One gram of water `(1 cm^3)` becomes `1671 cm^3` of steam when boiled at a constant pressure of 1 atm `(1.013xx10^5Pa)`. The heat of vaporization at this pressure is `L_v=2.256xx10^6J//kg`. Compute (a) the work done by the water when it vaporizes and (b) its increase in internal energy.

Explanation

To solve the problem, we need to compute two things: (a) the work done by the water when it vaporizes and (b) the increase in internal energy.

### Step-by-Step Solution:

**Given Data:**
- Mass of water, \( m = 1 \text{ gram} = 1 \times 10^{-3} \text{ kg} \)
- Initial volume of water, \( V_i = 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3 \)
- Final volume of steam, \( V_f = 1671 \text{ cm}^3 = 1671 \times 10^{-6} \text{ m}^3 \)
- Pressure, \( P = 1.013 \times 10^5 \text{ Pa} \)
- Heat of vaporization, \( L_v = 2.256 \times 10^6 \text{ J/kg} \)

### (a) Work Done by the Water When It Vaporizes

1. **Calculate the Change in Volume (\( \Delta V \)):**
\[ \Delta V = V_f - V_i = (1671 - 1) \times 10^{-6} \text{ m}^3 = 1670 \times 10^{-6} \text{ m}^3 \]

2. **Calculate the Work Done (\( W \)):**
The work done by the system at constant pressure is given by:
\[ W = P \Delta V \]
Substituting the values:
\[ W = (1.013 \times 10^5 \text{ Pa}) \times (1670 \times 10^{-6} \text{ m}^3) \]
\[ W = 169.171 \text{ J} \approx 169 \text{ J} \]

### (b) Increase in Internal Energy

1. **Calculate the Heat Added (\( Q \)):**
The heat added during the phase change is given by:
\[ Q = m \cdot L_v \]
Substituting the values:
\[ Q = (1 \times 10^{-3} \text{ kg}) \times (2.256 \times 10^6 \text{ J/kg}) = 2260.256 \text{ J} \]

2. **Apply the First Law of Thermodynamics:**
The first law states:
\[ \Delta U = Q - W \]
Substituting the values:
\[ \Delta U = 2260.256 \text{ J} - 169 \text{ J} = 2091.256 \text{ J} \approx 2087 \text{ J} \]

### Final Answers:
- (a) Work done by the water when it vaporizes: **169 J**
- (b) Increase in internal energy: **2087 J**