressures:
\[ P_{\text{total}} = P_{\text{O}_2} + P_{\text{H}_2O} \]
Where:
- \(P_{\text{total}} = 760\) torr
- \(P_{\text{H}_2O} = 355\) torr

Now, we can find the partial pressure of oxygen:
\[ P_{\text{O}_2} = P_{\text{total}} - P_{\text{H}_2O} = 760 - 355 = 405 \text{ torr} \]

### Step 3: Calculate the partial pressure of oxygen in the second vessel
Using Boyle's Law, which states that \(P_1V_1 = P_2V_2\), we can find the new pressure of oxygen when the gas is transferred to the second vessel.

Let:
- \(P_1 = P_{\text{O}_2} = 405\) torr (in the first vessel)
- \(V_1 = 100\) ml (volume of the first vessel)
- \(V_2 = 50\) ml (volume of the second vessel)

Now, we can calculate \(P_2\) (the partial pressure of oxygen in the second vessel):
\[ P_2 = \frac{P_1 \times V_1}{V_2} = \frac{405 \times 100}{50} = 810 \text{ torr} \]

### Step 4: Determine the total pressure in the second vessel
In the second vessel, the partial pressure of water vapor remains unchanged because the temperature is constant and we are neglecting any condensation. Therefore:
\[ P_{\text{H}_2O} = 355 \text{ torr} \]

Now, we can find the total pressure in the second vessel:
\[ P_{\text{total}} = P_{\text{O}_2} + P_{\text{H}_2O} = 810 + 355 = 1165 \text{ torr} \]

### Summary of Results
- Partial pressure of oxygen in the first vessel: \(405\) torr
- Partial pressure of oxygen in the second vessel: \(810\) torr
- Total pressure in the second vessel: \(1165\) torr