The Reflection on Work Done in Freezing Water at 0°C

What is the work done when 1.0 mol of H₂O is frozen at 0°C and 1.0 atm?

How can the work done during a phase change be calculated using the given information?

Main Answer:

The work done when 1.0 mol of H₂O is frozen at 0°C and 1.0 atm is calculated using the equation for work done in a gas expansion or compression.

When water freezes at 0°C and 1.0 atm, it undergoes a phase change from liquid to solid (ice). During this process, the volume of the substance changes, and work is done against external pressure. The equation to calculate work done during a phase change is:

W = -PΔV

Where:

W is the work done (in Joules)

P is the external pressure (in atm)

ΔV is the change in volume (in liters)

Given that 1.0 mol of H₂O is frozen and the density of ice at 0°C is 0.917 g/mL, we can calculate the change in volume (ΔV) as follows:

ΔV = mass / density = 1.0 mol x 18.015 g/mol / 0.917 g/mL = 18.62 mL

Since 1 mL is equal to 1 cm³, we have ΔV = 18.62 cm³ = 0.01862 L.

Given that the external pressure (P) is 1.0 atm, we can now calculate the work (W):

W = -PΔV = -1.0 atm x 0.01862 L = -0.01862 L atm

Converting L atm to Joules (since 1 L atm = 101.325 J):

W = -0.01862 L atm x 101.325 J/L atm = -1.886 J

The negative sign indicates that work is done on the system during the phase change.