540 calories of heat convert 1 cubic centimeter of water at C o100 into 1671 cubic centimeter of steam at C o100 at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly

Options

A. 540 cal
B. 40 cal
C. Zero cal
D. 500 cal

Answer

To solve the problem, we need to calculate the work done against atmospheric pressure when converting 1 cubic centimeter of water at 100°C into 1671 cubic centimeters of steam at the same temperature and pressure of one atmosphere.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume of water, \( V_1 = 1 \, \text{cm}^3 \)
- Final volume of steam, \( V_2 = 1671 \, \text{cm}^3 \)
- Pressure, \( P = 1 \, \text{atm} = 1.013 \times 10^5 \, \text{N/m}^2 \)

2. **Calculate the Change in Volume:**
\[ \Delta V = V_2 - V_1 = 1671 \, \text{cm}^3 - 1 \, \text{cm}^3 = 1670 \, \text{cm}^3 \]
Convert the volume from cubic centimeters to cubic meters:
\[ \Delta V = 1670 \, \text{cm}^3 \times 10^{-6} \, \text{m}^3/\text{cm}^3 = 1.67 \times 10^{-3} \, \text{m}^3 \]

3. **Calculate the Work Done:**
The work done against atmospheric pressure is given by the formula:
\[ W = P \Delta V \]
Substituting the values:
\[ W = (1.013 \times 10^5 \, \text{N/m}^2) \times (1.67 \times 10^{-3} \, \text{m}^3) \]

4. **Perform the Calculation:**
\[ W = 1.013 \times 10^5 \times 1.67 \times 10^{-3} = 169.171 \, \text{J} \]

5. **Convert Work Done from Joules to Calories:**
Since \( 1 \, \text{calorie} = 4.184 \, \text{J} \):
\[ W = \frac{169.171 \, \text{J}}{4.184 \, \text{J/cal}} \approx 40.5 \, \text{cal} \]

### Final Result:
The work done against atmospheric pressure is approximately **40.5 calories**.