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

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the initial conditions
- The vapor pressure of water at \(80^\circ C\) is given as \(355\) torr.
- The total pressure in the \(100 \, \text{ml}\) vessel containing water-saturated oxygen is \(760\) torr.

### Step 2: Calculate the partial pressure of oxygen in the \(100 \, \text{ml}\) vessel
Using Dalton's Law of Partial Pressures:
\[ P_{\text{total}} = P_{\text{O}_2} + P_{\text{H}_2O} \]
Where:
- \(P_{\text{total}} = 760 \, \text{torr}\)
- \(P_{\text{H}_2O} = 355 \, \text{torr}\)

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

### Step 3: Transfer the gas to a \(50 \, \text{ml}\) vessel
Since the volume is halved from \(100 \, \text{ml}\) to \(50 \, \text{ml}\), we can use Boyle's Law, which states that pressure is inversely proportional to volume at constant temperature:
\[ P_1 V_1 = P_2 V_2 \]
Let:
- \(P_1 = 405 \, \text{torr}\) (initial pressure of oxygen)
- \(V_1 = 100 \, \text{ml}\)
- \(V_2 = 50 \, \text{ml}\)

We need to find \(P_2\):
\[ 405 \, \text{torr} \times 100 \, \text{ml} = P_2 \times 50 \, \text{ml} \]
\[ P_2 = \frac{405 \, \text{torr} \times 100 \, \text{ml}}{50 \, \text{ml}} = 810 \, \text{torr} \]

### Step 4: Calculate the partial pressure of water vapor in the \(50 \, \text{ml}\) vessel
Since we are told to neglect the volume of any water that condenses, the vapor pressure of water remains the same:
\[ P_{\text{H}_2O} = 355 \, \text{torr} \]

### Step 5: Calculate the total pressure in the \(50 \, \text{ml}\) vessel
Now we can find the total pressure in the \(50 \, \text{ml}\) vessel:
\[ P_{\text{total}} = P_{\text{O}_2} + P_{\text{H}_2O} = 810 \, \text{torr} + 355 \, \text{torr} = 1165 \, \text{torr} \]

### Final Results:
- Partial pressure of oxygen, \(P_{\text{O}_2} = 810 \, \text{torr}\)
- Partial pressure of water vapor, \(P_{\text{H}_2O} = 355 \, \text{torr}\)
- Total pressure, \(P_{\text{total}} = 1165 \, \text{torr}\)