When 100 ml of `O_2-O_3` mixture was passed through turpentine oil, there was reduction of volume by 20 ml. If 100 ml of such a mixture is heated, what will be the increase in the volume?

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the initial conditions
We start with a mixture of oxygen (O₂) and ozone (O₃) totaling 100 ml. When this mixture is passed through turpentine oil, 20 ml of the mixture is absorbed, which we will assume is entirely ozone (O₃).

### Step 2: Determine the volumes of O₂ and O₃
Since 20 ml of the mixture is absorbed, the remaining volume of the gas mixture is:
- Volume of O₃ = 20 ml (absorbed)
- Volume of O₂ = 100 ml - 20 ml = 80 ml

### Step 3: Analyze the reaction upon heating
When the ozone (O₃) is heated, it decomposes into oxygen (O₂) according to the reaction:
\[ 2 O_3 \rightarrow 3 O_2 \]
This means that for every 2 moles of O₃, we produce 3 moles of O₂.

### Step 4: Calculate the increase in volume from O₃ to O₂
From the volume of O₃ that we have (20 ml), we can find out how much O₂ will be produced:
- According to the stoichiometry of the reaction, 2 volumes of O₃ produce 3 volumes of O₂.

Using the ratio:
\[ \text{Volume of O₂ produced} = \left(\frac{3}{2}\right) \times \text{Volume of O₃} \]
Substituting the volume of O₃:
\[ \text{Volume of O₂ produced} = \left(\frac{3}{2}\right) \times 20 \text{ ml} = 30 \text{ ml} \]

### Step 5: Calculate the total volume after heating
After heating, we have:
- Volume of O₂ originally present = 80 ml
- Volume of O₂ produced from O₃ = 30 ml

Thus, the total volume after heating is:
\[ \text{Total Volume} = 80 \text{ ml} + 30 \text{ ml} = 110 \text{ ml} \]

### Step 6: Determine the increase in volume
The increase in volume due to the heating process is:
\[ \text{Increase in Volume} = \text{Final Volume} - \text{Initial Volume} \]
\[ \text{Increase in Volume} = 110 \text{ ml} - 100 \text{ ml} = 10 \text{ ml} \]

### Final Answer
The increase in volume when the 100 ml of the O₂-O₃ mixture is heated is **10 ml**.

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