## Chapter 3

## Q. 3.3

Air is compressed from an initial state of 1 bar and 298.15 K to a final state of 3 bar and 298.15 K by three different mechanically reversible processes in a closed system:

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

(b) Isothermal compression.

(c) Adiabatic compression followed by cooling at constant volume.

These processes are shown in the figure. We assume air to be in its ideal-gas state, and assume constant heat capacities, C^{Ig}_{V} = 20.785 and C^{Ig}_{P} = 29.100 J·mol^{−1}·K^{−1}. Calculate the work required, heat transferred, and the changes in internal energy and enthalpy of the air for each process.

## Step-by-Step

## Verified Solution

Choose the system as 1 mol of air. The initial and final states of the air are identical with those of Ex. 2.7. The molar volumes given there are

V^{ig}_{1} = 0.02479 m^{3} V^{ig}_{2} = 0.008263 m^ {3}

Because T is the same at the beginning and end of the process, in all cases,

ΔU^{ig} = Δ H^{ig} = 0

(a) The process here is exactly that of Ex. 2.7(b), for which:

Q = −4958 J and W = 4958 J

(b) Equation (3.20) for isothermal compression applies.The appropriate value of R here (from Table A.2 of App. A) is R = 8.314 J· mol^{−1} · K^{−1} .

\varrho = – W RT In \frac{V^{ig}_{2} }{V^{ig}_{1} } = RT In \frac{P_{1} }{P_{2}} ( const T ) (3.20)

**Table A.2: Values of the Universal Gas Constant**

R = 8.314 J·mol^{−1} ·K^{−1} = 8.314 m³ ·Pa·mol^{−1} ·K^{−1}

= 83.14 cm³ ·bar·mol^{−1} ·K^{−1}= 8314 cm³ ·kPa·mol^{−1} ·K^{−1}

= 82.06 cm³ ·(atm) ·mol^{−1 }·K^{−1} = 62,356 cm³ ·(torr) ·mol^{−1} ·K^{−1}

= 1.987 (cal)·mol^{−1}· K^{−1} = 1.986 ( Btu ) (lb mole)^{−1} (R)^{−1}

= 0.7302 (ft)³ ( atm ) (lb mol)^{−1} (R)^{−1}= 10.73 (ft)³ ( psia ) (lb mol)^{−1} (R)^{−1}

= 1545 (ft) ( lb_{f} ) (lb mol)^{−1} ( R )^{−1}

Q = −W = RT ln \frac{P_{1} }{P_{2}} = ( 8.314 ) ( 298.15 ) ln\frac{1}{3} = -2723 J

(c) The initial step of adiabatic compression takes the air to its final volume of 0.008263 m³. By Eq. (3.23a), the temperature at this point is:

T(V^{ig} )^{-1} = const (3.23a)

T^{‘} = T_{1} \left(\frac{V^{ig}_{1} }{V^{ig}_{2}} \right) ^{\gamma – 1} = ( 298.15 )\left(\frac{0.02479}{0.008263} \right) ^{0.4} = = 462.69 K

For this step, Q = 0, and by Eq. (3.25), the work of compression is:

W =\Delta U^{ig}_{}= C^{ig}_{V} \Delta T (3.25)

W = C^{ig}_{V} \Delta T = C^{ig}_{V}(T^{‘} – T_{1}) =( 20.785 )( 462.69 − 298.15 ) = 3420 J

For the constant-volume step, no work is done; the heat transfer is:

Q = Δ U^{ig} = C^{ig}_{V} ( T_{2} − T′ ) = 20.785 ( 298.15 − 462.69 ) = −3420 J

Thus for process (c),

W = 3420 J and Q = −3420 J

Although the property changes ΔU^{ig} and ΔH^{ig} are zero for each process, Q and W are path-dependent, and here Q = −W. The figure shows each process on a PVig diagram. Because the work for each of these mechanically reversible processes is given by W = \int_{}^{}{Pd} V^{ig} , the work for each process is proportional to the total area below the paths on the PVig diagram from 1 to 2. The relative sizes of these areas correspond to the numerical values of W.