Question

A volume of air (assumed to be an ideal gas) is first cooled without changing its volume and then expanded without changing its pressure, as shown by path abc in Fig. . (a) How does the final temperature of the gas compare with its initial temperature? (b) How much heat does the air exchange with its surroundings during process abc? Does the air absorb heat or release heat during this process? Explain. (c) If the air instead expands from state a to state c by the straight-line path shown, how much heat does it exchange with its surroundings?

Short answer

(a) $T_c=T_a$

(b) $Q_{abc}=4kJ$

(c)$Q_{ac}=8kJ$

Step-by-step solution

Step 1:

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

(a) Calculate the final temperature using gas equations

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

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

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.

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

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

(b) Calculate the heat

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 ${\mathrm{\Delta U}}_{\mathrm{ab}}$depends on the change in temperature which in this case is zero.

$$\begin{gathered} {\mathrm{\Delta U}}_{\mathrm{ab}}={\mathrm{nC}}_{\mathrm{V}}{\mathrm{\Delta T}}_{\mathrm{ab}} \\ ={\mathrm{nC}}_{\mathrm{V}}\times 0 \\ =0 \end{gathered}$$

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.

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

(c) Calculate the heat energy in case of expansion

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

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

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.