3 moles of an ideal gas is compressed from `30 dm^3` to `20 dm^3` against a constant pressure of `3.039 xx 10^5 Nm^(-2)`.The work done in calories is `(1 J =0.239 cal )`

Options

A. `+ 30.39 kcal `
B. `+ 72.63 kcal`
C. `+ 726.3 kcal `
D. `+303. 9 kcal `

Explanation

To solve the problem, we will calculate the work done during the compression of the gas using the formula for work done on a gas at constant pressure.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (N) = 3 moles
- Initial volume (V1) = 30 dm³
- Final volume (V2) = 20 dm³
- External pressure (P) = 3.039 x 10^5 N/m²

2. **Calculate the Change in Volume (ΔV):**
\[ \Delta V = V2 - V1 = 20 \, \text{dm}^3 - 30 \, \text{dm}^3 = -10 \, \text{dm}^3 \]

3. **Calculate the Work Done (W):**
The formula for work done on a gas at constant pressure is given by:
\[ W = -P \Delta V \]
Substituting the values:
\[ W = - (3.039 \times 10^5 \, \text{N/m}^2) \times (-10 \, \text{dm}^3) \]
Note: 1 dm³ = 0.001 m³, so we convert dm³ to m³:
\[ W = (3.039 \times 10^5) \times (10 \times 0.001) = 3.039 \times 10^5 \times 0.01 = 3.039 \times 10^3 \, \text{J} \]

4. **Convert Work Done from Joules to Calories:**
We know that 1 J = 0.239 cal. Therefore:
\[ W_{\text{cal}} = W_{\text{J}} \times 0.239 = 3.039 \times 10^3 \, \text{J} \times 0.239 \, \text{cal/J} \]
\[ W_{\text{cal}} = 726.3 \, \text{cal} \]

5. **Convert Calories to Kilocalories:**
Since 1 kilocalorie (kcal) = 1000 calories:
\[ W_{\text{kcal}} = \frac{726.3 \, \text{cal}}{1000} = 0.7263 \, \text{kcal} \]

### Final Answer:
The work done in calories is approximately **726.3 kcal**.