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 step by step, we will first calculate the work done by the water when it vaporizes and then find the increase in internal energy.

### Step 1: Calculate the Work Done During Vaporization

The work done (W) during the vaporization of water can be calculated using the formula:

\[ W = P \Delta V \]

Where:
- \( P \) is the pressure (in Pascals)
- \( \Delta V \) is the change in volume (in cubic meters)

Given:
- Pressure, \( P = 1.013 \times 10^5 \, \text{Pa} \)
- Initial volume of water, \( V_i = 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \)
- Final volume of steam, \( V_f = 1671 \, \text{cm}^3 = 1671 \times 10^{-6} \, \text{m}^3 \)

Now, calculate the change in volume:

\[ \Delta V = V_f - V_i \]
\[ \Delta V = (1671 \times 10^{-6}) - (1 \times 10^{-6}) \]
\[ \Delta V = 1670 \times 10^{-6} \, \text{m}^3 \]

Now substitute the values into the work done formula:

\[ W = P \Delta V \]
\[ W = (1.013 \times 10^5) \times (1670 \times 10^{-6}) \]

Calculating this gives:

\[ W = 169.1711 \, \text{J} \]

Rounding this, we find:

\[ W \approx 169 \, \text{J} \]

### Step 2: Calculate the Heat Absorbed During Vaporization

The heat absorbed (Q) during the phase change can be calculated using the formula:

\[ Q = mL_v \]

Where:
- \( m \) is the mass of the water (in kg)
- \( L_v \) is the latent heat of vaporization (in J/kg)

Given:
- Mass of water, \( m = 1 \, \text{g} = 1 \times 10^{-3} \, \text{kg} \)
- Latent heat of vaporization, \( L_v = 2.256 \times 10^6 \, \text{J/kg} \)

Now substitute the values into the heat formula:

\[ Q = (1 \times 10^{-3}) \times (2.256 \times 10^6) \]
\[ Q = 2256 \, \text{J} \]

### Step 3: Calculate the Increase in Internal Energy

According to the first law of thermodynamics, the change in internal energy (\( \Delta U \)) is given by:

\[ \Delta U = Q - W \]

Substituting the values we calculated:

\[ \Delta U = 2256 \, \text{J} - 169 \, \text{J} \]
\[ \Delta U = 2087 \, \text{J} \]

### Final Answers:
(a) The work done by the water when it vaporizes is approximately **169 J**.
(b) The increase in internal energy is **2087 J**.