## Chapter 2

## Q. 2.7

Air at 1 bar and 298.15 K is compressed to 3 bar and 298.15 K by two different closed-system mechanically reversible processes:

(a) Cooling at constant pressure followed by heating at constant volume.

(b) Heating at constant volume followed by cooling at constant pressure.

Calculate the heat and work requirements and ΔU and ΔH of the air for each path. The following heat capacities for air may be assumed independent of temperature:

C_{V} = 20.785 and C_{P}= 29.100 J· mol^{-1}·K^{-1}

Assume also that air remains a gas for which PV/T is a constant,regardless of the changes it undergoes. At 298.15 K and 1 bar the molar volume of air is 0.02479 m^{3} · mol^{−1}.

## Step-by-Step

## Verified Solution

In each case take the system as 1 mol of air contained in an imaginary piston/cylinder arrangement. Because the processes are mechanically reversible, the piston is imagined to move in the cylinder without friction. The final volume is:

V_{2} = V_{1} \frac{P_{1} }{P_{2} } = 0.02479(\frac{1}{3} ) = 0.008263 m^{3}.

The two paths are shown on the V vs. P diagram of Fig. 2.3(I) and on the T vs. P diagram of Fig. 2.3(II).

(a) During the first step of this path, air is cooled at a constant pressure of 1 bar until the final volume of 0.008263 m³ is reached. The temperature of the air at the end of this cooling step is:

T^{‘} = T_{1} \frac{V_{2} }{V_{1} } = 298.15\left(\frac{0.008263}{0.02479} \right) = 99.38 K

Thus, for the first step,

Q = ΔH = C P ΔT = (29.100) (99.38 − 298.15) = −5784 J

W = − P ΔV = − 1 \times 10^{ 5} Pa \times (0.008263 − 0.02479) m^{ 3} = 1653 J

ΔU = ΔH − Δ(PV) = ΔH − P ΔV = − 5784 + 1653 = − 4131 J

The second step is at constant V_{2 } with heating to the final state. Work W = 0, and for this step:

ΔU = Q = C_{V } ΔT = (20.785) (298.15 − 99.38) = 4131 J

V ΔP = 0.008263 m^{3} \times (2 \times 10^{5} ) Pa = 1653 J

ΔH = ΔU + Δ(PV) = ΔU + V ΔP = 4131 + 1653 = 5784 J

For the overall process:

Q = −5784 + 4131 = −1653 J

W = 1653 + 0 = 1653 J

ΔU = − 4131 + 4131 = 0

ΔH = − 5784 + 5784 = 0

Notice that the first law, ΔU = Q + W, applied to the overall process is satisfied.

(b) Two different steps of this path produce the same final state of the air. In the first step air is heated at a constant volume equal to V_{1 } until the final pressure of 3 bar is reached. The air temperature at the end of this step is:

T^{‘} = T_{1} \frac{P_{2} }{P_{1} } = 298.15\left(\frac{3}{1} \right) =894.45 K

For this first constant-volume step, W = 0, and

Q = ΔU = C_{V} ΔT = ( 20.785 ) ( 894.45 − 298.15 ) = 12,394 J

V ΔP = ( 0.02479) ( 2 \times 10^{5} ) = 4958 J

ΔH = ΔU + V ΔP = 12,394 + 4958 = 17,352 J

In the second step air is cooled at P = 3 bar to its final state:

Q = ΔH = C_{P} ΔT = ( 29.10 ) ( 298.15 − 894.45 ) = − 17,352 J

W = −P ΔV = − ( 3 \times 10^{5})( 0.008263 − 0.02479 ) = 4958 J

ΔU = ΔH − Δ ( PV ) = ΔH − P ΔV = − 17,352 + 4958 = − 12,394 J

For the two steps combined,

Q = 12,394 − 17,352 = − 4958 J

W = 0 + 4958 = 4958 J

Δ U = 12,394 − 12,394 = 0

ΔH = 17,352 − 17,352 = 0

This example illustrates that changes in state functions (ΔU and ΔH) are independent of path for given initial and final states. On the other hand, Q and W depend on the path. Note also that the total changes in ΔU and ΔH are zero. This is because the input information provided makes U and H functions of temperature only, and T_{1} = T_{2}. While the processes of this example are not of practical interest, state-function changes (ΔU and ΔH) for actual flow processes are calculated as illustrated in this example for processes that are of practical interest. This is possible because the state-function changes are the same for a reversible process, like the ones used here, as for a real process that connects the same states.