Question

A thermodynamic system is taken from state a to state c in Fig. P19.38 along either path abc or path adc. Along path abc, the work W done by the system is 450 J. Along path adc, W is 120 J . The internal energies of each of the four states shown in the figure are ${\mathbf{U}}_{\mathbf{a}}\mathbf{=}\mathbf{150}\mathbf{}\mathbf{J}\mathbf{,}{\mathbf{U}}_{\mathbf{b}}\mathbf{=}\mathbf{240}\mathbf{}\mathbf{J}\mathbf{,}{\mathbf{U}}_{\mathbf{c}}\mathbf{=}\mathbf{680}\mathbf{}\mathbf{J}$and ${\mathbf{U}}_{\mathbf{d}}\mathbf{=}\mathbf{330}\mathbf{}\mathbf{J}$. Calculate the heat flow for each of the four processes ab, bc, ad, and dc. In each process, does the system absorb or liberate heat?

Short answer

Theamount of heat in segment:

ab: +90 J

bc: +890 J

ad: 300 J

dc: 350 J

Step-by-step solution

Identification of the given data.

Universal gas constant $\mathrm{R}=8.315\frac{\mathrm{J}}{\mathrm{mol},\mathrm{K}}$.

From the P-V diagram,

The internal energies of the segments: ab, bc, ad, dc are:-

${\mathrm{U}}_{\mathrm{a}}=150\mathrm{J},{\mathrm{U}}_{\mathrm{b}}=240\mathrm{J},{\mathrm{U}}_{\mathrm{c}}=680\mathrm{J}$ and ${\mathrm{U}}_{\mathrm{d}}=330\mathrm{J}$

The work done along the paths: abc and adc are:-

${\mathrm{W}}_{\mathrm{abc}}=450\mathrm{J}\mathrm{and}{\mathrm{W}}_{\mathrm{adc}}=120\mathrm{J}$

Concept of First Law of Thermodynamics.

The first law of thermodynamics reiterates or supports the law of conservation of energy in thermodynamic systems or in terms of thermodynamic macro parameters which states that energy can only be transformed to other forms and not be created or destroyed.

Mathematically,

$\mathbf{∆}\mathbf{U}\mathbf{=}\mathbf{Q}\mathbf{-}\mathbf{W}$

(1)

Here, $∆\mathrm{U}$is the internal energy, $\mathrm{Q}$is the heat energy in or out the system and $\mathrm{W}$is the work done on or by the system.

Determine the amount of heat in the segments ab, bc, ad and dc.

Rewrite equation (1) as below.

$\mathrm{Q}=∆\mathrm{U}+\mathrm{W}$

The segment wise heat energies are calculated below,

For the segment ab:

W = 0 , as the volume remains constant.

So, the heat is equal to the change in internal energy between point a and b,

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{ab}}=∆\mathrm{U} \\ =240\mathrm{J}-150\mathrm{J} \\ =90\mathrm{J} \end{gathered}$$

For the segment bc:

${\mathrm{W}}_{\mathrm{bc}}=450\mathrm{J}$

Now define the amount heat as below.

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{bc}}=∆\mathrm{U}+\mathrm{W} \\ =\left(680-240\right)\mathrm{J}+450\mathrm{J} \\ =890\mathrm{J} \end{gathered}$$

For the segment ad:

${\mathrm{W}}_{\mathrm{ad}}=120\mathrm{J}$

So, the heat will be

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{ad}}=∆\mathrm{U}+\mathrm{W} \\ =\left(330-150\right)\mathrm{J}+120\mathrm{J} \\ =300\mathrm{J} \end{gathered}$$

For the segment dc:

$\mathrm{W}=0$, as the volume remains constant

So, the heat is equal to the change in internal energy between point d and c,

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{dc}}=∆\mathrm{U} \\ =680\mathrm{J}-330\mathrm{J} \\ =350\mathrm{J} \end{gathered}$$

Hence, the heat energy are calculated for the four segments.