At 273 K one atm, 'a' litre of `N_(2)O_(4)` decomposes to `NO_(2)` as :
`N_(2)O_(4)(g) hArr 2NO_(2)(g)`. To what extent has the decomposition proceeded when he original volume is `25%` less then that of existing volume ?
[Report your answer up to decimal places.]

Answer

To solve the problem, we need to analyze the decomposition of \( N_2O_4 \) into \( NO_2 \) and determine the extent of decomposition when the original volume is 25% less than the existing volume.

### Step-by-Step Solution:

1. **Define Initial Volume**:
Let the initial volume of \( N_2O_4 \) be \( V_0 = a \) liters.

2. **Decomposition Reaction**:
The reaction is given as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]
From the stoichiometry of the reaction, 1 mole of \( N_2O_4 \) produces 2 moles of \( NO_2 \).

3. **Volume Change**:
If \( x \) liters of \( N_2O_4 \) decompose, then:
- Volume of \( N_2O_4 \) left = \( a - x \)
- Volume of \( NO_2 \) produced = \( 2x \)

Therefore, the total volume after decomposition becomes:
\[ V_{total} = (a - x) + 2x = a + x \]

4. **Condition Given**:
It is given that the existing volume is 25% more than the original volume. Thus, we can express this mathematically:
\[ V_{total} = a + x = 0.75a \]

5. **Setting Up the Equation**:
Rearranging the equation gives:
\[ x = 0.75a - a = -0.25a \]
This means that:
\[ x = 0.25a \]

6. **Extent of Decomposition**:
The extent of decomposition (denoted as \( \alpha \)) can be calculated as the ratio of the volume of \( N_2O_4 \) that has decomposed to the initial volume:
\[ \alpha = \frac{x}{a} = \frac{0.25a}{a} = 0.25 \]

7. **Convert to Percentage**:
To express \( \alpha \) as a percentage:
\[ \alpha \times 100 = 0.25 \times 100 = 25\% \]

### Final Answer:
The extent of decomposition when the original volume is 25% less than the existing volume is **25.0%**.