Question 2.13.1: Ten liters of a monatomic ideal gas at 25° C and 10 atm pres...

The size of the system must first be calculated. From consideration of the initial state of the system (the point a in Figure 2.6),

n = the number of moles =$\frac{P_{a}V_{a}}{RT_{a}} =\frac{10\times 10}{0.08206\times 298} = 4.09$

a. The isothermal reversible expansion . The state of the gas moves from a to b along the 298-degree isotherm. Along any isotherm, the product PV is constant:

$V_b =\frac{P_aV_a}{P_b} =\frac{10\times 10}{1} = 100$ liters

For an ideal gas undergoing an isothermal process, ΔU = 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 kJ of work and absorbs 23.3 kJ of heat from the constant-temperature surroundings.

Since for an ideal gas, H is a function only of temperature , $\Delta H^{\prime }_{(a\rightarrow b)} =0$; that is,

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 times, given by $PV^\gamma$ = constant, and the final state is the point c in the diagram. The volume$V_c$ is obtained from $P_aV^{\gamma }_{a} = P_cV^{\gamma }_{c}$ as

$V_c =\left\lgroup\frac{10\times 10^{5/3} }{1} \right\rgroup ^{3/5} = 39.8$ liters

and

$T_c =\frac{P_cV_c}{nR} =\frac{1\times 39.8}{4.09\times 0.08206} =119$ degrees

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  kJ.

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 ).

$\Delta U^{\prime }_{(a\rightarrow e)} =0$, as this is an isothermal change of state

$\Delta U^{\prime }_{(e\rightarrow c)} =q_v(\Delta V =0$, and hence, w=0)

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

Thus,

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 ).

$\Delta U^{\prime }_{(a\rightarrow d)} =q_v(\Delta V =0$, and hence, w=0)

$=\int_{a}^{d}{nc_vdT }$ , and as the state d lies in the 119-degree isotherm, then

$\Delta U^{\prime }_{(d\rightarrow c)}=0$, as this is an isothermal process, and hence,

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 ).

$\Delta U^{\prime }_{(a\rightarrow d)}=0$, as this process is isothermal

$\Delta U^{\prime }_{(b\rightarrow c)}=q_p-w$, and as $P_b=P_c$, then$w=P_b(V_c-V_b)$

Since $c_v=1.5  R$ and $c_p-c_v=R$, then $c_p-c_v=R$; and as 1 liter atm equals 101.3 J,

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 ).

$\Delta U^{\prime }_{(a\rightarrow f)} = q_v(V_a=V_f$, and hence,$w=0)$

From the ideal gas law,

That is, the state f lies on the 30-degree isotherm. Thus,

Thus,

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

From the ideal gas law,

$T_g\frac{P_gV_g}{nR}=\frac{10\times 39.8}{4.09\times 0.08206}=1186$ degrees

and hence, the state g lies on the 1186-degree isotherm. Thus,

Thus,

The value of $\Delta U^{\prime }_{({a\rightarrow c})}$ 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 (Figure 2.6). The enthalpy change is most simply calculated from the consideration of a path which involves an isothermal portion, over which $\Delta H^\prime =0$, and an isobaric portion, over which $\Delta H^\prime =q_p=\int{nc_pdT}$ . 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 kJ. In the case of the reversible adiabatic path, q = 0, and hence, w = +9.12 kJ. 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