Answer :
The critical point of water in terms of variation of pressure and volume: At the critical point of water, the liquid-vapor phase boundary ends. There is no distinction between the two phases. This point is found at a temperature of 647 K and a pressure of 22.064 MPa.
At the critical point, the densities of the liquid and vapor become identical. Thus, the critical point represents the endpoint of the water’s condensation line and the beginning of its vaporization line. The critical point of water can be explained in terms of variation in its pressure and volume by considering the concept of the compressibility factor (Z). For water, Z is found to be 1 at the critical point.
For gases, the expansivity in isobaric processes, ap, is given by 1 dv = ap V dT. We know, for an ideal gas, PV=nRT ... [Equation 1]
We also know that V/n=RT/P … [Equation 2]
So, V = nRT/P ... [Equation 3]
Taking differentials of Equation 3, we get:
dV= (dRT)/P – (nRdT)/P … [Equation 4]
Equating the right-hand side of Equation 4 to Equation 1, we get:1 dv= (dRT)/P – (nRdT)/P … [Equation 5]
Therefore, ap = 1/V (dV/dT) at constant pressure.
Substituting Equation 3 in Equation 5, we get:1 dv= (dR/P) (T/V) – (R/P) dT… [Equation 6]
For an ideal gas, PV=nRT
Therefore, PV/T = nR
Substituting this value of nR in Equation 6 and simplifying, we get ap = 1/Tр, where р is the pressure of the gas.
This shows that for an ideal gas, the expansivity in isobaric processes, ap, is inversely proportional to temperature. Hence, for an ideal gas, ap T р.
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