91Ó°ÊÓ

A gas in which there is typically no molecular attraction: a gas that perfectly satisfies the ideal-gas law.

The initial and final temperature are shown on the figure at point a and c respectively. So,

The pressure at point a, $P_a=3×{10}^{5}Pa$

The volume at point a, ${\mathrm{V}}_{\mathrm{a}}=0.02{\mathrm{m}}^3$

The pressure at point c, $P_c={10}^{5}Pa$

The volume at point c,$V_c=0.06m^3$

The initial and final states already given, so use the ideal gas equation to obtain a comparison between the temperatur ${\mathrm{T}}_{\mathrm{C}}\mathrm{and}{\mathrm{T}}_{\mathrm{a}}$.

$$\begin{gathered} \frac{P_a{V}_a}{T_a}=\frac{P_c{V}_c}{T_c} \\ \frac{T_c}{T_a}=\frac{P_c{V}_c}{P_a{V}_a} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} \frac{T_c}{T_a}=\frac{{10}^5×0.06}{3×{10}^5×0.02} \\ =1 \\ {T}_c=T_a \end{gathered}$$

The final temperature equals the initial temperature which indicates that the temperature is constant during the process.

The heat flow during the process abc is given by $Q_{abc}=Q_{ab}+Q_{bc}$.

For the , $Q_{ab}$it is given by the first law of thermodynamic.

As the volume is constant during the process ab, the work done will be zero. Also in part-a, we concluded that the temperature is constant during the process and as we know ${\mathrm{Δ\pm «}}_{\mathrm{ab}}$depends on the change in temperature which in this case is zero.

$$\begin{gathered} {\mathrm{Δ\pm «}}_{\mathrm{ab}}={\mathrm{nC}}_{\mathrm{V}}{\mathrm{Δ°Õ}}_{\mathrm{ab}} \\ ={\mathrm{nC}}_{\mathrm{V}}×0 \\ =0 \end{gathered}$$

Therefore,

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{ab}}=∆{\mathrm{U}}_{\mathrm{ab}}+{\mathrm{W}}_{\mathrm{ab}} \\ =0+0 \\ =0 \end{gathered}$$

For ${\mathrm{Q}}_{bc}$, it is given by ${\mathrm{Q}}_{\mathrm{bc}}=∆{\mathrm{U}}_{\mathrm{bc}}+{\mathrm{W}}_{\mathrm{bc}}$.

As the pressure is constant, the work done is given by

$$\begin{gathered} {\mathrm{W}}_{\mathrm{bc}}={\mathrm{P}}_{\mathrm{b}}\left({\mathrm{V}}_{\mathrm{c}}-{\mathrm{V}}_{\mathrm{b}}\right) \\ ={10}^5\left(0.06-0.02\right) \\ =4\mathrm{kJ} \end{gathered}$$

So, the heat energy is given by

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{bc}}=∆{\mathrm{U}}_{\mathrm{bc}}+{\mathrm{W}}_{\mathrm{bc}} \\ =0+4 \\ =4\mathrm{kJ} \end{gathered}$$

Now, after calculating and we can plug their values into the equation to calculate the heat flow during the process abc.

$$\begin{gathered} {\mathrm{Q}}_{a\mathrm{bc}}={\mathrm{U}}_{ab}+{\mathrm{W}}_{\mathrm{bc}} \\ =0+4 \\ =4\mathrm{kJ} \end{gathered}$$

To calculate the heat in the path ac, we must determine the work done ${\mathrm{W}}_{\mathrm{ac}}\mathrm{and}∆{\mathrm{U}}_{\mathrm{ac}}$

role="math" localid="1668082244899" ${\mathrm{Q}}_{\mathrm{ac}}={\mathrm{W}}_{\mathrm{ac}}+∆{\mathrm{U}}_{\mathrm{ac}}$

Here is zero. The work done will be

$$\begin{gathered} {\mathrm{W}}_{\mathrm{ac}}=\frac{1}{2}\left({\mathrm{P}}_{\mathrm{a}}+{\mathrm{P}}_{\mathrm{c}}\right)\left({\mathrm{V}}_{\mathrm{c}}+{\mathrm{P}}_{\mathrm{a}}+{\mathrm{V}}_{\mathrm{a}}\right) \\ =\frac{1}{2}\left(3×{10}^5+{10}^5\right)\left(0.06-0.02\right) \\ =8\mathrm{kJ} \end{gathered}$$

Now, we will put the values of and into the equation to get heat flow.

$$\begin{gathered} {\mathrm{Q}}_{ac}={\mathrm{W}}_{\mathrm{bc}}+∆{\mathrm{U}}_{\mathrm{ac}} \\ =8+0 \\ =8\mathrm{kJ} \end{gathered}$$

The heat flow is positive which means the air absorb heat during the process from the surroundings.