Calculate the work done (in joules) when 5.0 mole of water is frozen at 0°C and 1.0 atm. The volumes of five moles of water and ice at 0°C are 0.0900 L and 0.0980 L, respectively.
To calculate the work done, we need to consider the change in volume and pressure during the freezing process. The work done is given by the equation:
Work done = - P ΔV
Where:
Work done is the energy transferred as a result of the change in volume and pressure (in joules).
P is the pressure (in pascals).
ΔV is the change in volume (in cubic meters).
First, we need to convert the given volumes from liters to cubic meters:
Volume of 5.0 moles of water = 0.0900 L = 0.0900 x 0.001 m^3 = 0.000090 m^3
Volume of 5.0 moles of ice = 0.0980 L = 0.0980 x 0.001 m^3 = 0.000098 m^3
Next, we calculate the change in volume:
ΔV = Volume of ice - Volume of water
= 0.000098 m^3 - 0.000090 m^3
= 0.000008 m^3
From the given information, the pressure (P) during the freezing process is 1.0 atm. We need to convert this to pascals:
1.0 atm = 1.01325 x 10^5 Pa
Now, we can calculate the work done:
Work done = - P ΔV
= - (1.01325 x 10^5 Pa) x (0.000008 m^3)
= - 8.11 joules
Therefore, the work done when 5.0 moles of water is frozen at 0°C and 1.0 atm is -8.11 joules. The negative sign indicates that work is done on the system (water), resulting in a decrease in volume.