A thermodynamic system is taken from state a to state c in Fig. Pl9 .42 along either path a b c

A thermodynamic system is taken from state a to state c in...

A thermodynamic system is taken from state a to state $c$ in Fig. $\mathrm{Pl9} .42$ along either path $a b c$ or path adc. Along path $a b c,$ the work $W$ done by the system is 450 $\mathrm{J}$ . Along path adc, Wis 120 $\mathrm{J}$ . The internal energies of each of the four states shown in the figure are $U_{a}=150 \mathrm{J}$ $U_{b}=240 \mathrm{J}, \quad U_{c}=680 \mathrm{J}, \quad$ and $U_{d}=330 \mathrm{J}$ . Calculate the heat flow $Q$ for each of the four processes $a b, b c, a d,$ and $d c .$ In each process, does the system absorb or liberate heat?

A thermodynamic system is taken from state a to state $c$ in Fig. $\mathrm{Pl9} .42$ along either
path $a b c$ or path adc. Along path $a b c,$ the work $W$ done by the system is 450 $\mathrm{J}$ . Along path adc, Wis 120 $\mathrm{J}$ . The internal energies of each of the four states shown in the figure are $U_{a}=150 \mathrm{J}$ $U_{b}=240 \mathrm{J}, \quad U_{c}=680 \mathrm{J}, \quad$ and $U_{d}=330 \mathrm{J}$ . Calculate the heat flow $Q$ for each of the four processes $a b, b c, a d,$ and $d c .$ In each process, does the system absorb or liberate heat?

Key Concepts

Sign Conventions in Thermodynamics

Proper use of sign conventions is critical when analyzing thermodynamic processes. Typically, work done by the system is considered positive, while work done on the system is negative; similarly, heat added to the system (absorbed) is taken as positive and heat released as negative. These conventions help in consistently applying the first law of thermodynamics and correctly interpreting the nature of energy exchanges in any process.

Heat Flow Calculation

Calculating the heat flow in any process involves manipulating the first law of thermodynamics. By rearranging the equation ?U = Q - W, one can determine the heat transfer as Q = ?U + W for each process. This calculation is fundamental for quantifying the energy exchange in thermodynamic processes and determining whether the system is absorbing heat or releasing it.

Path Dependence in Thermodynamic Processes

The quantities of work and heat are path-dependent, meaning that they can vary even when a system starts and ends at the same state. This concept is important in thermodynamics because it explains how different procedures (or paths) to change a system's state can lead to significantly different energy exchanges, even though the change in a state function such as internal energy remains constant.

State Functions vs Path Functions

State functions, such as internal energy, depend only on the current state of a system and not on the path taken to reach that state. In contrast, path functions like heat and work depend on the specific process or path followed between two states. Recognizing this distinction is essential for analyzing thermodynamic processes and understanding why different paths between the same states can yield different heat and work values.

First Law of Thermodynamics

This principle states that the change in a system's internal energy is equal to the heat added to the system minus the work done by the system. It is a statement of energy conservation applied to thermodynamic processes. Understanding this law is crucial because it relates the internal energy, heat transfer, and work in any process, irrespective of the specific pathway taken by the system.

Transcript

00:01 So, from the problem given figure, we can observe the pressure volume diagram, which shows thermodynamic processes, and we are asked to evaluate the heat flow for each of the process.

00:16 Now, for every single one.

00:19 So we have four processes.

00:21 We have process from a to b, we have a process from b to c, we have process from a to c, we have also given the work done in state abc and in the state adc.

00:41 Also we are given every single internal energy state for the point a, for the point b, for the point c and for the point d.

00:53 So from the first law of thermodynamics for every part of this problem now this problem is a singular problem but he has it has parts where we have to evaluate every single process given in the figure so for we will use the first law of thermodynamics for every transition for every process so it will be that q equals delta u plus w and delta u equals u final minus u initial so this will be used for every process.

01:48 So here we're given and we will start from the process a to b now.

01:57 In this case, first we will evaluate for every process we will first evaluate the change in internal energy and then we will evaluate the work done or we will already have given the work done.

02:12 So and then we'll simply simply.

02:15 Add them together to evaluate the q, the heat corresponding.

02:21 So for a to b, first we need to find out the change in internal energy, which is going to be delta u .a, b, and this is equal ub minus ua.

02:38 We are given this and we are given this term.

02:41 So this is 240 joules minus 150 joules.

02:47 And this equals to 90 joules.

02:54 Also, the work done from a to b is going to be zero since there will be no change in the volume here.

03:04 So the work done is zero and we can therefore state that q equals qa b actually equals delta uab.

03:21 Hue ab equals 90 joules.

03:27 This is for this part of the process.

03:30 This is the first process given on the pv diagram.

03:33 So for the next process which is from b to c the work is not zero.

03:43 So we can start by evaluating the internal energy change corresponding to this process...

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