Answer:
Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C
Explanation:
Derivation from state postulate
Using the state postulate, take the specific entropy, , for a homogeneous substance to be a function of specific volume and temperature .
ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT
During a phase change, the temperature is constant, so
ds = (partial s/partial v)(T) dv
Using the appropriate Maxwell relation gives
ds = (partial P/partial T)(v) dv
s(β) – s(aplαha) = dP/dT (v(β) – v(α))
dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv
Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.
For a closed system undergoing an internally reversible process, the first law is
du = δq – δw = Tds - Pdv
Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have
du + Pdv = dh Tds,
ds = dh/T,
Δs = Δh/T = L/T
After substitution of this result into the derivative of the pressure, one finds
dp/dT = L/TΔv
<u>This last equation is the Clapeyron equation.</u>
a)
(dP/dT) = dH/TdV => dP/dlnT = dH/dV
=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]
= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]
= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa
=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa
or
T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)
b) if the pressure in Denver is 84.6 kPa:
T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]
≅ 273.15 = 0°C T₁(freezing) essentially no change
