One mole of an ideal gas is compressed from 500 `cm^(3)` against a constant pressure of `1.216 xx 10^(5)`Pa. The work involved in the process is 35.50 J. calculate the final volume .

Explanation

To solve the problem step by step, we will calculate the final volume of the gas after compression.

### Given Data:
- Number of moles of gas (n) = 1 mole
- Initial volume (V1) = 500 cm³ = 500 × 10⁻⁶ m³ (since 1 m³ = 10⁶ cm³)
- External pressure (P_external) = 1.216 × 10⁵ Pa
- Work done (W) = 35.50 J

### Step 1: Write the formula for work done during compression
The work done on the gas during an irreversible process at constant pressure is given by the formula:
\[ W = -P_{\text{external}} (V_2 - V_1) \]

### Step 2: Rearrange the formula to find the final volume (V2)
We can rearrange the formula to solve for \( V_2 \):
\[ V_2 = V_1 + \frac{-W}{P_{\text{external}}} \]

### Step 3: Substitute the values into the equation
Now, we can substitute the known values into the rearranged formula:
- \( V_1 = 500 \times 10^{-6} \, \text{m}^3 \)
- \( W = 35.50 \, \text{J} \)
- \( P_{\text{external}} = 1.216 \times 10^5 \, \text{Pa} \)

Substituting these values:
\[ V_2 = 500 \times 10^{-6} + \frac{-35.50}{1.216 \times 10^5} \]

### Step 4: Calculate the work term
First, calculate \( \frac{-35.50}{1.216 \times 10^5} \):
\[ \frac{-35.50}{1.216 \times 10^5} \approx -2.917 \times 10^{-4} \, \text{m}^3 \]

### Step 5: Add the initial volume and the work term
Now, we can calculate \( V_2 \):
\[ V_2 = 500 \times 10^{-6} - 2.917 \times 10^{-4} \]
\[ V_2 = 500 \times 10^{-6} - 291.7 \times 10^{-6} \]
\[ V_2 = (500 - 291.7) \times 10^{-6} \]
\[ V_2 = 208.3 \times 10^{-6} \, \text{m}^3 \]

### Step 6: Convert the final volume back to cm³
To convert \( V_2 \) back to cm³:
\[ V_2 = 208.3 \times 10^{-6} \, \text{m}^3 \times 10^6 \, \text{cm}^3/\text{m}^3 \]
\[ V_2 = 208.3 \, \text{cm}^3 \]

### Final Answer:
The final volume \( V_2 \) is approximately **208.3 cm³**.

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