Relation between the volume of gas `(2)` that dissolves in a fixed volume of solvent `(1)` and the partial pressure of gas`(2)` is (`n_(t)` = total moles, `K_(1)`and `K_(2)` are Henry's constants)

Options

A. `n_(1)RT//n_(2)`
B. `n_(2)RT//K_(2)p_(2)`
C. `n_(t)RT//K_(2)`
D. None of these

Explanation

To derive the relationship between the volume of gas (V₂) that dissolves in a fixed volume of solvent (V₁) and the partial pressure of gas (P₂), we can follow these steps:

### Step 1: Understanding Henry's Law
According to Henry's Law, the partial pressure of a gas (P₂) is directly proportional to its mole fraction (x₂) in the solution. This can be expressed as:
\[ P_2 = K_2 \cdot x_2 \]
where \( K_2 \) is Henry's constant for the gas.

### Step 2: Expressing Mole Fraction
The mole fraction \( x_2 \) of the gas in the solution can be defined as:
\[ x_2 = \frac{n_2}{n_1 + n_2} \]
where:
- \( n_2 \) = moles of gas dissolved
- \( n_1 \) = moles of solvent

### Step 3: Substituting Mole Fraction into Henry's Law
Substituting the expression for mole fraction into Henry's Law gives:
\[ P_2 = K_2 \cdot \frac{n_2}{n_1 + n_2} \]

### Step 4: Using the Ideal Gas Law
The ideal gas law states:
\[ PV = nRT \]
For the gas, we can express its volume (V₂) as:
\[ P_2 V_2 = n_2 R T \]

### Step 5: Relating Partial Pressure to Volume
We can substitute \( P_2 \) from Henry's Law into the ideal gas law:
\[ \left( K_2 \cdot \frac{n_2}{n_1 + n_2} \right) V_2 = n_2 R T \]

### Step 6: Neglecting n₂
Since the amount of gas dissolved (n₂) is very small compared to the solvent (n₁), we can neglect \( n_2 \) in the denominator:
\[ P_2 \approx K_2 \cdot \frac{n_2}{n_1} \]

### Step 7: Rearranging the Equation
Substituting this approximation back into the ideal gas equation gives:
\[ \left( K_2 \cdot \frac{n_2}{n_1} \right) V_2 = n_2 R T \]

### Step 8: Canceling n₂
Dividing both sides by \( n_2 \) (assuming \( n_2 \neq 0 \)):
\[ K_2 \cdot \frac{V_2}{n_1} = R T \]

### Step 9: Solving for V₂
Rearranging this gives us the relationship between the volume of gas and the partial pressure:
\[ V_2 = \frac{n_1 R T}{K_2} \]

### Final Result
Thus, the relationship between the volume of gas that dissolves in a fixed volume of solvent and the partial pressure of the gas is:
\[ V_2 = \frac{n_1 R T}{K_2} \]