EAS 6140 Thermodynamics
Worksheet 7– Combined 1st and 2nd Law
14. Write the first and second law combined in enthalpy form (intensive)
dh  Td  vdp
15. The Gibbs energy is defined as g = u - T pv
Write an expression for the Gibbs energy in differential form
dg  du  ( dT  Td )  pdv  vdp
16. Using the expressions in #14 and #15, write an expression for the Gibbs energy in
differential form that has natural independent variables T, p.
dg   dT  vdp
du= T d – p dv
dh= T d + v dp
dg= –  dT + v dp
12. Write the natural independent variables for:
internal energy:
entropy and volume
enthalpy:
entropy and pressure
Gibbs function:
temperature and pressure
13. The conditions for thermodynamic equilibrium are:
a) at constant T and p:
dg = 0
b) at constant  and p:
dh = 0
c) at constant  and V:
du = 0
17. Match the thermodynamic equilibrium conditions:
___c_____ constant , v
a. dg = 0
___b_____ constant , p
b. dh = 0
___a_____ constant T, p
c. du = 0
Potential temperature and the second law
We previously derived equation (2.22):
T2
T1
=
p2
p1
Rc
p
12. Which of the following equations were used in this derivation (circle all that apply)
a) first law of thermodynamics
b) second law of thermodynamics
c) ideal gas law
d) hydrostatic equation
13. Which of the following assumptions were made in this derivation (circle all that apply)
a) adiabatic
b) isothermal
c) isobaric
d) reversible
The potential temperature, , for the atmosphere is defined as
p
 = T p0
Rc
p
where po is 1000 hPa (mb). The potential temperature may be looked upon as the temperature of
a sample of gas would have if it were compressed (or expanded adiabatically from a given state p
and T to a reference pressure of 1000 hPa (mb). Refer to p 66.
14. Take the log of the potential energy equation, then write in differential form (i.e. in the left
hand side you will have d(ln)
ln   ln T 
R
R
ln p0  ln p
cp
cp
 d (ln  )  d (ln T ) 
R
d (ln p)
cp
15. Write the first and second law combined in enthalpy form for an ideal gas (2.3.2)
dh  Td  vdp
16. From the expressions in #14 and #15, write an expression that relates entropy to potential
temperature.
dh  Td  vdp  v 
RT
p
RT
dp  c p dT
p
d  c p d (ln T )  Rd (ln p)
dh  Td 
d (ln  )  d (ln T ) 
R
d (ln p)
cp
c p d (ln  )  c p d (ln T )  Rd (ln p)
 d  c p d (ln  )
17. If potential temperature remains constant, the entropy (increases, decreases, remains the same).
Remains the same
18. The potential temperature remains constant for a sample of ideal gas under dry adiabatic
ascent/descent (yes, no, sometimes)
Yes
19. The potential temperature of a parcel of air with T=288 K at p=900 hPa (mb) is
(greater than, less than, equal to) 288 K.
Greater than
EAS 6140 Thermodynamics of Atmospheres and Oceans
Thermodynamic Charts
Important Note – Because in viewing these charts you might see a slightly different
value, this could lead to differences in calculation results. So use the numbers here in a
relative comparison to your findings.
Using the Stuve (or Skew-T) diagram, read off the temperature and determine the
potential temperature at each of the following levels
Level
1000 mb
TFFR sounding
T

27
300
OAK sounding
T

20
293
900 mb
20
302
22
302.12
700 mb
10
313.4
9
312.28
500 mb
-7
324.32
-9
321.88
Estimate dT/dz and d/dz for the following layers
Level
TFFR sounding
T

1000-900 mb
-10.2ºC/km 2.9 ºC/km
OAK sounding
T

+3.0075ºC/km 13.7 ºC/km
700-500 mb
-6.69ºC/km
-6.28ºC/km 4.04 ºC/km
3.56 ºC/km
Determine the change in T and  of an air parcel initially at 900 mb if subjected to
TFFR sounding
OAK sounding
T

T

a. adiabatic lifting of 100 mb
-10
0
-10
0
b. adiabatic lowering of 100 mb
+10
0
+10
0
c. radiative heating of 10oC
+10
+10
+10
+10
d. radiative cooling of 10oC
-10
-10
-10
-10
Sketch the following paths on the TFFR diagram
a) A parcel of air is lifted dry adiabatically from an initial height of 900 mb to a height of
700 mb. The parcel is then lowered back down to the initial pressure. Label this path A.
Is this path reversible? yes
b) A parcel of air is lifted dry adiabatically from an initial height of 900 mb to a height of
700 mb. The parcel is then heated radiatively by 10oC at constant pressure. The parcel is
then lowered back down to the initial pressure.Label this path B. Is this path
reversible?no