THE THERMODYNAMICS OF SIMPLE SYSTEMS
3.11 ADIABATIC WORK
Because partial derivatives like (∂U/∂T)p, and so on, can be handled just as though they were algebraic variables, it is possible to develop quite an arsenal of equa- tions relating the first law quantities described so far and to expand them to include other variables (Klotz and Rosenberg, 2008). An important concept is that of adi- abatic (perfectly insulated) work done on or by a gas. The work dw behaves like a thermodynamic function because the path has been specified by setting q =0.
Now dU = dq+dw=p dV for a system restricted to pressure–volume work. The energy U is a state variable U = f (V,T ) for one mole, so
dU = ∂U
∂V
T
dV+ ∂U
∂T
V
d T and
∂U
∂V
T
dV+ ∂U
∂T
V
d T+p dV=0
The first term above drops out if we consider expansion of an ideal gas because the functional dependence on V disappears (Joule experiment). Also, (∂U/∂T)V =CV
ADIABATIC WORK 51 and p=RT
V so
CVd T + RT
V dV=CV
d T
T +RdV V =0 Since Cp−CV =R, we obtain
CV
d T
T +RdV V = d T
T +Cp−CV
CV
dV V = d T
T +(γ−1)dV V =0 Integrating between limits, with a little algebraic manipulation, yields
T Vγ−1 =k and pVγ =k where k and kare constant.
The expression pVγ =k looks like Boyle’s law except for the parameterγ, which is always greater than 1. The presence ofγ causes the pressure p=k/Vγ to be lower at any point during the expansion than it is during the isothermal (Boyle’s law) expansion (upper curve in Fig. 3.7).
The difference between the two expansions is in the heat that flows or does not flow into the system to maintain T=const. In the isothermal case heat transfer is allowed.
Heat is not allowed into the system in the adiabatic case where dq=0 by definition.
Without a compensating heat flow, the adiabatic system cools during expansion and the pressure is always lower at any specified volume than it is in the isothermal case ( p=k/Vγ, γ >1.0). The entire pV curve falls below the isothermal curve in Fig. 3.7.
p
V V1 V2
FIGURE 3.7 Two expansions of an ideal gas. The upper curve is isothermal and the lower curve is adiabatic. The adiabatic expansion does less work because there is no heat flow into the system.
PROBLEMS AND EXAMPLE Example 3.1 Line Integrals
What is the line integral of the function f (x,y)=x y over the parabolic curve y= f (x)=x2/2 from (x,y)=(0,0) to 1,12
?
Solution 3.1 One way of writing a line integral of the function I =
C f (x,y) ds over the curve C is to specify y = f (x) and ds=
1+
d y dx
21/2
dx. For example, integrating the function f (x,y)=x y over the parabolic curve y= f (x)=x2/2 from (x,y)=(0,0) to 1,12
. We have (Steiner, 1996) f (x,y)=x y=x
x2 2
=x3 2 and
dy dx =
d x2
2
dx =x Thus,
I =
C
f (x, f (x))
1+ dy
dx 21/2
dx= 1
0
x3
2 1+x21
2dx
= 1 2
1
0
x6+x81/2 dx
where the limits of integration are the limits on x. Integration by MathcadC gives 1
2 1
0
x6+x8.5
dx=0.161
Problem 3.1
One expression of a line integral is
C
F (x,y) dx+G(x,y) dy
where the subscript C indicates a line (or curve) integral. If F (x,y)= −y, G(x,y)= x y, and the line is the diagonal from x=1 to y=1 (Fig. 3.8). Carry out the integration.
PROBLEMS AND EXAMPLE 53 y
x 1
1
FIGURE 3.8 C=Diagonal along x =1 to y=1.
Problem 3.2
If F (x,y)= −y and G(x,y)=x y, evaluate the line integral over the quarter-circular arc from x =1 to y=1 (Fig. 3.9). Notice that the beginning and end points are the same and the functions are the same as in the previous problem but the path is different.
y
x 1
1
FIGURE 3.9 C=Quarter-circular arc.
Problem 3.3
A mass m of 20.0 kg is raised to a height h of 20.0 m and allowed to drop. Ignoring air resistance, what is its speed when it hits the ground? What is its kinetic energy?
Problem 3.4
A 20-kg rock was carried up a hill that is 20 m high.
(a) What was the energy increase in the rock?
(b) What amount of work was done on the rock when carrying it up the hill?
(c) What is the kinetic energy gained by the rock when it rolls down the hill?
Problem 3.5
In a Joule experiment, two 20.3-kg weights fell 1.524 m to drive a paddle wheel immersed in 6.31 kg of water. The experiment was repeated 20 times, after which the temperature of the water bath was found to have risen by 0.352 K. What is the mechanical equivalent of heat in J K−1according to this experiment?
Problem 3.6
Show that dw is an inexact differential for one mole of an ideal gas undergoing a reversible expansion.
Problem 3.7
The definition of the calorie is that amount of heat that is necessary to raise 1 g of water 1 K. As it stands, this definition is approximately valid for temperature changes not too far from room temperature. Use this approximate definition to predict the final temperature of 200 g of water at 283 K mixed with 450 g of water at 350 K.
Problem 3.8
Show that, for an ideal gas with constant CV, we have T2
T1 = V1
V2
R
CV
Problem 3.9
From the kinetic theory of gases, we get the expression U =32RT for the energy of an ideal monatomic gas. Show that dU is an exact differential.
Problem 3.10
(a) How much energy is required to heat 1.0 mol of an ideal monatomic gas at constant V from 25.0 to 75.0 K?
(b) What is the heat input if the process is carried out at constant pressure?
Problem 3.11
A 10.0-g piece of iron was heated to 100.0◦C by immersing it in boiling water and then quickly transferring it to an insulated beaker containing 1000 g of water at 25◦C. What was the final temperature of the water? The specific heat of iron is 0.449 Jg−1K−1. The specific heat of water is 4.184 Jg−1K−1.
PROBLEMS AND EXAMPLE 55 Problem 3.12
The ratioγ =Cp/CV is 1.40 for nitrogen, N2. The speed of sound in air (mostly nitrogen) is said to be
vsound∼= γRT
M
where M is the molar mass of nitrogen. Find the value of vsoundat 273 K and compare it with the experimental value of 334 m s−1.