Three moles of an ideal gas are taken around cycle acb shown in Fig. P19.42 . For this gas, ${\mathit{C}}_{\mathbf{p}}\mathbf{=}\mathbf{29}\mathbf{.}\mathbf{1}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{.}\mathit{K}$. Process ac is at constant pressure, process ba is at constant volume, and process cb is adiabatic. The temperatures of the gas in states a, c, and b are ${\mathit{T}}_{\mathbf{a}}\mathbf{=}\mathbf{300}\mathit{K}\mathbf{,}{\mathit{T}}_{\mathbf{c}}\mathbf{=}\mathbf{492}\mathit{K}$and ${\mathit{T}}_{\mathbf{b}}\mathbf{=}\mathbf{600}\mathit{K}$. Calculate the total work for the cycle.

For path ac, the pressure is constant while the volume is changing. So, the work done will be $W_{ac}=p∆V$. But there is no information for pressure and volume.

According to the ideal gas equation,

$\mathit{p}\mathbf{∆}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathbf{∆}\mathit{T}$.

Consider the given data as below.

The temperature, $T_c=492K$

The temperature, $T_a=300K$

So, work done will be

localid="1664343178081" $$\begin{gathered} W_{ac}=nR∆T \\ =nR\left(T_c-T_a\right) \\ =3\times 8.314\times \left(492-300\right) \\ =4.79kJ \end{gathered}$$

At path cb, the process is adiabatic where $∆Q=0$. To calculate the work done, we will use the first law of thermodynamics, where $W=Q-∆U$. But there is no heat involved, so, $W_{cb}=-∆U$.

Here, change in internal energy is given by $∆U=n{C}_v∆T$.

The relation between molar heat capacities is

${\mathit{C}}_{\mathbf{p}}\mathbf{-}{\mathit{C}}_{\mathbf{v}}\mathbf{=}\mathit{R}$.

$$\begin{gathered} W_{cb}=-∆U_{cb} \\ =-n{C}_v∆T_{cb} \\ =-n\left(C_p-R\right)\left(T_b-T_c\right) \\ =-3\left(29.1-8.314\right)\left(600-492\right) \\ {W}_{cb}=-6.73kJ \end{gathered}$$

In the path ba, the volume is constant. So, the work will be zero.

The total work is given by

$$\begin{gathered} W_t=W_{ac}+W_{cb}+W_{ba} \\ =4.79-6.73+0 \\ =-1.95kJ \end{gathered}$$

Hence, the work is $-1.95KJ$. Since, the work done is negative, the work is done on the gas.