A mole of liquid water is vaporized at 100 degree C and 1 atm. Th heat of vaporization is 40.7 kJ/mol. What are the values of Wrev, q, detaU and deta H?
I'm not sure what is the volume because the pressure it not change so I so confort can I use Wrev = -nRT.ln(V2/V1) or Wrev = -nRT.ln(P1/P2)
To find the values of Wrev, q, ΔU, and ΔH, let's go through each step.
1. Wrev (reversible work):
The process of vaporizing water is an isothermal expansion, which means the temperature remains constant throughout. However, the pressure is also constant at 1 atm, so we cannot use the equation Wrev = -nRT ln(V2/V1) directly because the volume is not changing. Instead, since the process is at constant pressure, we can use the formula Wrev = -PΔV, where ΔV is the change in volume. However, since the volume doesn't change during vaporization, ΔV = 0, which means the value of Wrev is 0.
2. q (heat):
Since the process is isothermal, the heat added or removed from the system is equal to the enthalpy change (ΔH). Therefore, q = ΔH.
3. ΔU (change in internal energy):
The change in internal energy (ΔU) is given by the equation ΔU = q - W. In this case, W is 0 (as we found in step 1), so ΔU = q.
4. ΔH (change in enthalpy):
To find ΔH, we can use the formula ΔH = ΔU + PΔV, where ΔV is the change in volume. However, since the volume doesn't change during vaporization, ΔV = 0. Thus, ΔH = ΔU.
Considering all these points, we can conclude that Wrev = 0, q = ΔU = ΔH, and the values for q, ΔU, and ΔH are all equal.