Boiling water: Suppose 1.0 g of water vaporizes isobarically at atmospheric pressure `(1.01xx10^5 Pa)`. Its volume in the liquid state is `V_i=V_(liqu i d)=1.0cm^3` and its volume in vapour state is `V_f=V_(vapour)=1671 cm^3`. Find the work done in the expansion and the change in internal energy of the system. Ignore any mixing of the stream and the surrounding air. Take latent heat of vaporization `L_v=2.26xx10^6J//kg`.

Explanation

To solve the problem step by step, we will calculate the work done during the vaporization of water and the change in internal energy of the system.

### Step 1: Calculate the Work Done in Expansion
The work done \( W \) during an isobaric process is given by the formula:

\[ W = P \times (V_f - V_i) \]

Where:
- \( P \) is the pressure (in Pascals)
- \( V_f \) is the final volume (in cubic meters)
- \( V_i \) is the initial volume (in cubic meters)

Given:
- \( P = 1.01 \times 10^5 \, \text{Pa} \)
- \( V_i = 1.0 \, \text{cm}^3 = 1.0 \times 10^{-6} \, \text{m}^3 \)
- \( V_f = 1671 \, \text{cm}^3 = 1671 \times 10^{-6} \, \text{m}^3 \)

Now, substituting the values into the work done formula:

\[ W = 1.01 \times 10^5 \, \text{Pa} \times \left(1671 \times 10^{-6} \, \text{m}^3 - 1.0 \times 10^{-6} \, \text{m}^3\right) \]

Calculating the difference in volume:

\[ V_f - V_i = 1671 \times 10^{-6} - 1.0 \times 10^{-6} = 1670 \times 10^{-6} \, \text{m}^3 \]

Now substituting back:

\[ W = 1.01 \times 10^5 \times 1670 \times 10^{-6} \]

Calculating \( W \):

\[ W = 1.01 \times 1670 \times 10^{-1} = 168.67 \, \text{J} \approx 169 \, \text{J} \]

### Step 2: Calculate the Change in Internal Energy
According to the first law of thermodynamics:

\[ Q = \Delta U + W \]

Where:
- \( Q \) is the heat added to the system
- \( \Delta U \) is the change in internal energy
- \( W \) is the work done by the system

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

\[ Q = m \times L_v \]

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

Given:
- \( m = 1.0 \, \text{g} = 1.0 \times 10^{-3} \, \text{kg} \)
- \( L_v = 2.26 \times 10^6 \, \text{J/kg} \)

Calculating \( Q \):

\[ Q = 1.0 \times 10^{-3} \, \text{kg} \times 2.26 \times 10^6 \, \text{J/kg} = 2260 \, \text{J} \]

Now substituting \( Q \) and \( W \) into the first law equation to find \( \Delta U \):

\[ 2260 = \Delta U + 169 \]

Rearranging to solve for \( \Delta U \):

\[ \Delta U = 2260 - 169 = 2091 \, \text{J} \]

### Final Answers
- Work done \( W \) = 169 J
- Change in internal energy \( \Delta U \) = 2091 J