An ideal gas is allowed to expand from 1L to 10 L against a constant external pressure of 1 bar. The work done in `x xx 10^(y)J`. The numerical value of x is_____.

Explanation

To solve the problem of calculating the work done by an ideal gas expanding from 1 L to 10 L against a constant external pressure of 1 bar, we can follow these steps:

### Step 1: Identify the initial and final volumes
- Initial volume (V1) = 1 L
- Final volume (V2) = 10 L

### Step 2: Calculate the change in volume
- Change in volume (ΔV) = V2 - V1 = 10 L - 1 L = 9 L

### Step 3: Convert the change in volume to cubic meters
- Since 1 L = 10^(-3) m³, we convert 9 L to cubic meters:
\[ ΔV = 9 \, \text{L} = 9 \times 10^{-3} \, \text{m}^3 \]

### Step 4: Identify the external pressure
- External pressure (P) = 1 bar

### Step 5: Convert the pressure to pascals
- Since 1 bar = 10^5 Pa, we have:
\[ P = 1 \, \text{bar} = 10^5 \, \text{Pa} \]

### Step 6: Calculate the work done using the formula for work in an irreversible process
- The work done (W) is given by:
\[ W = -P \times ΔV \]
- Substituting the values we have:
\[ W = - (10^5 \, \text{Pa}) \times (9 \times 10^{-3} \, \text{m}^3) \]

### Step 7: Perform the calculation
- Calculate the work done:
\[ W = - (10^5) \times (9 \times 10^{-3}) = - 9 \times 10^{2} \, \text{J} \]

### Step 8: Express the work done in the required format
- The work done can be expressed as:
\[ W = -9 \times 10^{2} \, \text{J} \]
- The question asks for the numerical value of \( x \) in the expression \( x \times 10^{y} \, \text{J} \). Here, \( x = 9 \) and \( y = 2 \).

### Final Answer
- The numerical value of \( x \) is **9**.