Example 2.6: The heat capacities for mercury are given as follows:

T/°C	CP/(cal·g^-1·°K^-1)	CV/(cal·g^-1·°K^-1)
-20	0.0335	0.0294
100	0.0327	0.0276

If one gram of mercury is heated at constant pressure of 1 atm from -20 °C to 100 °C, compute the change in internal energy, enthalpy, entropy and volume.

Solution: T₁ = 253.15 °K, T₂ = 373.15 °K

With two data points, an average value of C_P and C_V can be found to be

$ Internal energy change, ΔU (Equation 2.22):

$ Enthalpy change, ΔH (Equation 2.24):

$ Entropy change, ΔS (Equation 2.25):

$ Volume change, ΔV (Equation 2.23):

(Pressure, P = 1 atm = 14.696 lb_f/in² = 2116.22 lb_f/ft²)