One gram of water changes from liquid to solid at a constant pressure of 1.00 atm and a constant temperature of 0°C. In the process, the volume changes from 0.60 cm3 to 1.14 cm3.
Find the work done on the water?
Find the change in the internal energy of the water?
Pressure*changevolume=work
pressure units: N/m^2, volume in dm^3, work in joules
To find the work done on the water and the change in internal energy, we need to use the formulas related to the first law of thermodynamics:
ΔU = Q - W,
where ΔU is the change in internal energy, Q is the heat transferred to the system, and W is the work done on the system.
1. Finding the work done on the water:
The work done on the water can be calculated using the formula:
W = PΔV
where W is the work done, P is the pressure, and ΔV is the change in volume.
Given:
P = 1.00 atm
ΔV = V2 - V1 = 1.14 cm^3 - 0.60 cm^3
First, convert the units to match:
1 atm = 101.325 Pa (approx)
1 cm^3 = 1x10^-6 m^3
So, P = 1.00 atm x 101.325 Pa/atm ≈ 101.325 Pa
Next, convert the volume to m^3:
ΔV = (1.14 cm^3 - 0.60 cm^3) x 1x10^-6 m^3/cm^3
Now, we can calculate the work done:
W = (101.325 Pa) x (1.14x10^-6 m^3 - 0.60x10^-6 m^3)
Calculate the numerical value of the work done.
2. Finding the change in internal energy:
Since the process is a phase change (from liquid to solid), it occurs at a constant temperature. According to the first law of thermodynamics, for an isothermal process (constant temperature), the change in internal energy is zero:
ΔU = Q - W
Since ΔU = 0 and we have already calculated W, we can use the equation to find Q:
Q = W
Therefore, the change in internal energy (ΔU) for the water is equal to zero.
Please note that you will need to substitute the appropriate numerical values into the equations to obtain the final answers for the work done and the change in internal energy.