Four gases like `H_(2), He, CH_(4)` and `CO_(2)` have Henry's constant values `(K_(H))` are 69.16, 144.97, 0.413 and 1.67. The gas which is more soluble in liquid is

Options

A. He
B. `CH_(4)`
C. `H_(2)`
D. `CO_(2)`

Answer

To determine which gas is more soluble in a liquid based on the given Henry's constant values, we can follow these steps:

### Step 1: Understand Henry's Law
Henry's Law states that the amount of gas that dissolves in a liquid at a given temperature is directly proportional to the partial pressure of that gas above the liquid. Mathematically, it can be expressed as:
\[ P = K_H \cdot x \]
where:
- \( P \) = partial pressure of the gas,
- \( K_H \) = Henry's constant,
- \( x \) = mole fraction of the gas in the solution.

### Step 2: Analyze the Relationship
From the equation, we can rearrange it to express the mole fraction \( x \):
\[ x = \frac{P}{K_H} \]
This shows that for a constant pressure \( P \), the mole fraction \( x \) (which indicates solubility) is inversely proportional to \( K_H \). This means that a lower value of \( K_H \) corresponds to a higher solubility of the gas in the liquid.

### Step 3: Compare the Henry's Constants
We have the following Henry's constant values for the gases:
- \( K_H \) for \( H_2 \) = 69.16
- \( K_H \) for He = 144.97
- \( K_H \) for \( CH_4 \) = 0.413
- \( K_H \) for \( CO_2 \) = 1.67

### Step 4: Identify the Lowest Henry's Constant
To find out which gas is more soluble, we need to identify the gas with the lowest Henry's constant value. From the values provided:
- The lowest \( K_H \) value is for \( CH_4 \) (Methane), which is 0.413.

### Step 5: Conclusion
Since \( CH_4 \) has the lowest Henry's constant, it is the most soluble gas in the liquid among the given options.

### Final Answer
The gas which is more soluble in liquid is **\( CH_4 \) (Methane)**.

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