If 250 mL of `N_(2)` over water at `30^(@)C` and a total pressure of 740 torr is mixed with 300 mL of Ne over water at `25^(@)C` and a total pressure of 780 torr, what will be the total pressure if the mixture is in a 500 mL vessel over water at `35^(@)C`.
(Given : Vapour pressure (Aqueous tension )of `H_(2)O` at `25^(@)C`, `30^(@)C` and `35^(@)C` are 23.8, 31.8 and 42.2 torr respectively. Assume volume of `H_(2)O(l)` is negligible in final vessel)

Options

A. 760 torr
B. 828.4 torr
C. 807.6 torr
D. 870.6 torr

Explanation

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

### Step 1: Calculate the moles of Nitrogen (N₂)

Given:
- Volume of N₂ = 250 mL = 0.250 L
- Total pressure of N₂ over water = 740 torr
- Temperature = 30°C = 303 K
- Vapor pressure of water at 30°C = 31.8 torr

First, we need to find the partial pressure of N₂:
\[ P_{N_2} = P_{total} - P_{H_2O} = 740 \, \text{torr} - 31.8 \, \text{torr} = 708.2 \, \text{torr} \]

Now, we can use the ideal gas law to calculate the moles of N₂:
\[ n = \frac{PV}{RT} \]
Where:
- \( P = 708.2 \, \text{torr} \) (convert to atm: \( \frac{708.2}{760} \))
- \( V = 0.250 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 303 \, \text{K} \)

Calculating:
\[ n_{N_2} = \frac{(708.2/760) \times 0.250}{0.0821 \times 303} \]

After calculating, we find:
\[ n_{N_2} \approx 0.00936 \, \text{mol} \]

### Step 2: Calculate the moles of Neon (Ne)

Given:
- Volume of Ne = 300 mL = 0.300 L
- Total pressure of Ne over water = 780 torr
- Temperature = 25°C = 298 K
- Vapor pressure of water at 25°C = 23.8 torr

First, we find the partial pressure of Ne:
\[ P_{Ne} = P_{total} - P_{H_2O} = 780 \, \text{torr} - 23.8 \, \text{torr} = 756.2 \, \text{torr} \]

Using the ideal gas law:
\[ n_{Ne} = \frac{PV}{RT} \]
Where:
- \( P = 756.2 \, \text{torr} \) (convert to atm: \( \frac{756.2}{760} \))
- \( V = 0.300 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 298 \, \text{K} \)

Calculating:
\[ n_{Ne} = \frac{(756.2/760) \times 0.300}{0.0821 \times 298} \]

After calculating, we find:
\[ n_{Ne} \approx 0.0122 \, \text{mol} \]

### Step 3: Calculate the total moles in the mixture

Now, we can find the total moles:
\[ n_{total} = n_{N_2} + n_{Ne} = 0.00936 + 0.0122 = 0.02156 \, \text{mol} \]

### Step 4: Calculate the total pressure in the final vessel

Given:
- Volume of the final vessel = 500 mL = 0.500 L
- Temperature = 35°C = 308 K
- Vapor pressure of water at 35°C = 42.2 torr

Using the ideal gas law for the total pressure:
\[ P = \frac{nRT}{V} \]
Where:
- \( n = 0.02156 \, \text{mol} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 308 \, \text{K} \)
- \( V = 0.500 \, \text{L} \)

Calculating:
\[ P_{gas} = \frac{0.02156 \times 0.0821 \times 308}{0.500} \]
After calculating, we find:
\[ P_{gas} \approx 1.09 \, \text{atm} \]

Convert to torr:
\[ P_{gas} \approx 1.09 \times 760 \approx 828.4 \, \text{torr} \]

### Step 5: Add the vapor pressure of water

Finally, we add the vapor pressure of water to find the total pressure:
\[ P_{total} = P_{gas} + P_{H_2O} = 828.4 \, \text{torr} + 42.2 \, \text{torr} = 870.6 \, \text{torr} \]

### Final Answer
The total pressure in the 500 mL vessel at 35°C is approximately **870.6 torr**.

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