Question

1 kg of water at 100°C is converted into steam at 100°C by boiling at atmospheric pressure. The volume of water changes from \(1.00 \times 10^{-3}\, \text{m}^3\) as a liquid to \(1.671 \times 10^{-3}\, \text{m}^3\) as steam. The change in internal energy of the system during the process will be:

• A. + 2476 kJ
• B. -2426 kJ
• C. -2090 kJ
• D. +2090 kJ

Hint:

When converting liquid to gas at constant pressure, the change in internal energy is mainly due to the latent heat of vaporization.

Answer 1

The Correct Option is B

The change in internal energy at constant pressure during a phase change can be calculated using the formula: \[ \Delta U = mL \] Where:
- \( m \) is the mass of the substance (1 kg),
- \( L \) is the latent heat of vaporization (\( 2257 \, \text{kJ/kg} \)).
Thus, the change in internal energy is: \[ \Delta U = 1 \times 2257 = 2257 \, \text{kJ} \] Now, the volume change from liquid to steam is: \[ \Delta V = V_2 - V_1 = (1.671 \times 10^{-3}) - (1.00 \times 10^{-3}) = 0.671 \times 10^{-3}\, \text{m}^3 \] Using the given atmospheric pressure of \( 1 \times 10^5 \, \text{Pa} \), we can calculate the work done: \[ W = P \Delta V = (1.0 \times 10^5) \times (0.671 \times 10^{-3}) = 67.1 \, \text{J} \] Thus, the total change in internal energy is: \[ \Delta U + W = 2257 + 67.1 = 2426.1 \, \text{kJ} \] So, the change in internal energy is approximately \( 2426 \, \text{kJ} \).