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.