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Making connections: macroscopic and microscopic

In thermodynamics, we often use the macroscopic picture when making calculations of how a system behaves, while the atomic and molecular picture gives underlying explanations in terms of averages and distributions. We shall see this again in later sections of this chapter. For example, in the topic of entropy, calculations will be made using the atomic and molecular view.

To get a better idea of how to think about the internal energy of a system, let us examine a system going from State 1 to State 2. The system has internal energy U 1 size 12{U rSub { size 8{1} } } {} in State 1, and it has internal energy U 2 size 12{U rSub { size 8{2} } } {} in State 2, no matter how it got to either state. So the change in internal energy Δ U = U 2 U 1 size 12{ΔU=U rSub { size 8{2} } - U rSub { size 8{1} } } {} is independent of what caused the change. In other words, Δ U size 12{ΔU} {} is independent of path . By path, we mean the method of getting from the starting point to the ending point. Why is this independence important? Note that Δ U = Q W size 12{ΔU=Q - W} {} . Both Q size 12{Q} {} and W size 12{W} {} depend on path , but Δ U size 12{ΔU} {} does not. This path independence means that internal energy U size 12{U} {} is easier to consider than either heat transfer or work done.

Calculating change in internal energy: the same change in U size 12{U} {} Is produced by two different processes

(a) Suppose there is heat transfer of 40.00 J to a system, while the system does 10.00 J of work. Later, there is heat transfer of 25.00 J out of the system while 4.00 J of work is done on the system. What is the net change in internal energy of the system?

(b) What is the change in internal energy of a system when a total of 150.00 J of heat transfer occurs out of (from) the system and 159.00 J of work is done on the system? (See [link] ).

Strategy

In part (a), we must first find the net heat transfer and net work done from the given information. Then the first law of thermodynamics ( Δ U = Q W size 12{ΔU=Q - W} {} ) can be used to find the change in internal energy. In part (b), the net heat transfer and work done are given, so the equation can be used directly.

Solution for (a)

The net heat transfer is the heat transfer into the system minus the heat transfer out of the system, or

Q = 40 . 00 J 25 . 00 J = 15 . 00 J. size 12{Q="40" "." "00"" J"-"25" "." "00"" J"="15" "." "00"" J"} {}

Similarly, the total work is the work done by the system minus the work done on the system, or

W = 10 . 00 J 4 . 00 J = 6 . 00 J. size 12{W="10" "." "00"" J"-4 "." "00"" J"=6 "." "00"" J"} {}

Thus the change in internal energy is given by the first law of thermodynamics:

Δ U = Q W = 15 . 00 J 6 . 00 J = 9 . 00 J. size 12{DU=Q-W="15" "." "00"" J"-6 "." "00"" J"=9 "." "00"" J"} {}

We can also find the change in internal energy for each of the two steps. First, consider 40.00 J of heat transfer in and 10.00 J of work out, or

Δ U 1 = Q 1 W 1 = 40 . 00 J 10 . 00 J = 30 . 00 J. size 12{DU rSub { size 8{1} } =Q rSub { size 8{1} } -W rSub { size 8{1} } ="40" "." "00"" J"-"10" "." "00"" J"="30" "." "00"" J"} {}

Now consider 25.00 J of heat transfer out and 4.00 J of work in, or

Δ U 2 = Q 2 W 2 = - 25 . 00 J ( 4 . 00 J ) = –21.00 J. size 12{DU rSub { size 8{2} } =Q rSub { size 8{2} } -W rSub { size 8{2} } "=-""25" "." "00"" J"- \( -4 "." "00"" J" \) "=-""21" "." "00"" J"} {}

The total change is the sum of these two steps, or

Δ U = Δ U 1 + Δ U 2 = 30 . 00 J + 21 . 00 J = 9 . 00 J. size 12{DU=DU rSub { size 8{1} } +DU rSub { size 8{2} } ="30" "." "00"" J"+ left (-"21" "." "00"" J" right )=9 "." "00"" J"} {}

Discussion on (a)

No matter whether you look at the overall process or break it into steps, the change in internal energy is the same.

Solution for (b)

Here the net heat transfer and total work are given directly to be Q = 150 . 00 J size 12{Q"=-""150" "." "00"" J"} {} and W = 159 . 00 J size 12{W"=-""159" "." "00"" J"} {} , so that

Δ U = Q W = 150 . 00 J ( 159 . 00 J ) = 9 . 00 J. size 12{DU=Q-W"=-""150" "." "00"" J"- \( -"159" "." "00"" J" \) =9 "." "00"" J"} {}

Discussion on (b)

A very different process in part (b) produces the same 9.00-J change in internal energy as in part (a). Note that the change in the system in both parts is related to Δ U size 12{ΔU} {} and not to the individual Q size 12{Q} {} s or W size 12{W} {} s involved. The system ends up in the same state in both (a) and (b). Parts (a) and (b) present two different paths for the system to follow between the same starting and ending points, and the change in internal energy for each is the same—it is independent of path.

Practice Key Terms 3

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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