Question 2.8: Calculate the internal energy and enthalpy changes resulting...

Because property changes are independent of process, calculations may be based on any process that accomplishes the change. Here, we choose a two-step, mechanically reversible process wherein 1 mol of air is (a) cooled at constant volume to the final pressure, and (b) heated at constant pressure to the final temperature. Of course, other paths could be chosen, and would yield the same result.

T_{1} = 5 + 273.15 = 278.15  K                     T_{2} = 60 + 273.15 = 333.15  K

With PV = kT, the ratio T/P is constant for step (a). The intermediate temperature between the two steps is therefore:

T′ = (278.15) (1 / 10) = 27.82 K

and the temperature changes for the two steps are:

ΔT_{a} = 27.82  −  278.15 = −250.33  K

ΔT_{b} = 333.15  −  27.82 = 305.33  K

For step (a), by Eqs. (2.17) and (2.14),

ΔU_{a}= C_{V} ΔT_{a} = ( 20.785 ) ( −250.33 ) = −5203.1  J

ΔH_{a} = ΔU_{a} + V Δ P_{a} = −5203.1  J + 2.312 \times  ​10^{−3}  m^{3}  \times  ​( −9 \times  ​10^{5} )  Pa = − 7283.9  J

For step (b), the final volume of the air is:

V_{2} = V_{1}\frac{p_{1}T_{2} }{p_{2}T_{1}} = 2.312 \times 10^{-3}\left(\frac{10 \times  333.15}{1 \times  278.15} \right) = 2.769 \times 10^{−2} m^{3}

By Eqs. (2.21) and (2.14),

Δ H_{b} = C_{P} Δ T_{b} = (29.100)(305.33) = 8885.1 J

Δ U_{b} = Δ H_{b}  − P ΔV_{b} = 8885.1 − (1 \times ​10^{ 5} ) (0.02769  −  0.00231) = 6347.1  J

For the two steps together,

ΔU = − 5203.1 + 6347.1 = 1144.0 J

 ΔH = − 7283.9 + 8885.1 = 1601.2 J

These values would be the same for any process that results in the same change of state^{9} .