Chapter 2, Problem 2.6P

Interpretation Introduction

Interpretation:

The numerical values for the missing quantities indicated by question marks.

Step	$\Delta U^t\text{/J}$	$Q\text{/J}$	$W/\text{J}$
12 23 34 42	$-200$ ? ? 4700	?   $-3800$   $-800$ ?	$-6000$ ?   $300$ ?
12341	?	?	$-1400$

Concept introduction:

First Law of Thermodynamics states that the energy can transfer from one form to another form but neither be created nor be destroyed. Thus, in an isolated system, total energy is constant.

The mathematical expression is given by:

  $\text{ΔU=Q+W}$

Where,

△U is change in internal energy
Q is heat
W is work done

Answer to Problem 2.6P

Step	$\Delta U^t\text{/J}$	$Q\text{/J}$	$W/\text{J}$
12 23 34 42	$-200$   $-4000$   $-500$ 4700	$5800$   $-3800$   $-800$   $200$	$-6000$   $-200$   $300$ 4500
12341	0	1000	$-1400$

Explanation of Solution

Given Data:

$\Delta U^t{}_{12}=-200$$W_{12}=-6000$$Q_{23}=-3800$$Q_{34}=-800$

$W_{34}=300$$\Delta U^t{}_{41}=4700$$W_{12341}=-1400$

This is a simple numeric question based on the formulae of First Law of Thermodynamics.

Formula given by first law of thermodynamics is:

  $\Delta U^t=Q+W$

Calculation:

From the First Law of Thermodynamics, we know that

$\Delta U^t=Q+W$

For the Step 1 to 2: $\Delta U^t{}_{12}=-200$$W_{12}=-6000$

$\begin{array}{c}{Q}_{12}=\Delta U^t{}_{12}-W_{12}\\ =-200-\left(-6000\right)\\ =5800\text{ J}\\ =5.8\times {10}^3\text{ J}\\ Q_{12}=5.8\times {10}^3\text{ J}\end{array}$

Step 3 to 4:

 $Q_{34}=-800$$W_{34}=300$

$\begin{array}{c}\Delta U^t{}_{34}=Q_{34}+W_{34}\\ =-800+\left(-300\right)\\ =-500\text{ J}\\ \Delta U^t{}_{34}=-500\text{ J}\end{array}$

Step 1 to 2 to 3 to 4 to 1:  Since $\Delta U^t$ is a state function, $\Delta U^t$ for a series of steps that leads back to initial state must be zero. Therefore, the sum of the $\Delta U^t$ values for all the steps must sum to zero.

$\begin{array}{c}\Delta U^t{}_{23}=-\Delta U^t{}_{12}-\Delta U^t{}_{34}-\Delta U^t{}_{41}\\ =200+500-4700\\ =-4000\text{ J}\\ \Delta U^t{}_{23}=-4000\text{ J}\end{array}$

Step 2 to 3:

 $\Delta U^t{}_{23}=-4000\text{ J}$$Q_{23}=-3800\text{ J}$

$\begin{array}{c}{W}_{23}=\Delta U^t{}_{23}-Q_{23}\\ =-4000+3800\\ =-200\text{ J}\\ W_{23}=-200\text{ J}\end{array}$

For a series of steps, the total work done is the sum of work done for each step,

$W_{12341}=-1400\text{ J}$

$\begin{array}{c}{W}_{12341}=W_{12}+W_{23}+W_{34}+W_{41}\\ W_{41}=W_{12341}-W_{12}-W_{23}-W_{34}\\ =-1400+6000+200-300\\ =-4500\text{ J}\\ =4.5\times {10}^{-3}\text{ J}\\ W_{41}=4.5\times {10}^{-3}\text{ J}\end{array}$

Step 4 to 1: $\Delta U^t{}_{41}=4700\text{ J}$$W_{41}=4500\text{ J}$

$\begin{array}{c}{Q}_{41}=\Delta U^t{}_{41}-W_{41}\\ =4700-4500\\ =200\text{ J}\\ Q_{41}=200\text{ J}\end{array}$

For the total process,

  $\begin{array}{c}\Delta U^t{}_{12}+\Delta U^t{}_{23}+\Delta U^t{}_{34}+\Delta U^t{}_{41}=\Delta U^t{}_{12341}\\ \Delta U^t{}_{12341}=-200-4000-500+4700\\ =0\text{ J}\\ \Delta U^t{}_{12341}=0\text{ J}\end{array}$

  $\begin{array}{c}{Q}_{12}+Q_{23}+Q_{34}+Q_{41}=Q_{12341}\\ Q_{12341}=5800-3800-800+200\\ =1000\text{ J}\\ Q_{12341}=1000\text{ J}\end{array}$

The required values for the given table are as follows,

$Q_{12}=5.8\times {10}^3\text{ J}$$\Delta U^t{}_{34}=-500\text{ J}$$\Delta U^t{}_{23}=-4000\text{ J}$$W_{23}=-200\text{ J}$

$W_{41}=4.5\times {10}^{-3}\text{ J}$$Q_{41}=200\text{ J}$$\Delta U^t{}_{12341}=0\text{ J}$$Q_{12341}=1000\text{ J}$

Conclusion

Step	$\Delta U^t\text{/J}$	$Q\text{/J}$	$W/\text{J}$
12 23 34 42	$-200$   $-4000$   $-500$ 4700	$5800$   $-3800$   $-800$   $200$	$-6000$   $-200$   $300$ 4500
12341	0	1000	$-1400$