External Work Calculation

Given:

• Mass of water, \( m = 1 \) g
• Volume of steam, \( V_{steam} = 1671 \) cm\(^3 = 1671 \times 10^{-6} \) m\(^3 \)
• Pressure, \( P = 1 \) atm \( = 1.013 \times 10^5 \) Pa
• Heat of vaporization, \( L = 2256 \) J/g

First, convert the mass of water to kg:

\( m = 1 \) g \( = 1 \times 10^{-3} \) kg

Initial volume of water is negligible compared to the volume of steam, so we can consider the change in volume to be approximately equal to the volume of the steam.

\( \Delta V = V_{steam} - V_{water} \approx V_{steam} = 1671 \times 10^{-6} \) m\(^3 \)

External work done during vaporization is given by:

\( W = P \Delta V \)

Substitute the values:

\( W = (1.013 \times 10^5 \text{ Pa}) \times (1671 \times 10^{-6} \text{ m}^3) \)

\( W = 1.013 \times 10^5 \times 1671 \times 10^{-6} \) J

\( W = 1.013 \times 1.671 \times 10^{-1} \times 10^5\)J

\(W = 169.28 \text{ J} \)

\(W \approx 169 \text{ J} \)

Therefore, the external work done is approximately 169 J. 🎉