An ideal gas is compressed at constant pressure of `10^(5)Pa` until its volume is halved. If the initial volume of the gas as `3.0 xx 10^(-2)m^(3)`, find the work done on the gas?

Explanation

To find the work done on the gas during the compression, we can use the formula for work done at constant pressure:

\[ W = P \times \Delta V \]

Where:
- \( W \) is the work done on the gas,
- \( P \) is the pressure,
- \( \Delta V \) is the change in volume.

### Step 1: Identify the initial and final volumes
Given:
- Initial volume \( V_i = 3.0 \times 10^{-2} \, m^3 \)
- The gas is compressed until its volume is halved, so:
\[ V_f = \frac{V_i}{2} = \frac{3.0 \times 10^{-2}}{2} = 1.5 \times 10^{-2} \, m^3 \]

### Step 2: Calculate the change in volume
The change in volume \( \Delta V \) is given by:
\[ \Delta V = V_f - V_i \]
Substituting the values:
\[ \Delta V = 1.5 \times 10^{-2} - 3.0 \times 10^{-2} = -1.5 \times 10^{-2} \, m^3 \]

### Step 3: Substitute the values into the work done formula
Given the pressure \( P = 10^5 \, Pa \), we can now calculate the work done:
\[ W = P \times \Delta V \]
\[ W = 10^5 \times (-1.5 \times 10^{-2}) \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ W = -10^5 \times 1.5 \times 10^{-2} = -1.5 \times 10^3 \, J \]

### Final Result
The work done on the gas is:
\[ W = -1500 \, J \]