NUMERICAL EXAMPLES

Ten liters of a monatomic ideal gas at 25°C and 10 atm pressure are expanded to a final pressure of 1 atm. The molar heat capacity of the gas at constant volume, C_v, is 3/2 R and is independent of temperature. Calculate the work done, the heat absorbed, and the change in U and in H for the gas if the process is carried out (1) isothermally and reversibly, and (2) adiabatically and reversibly. Having determined the final state of the gas after the reversible adiabatic expansion, verify that the change in U for the process is independent of the path taken between the initial and final states by considering the process to be carried out as

i. An isothermal process followed by a constant-volume process ii. A constant-volume process followed by an isothermal process iii. An isothermal process followed by a constant-pressure process iv. A constant-volume process followed by a constant-pressure process v. A constant-pressure process followed by a constant-volume process

The size of the system must first be calculated. From consideration of the initial state of the system (the point a in Fig. 2.3)

(a) The isothermal reversible expansion. The state of the gas moves from a to b along the 298 degrees isotherm. As, along any isotherm, the product PV is constant,

5. For an ideal gas, the internal energy U is a function only of temperature, and

34 Introduction to the Thermodynamics of Materials

Figure 2.3 The five process paths considered in the numerical example.

For an ideal gas undergoing an isothermal process, U=0 and hence, from the First Law,

Thus in passing from the state a to the state b along the 298 degree isotherm, the system performs 23.3 kilojoules of work and absorbs 23.3 kilojoules of heat from the constant-temperature surroundings.

(b) The reversible adiabatic expansion. If the adiabatic expansion is carried out reversibly, then during the process the state of the system is, at all time, given by PV^=constant, and the final state is the point c in the diagram. The volume V_C is obtained

from as

and

The point c thus lies on the 119 degree isotherm. As the process is adiabatic, q=0 and hence

The work done by the system as a result of the process equals the decrease in the internal energy of the system=9.13 kilojoules.

(i) An isothermal process followed by a constant-volume process (the path a → e → c;

that is, an isothermal change from a to e, followed by a constant-volume change from e to c).

and as the state e lies on the 298 degree isotherm then

As, for an ideal gas, H is a function only of temperature, then H_(a→b)=0; that is,

36 Introduction to the Thermodynamics of Materials

(ii) A constant-volume process followed by an isothermal process (the path a → d → c;

that is, a constant-volume change from a to d, followed by an isothermal change from d to c).

(iii) An isothermal process followed by a constant-pressure process (the path a → b → c;

that is, an isothermal change from a to b, followed by a constant-pressure change from b to c).

As C_v=1.5 R and C_p–C_v=R, then C_p=2.5 R; and as 1 liter atm equals 101.3 joules,

Thus

(iv) A constant-volume process followed by a constant-pressure process (the path a → f → c; that is, a constant-volume change from a to f, followed by a constant-pressure change from f to c).

Thus

From the ideal gas law

i.e., the state f lies on the 30 degrees isotherm. Thus

Thus

(v) A constant-pressure process followed by a constant-volume process (the path a → g → c; i.e., a constant-pressure step from a to g, followed by a constant-volume step from g to c).

From the ideal gas law

and hence the state g lies on the 1186 degrees isotherm. Thus

Thus

38 Introduction to the Thermodynamics of Materials

The value of U_(a→c) is thus seen to be independent of the path taken by the process between the states a and c.

The change in enthalpy from a to c. The enthalpy change is most simply calculated from consideration of a path which involves an isothermal portion over which H=0 and an isobaric portion over which H=q_p=nc_pdT. For example, consider the path a → b → c

and hence

or alternatively

in each of the paths (i) to (v) the heat and work effects differ, although in each case the difference q–w equals
9.12 kilojoules. In the case of the reversible adiabatic path, q=0 and hence w=+9.12 kilojoules. If the processes (i) to (v) are carried out reversibly, then For path (i) q=
9.12+the area aeih

For path (ii) q=
9.12+the area dcih

For path (iii) q=
9.12+the area abjh
the area cbji For path (iv) q=
9.12+the area fcih

For path (v) q=
9.12+the area agih

PROBLEMS

2.1 An ideal gas at 300 K has a volume of 15 liters at a pressure of 15 atm. Calculate (1) the final volume of the system, (2) the work done by the system, (3) the heat entering or leaving the system, (4) the change in the internal energy, and (5) the change in the enthalpy when the gas undergoes

a. A reversible isothermal expansion to a pressure of 10 atm b. A reversible adiabatic expansion to a pressure of 10 atm

The constant volume molar heat capacity of the gas, C_v, has the value 1.5 R.

2.2 One mole of a monatomic ideal gas, in the initial state T=273 K, P=1 atm, is subjected to the following three processes, each of which is conducted reversibly:

b. Then a doubling of its pressure at constant volume,

c. Then a return to the initial state along the path P=6.64310
⁴V²+ 0.6667.

Calculate the heat and work effects which occur during each of the three processes.

2.3 The initial state of a quantity of monatomic ideal gas is P=1 atm, V=1 liter, and T=373 K. The gas is isothermally expanded to a volume of 2 liters and is then cooled at constant pressure to the volume V. This volume is such that a reversible adiabatic compression to a pressure of 1 atm returns the system to its initial state. All of the changes of state are conducted reversibly. Calculate the value of V and the total work done on or by the gas.

2.4 Two moles of a monatomic ideal gas are contained at a pressure of 1 atm and a temperature of 300 K. 34,166 joules of heat are transferred to the gas, as a result of which the gas expands and does 1216 joules of work against its surroundings. The process is reversible. Calculate the final temperature of the gas.

2.5 One mole of N₂ gas is contained at 273 K and a pressure of 1 atm. The addition of 3000 joules of heat to the gas at constant pressure causes 832 joules of work to be done during the expansion. Calculate (a) the final state of the gas, (b) the values of U and H for the change of state, and (c) the values of C_v and C_p for N₂. Assume that nitrogen behaves as an ideal gas, and that the above change of state is conducted reversibly.

2.6 Ten moles of ideal gas, in the initial state P₁=10 atm, T₁=300 K, are taken round the following cycle:

a. A reversible change of state along a straight line path on the P–V diagram to the state P=1 atm, T=300 K,

b. A reversible isobaric compression to V=24.6 liters, and c. A reversible constant volume process to P=10 atm.

How much work is done on or by the system during the cycle? Is this work done on the system or by the system?

2.7 One mole of an ideal gas at 25°C and 1 atm undergoes the following reversibly