1. Consider two parcels of dry air that rise through the atmosphere from the surface (at
1000 hPa) to 800 hPa, maintaining mechanical equilibrium with the environment. Parcel
A rises adiabatically, and Parcel B rises isothermally.
(a) If the specific volume of both parcels at the surface is 0.8 m3, what is the specific
volume of each parcel at 800 hPa? Comment on whether or not (and why) your
calculation of the relative sizes of the parcels’ specific volumes makes physical sense to
you.
For the parcel rising adiabatically, one of Poisson’s equations can be used
p s v s   p800 v800 
 p
v800  v s  s
 p800
1


 
 1000  0.714

 0.8m 3 
 0.938m 3

800



For the parcel rising isothermally, the ideal gas law provides the constraint:
v800 
RT sfc
p800
.
The surface temperature is calculated using the ideal gas law applied at the surface,


p s v s 0.8m 3 1000 x10 2 Pa
Tsfc 

 278.7456K
R
287 J / K  kg
and then use the above equation:
v800 
RTsfc 287J / K  kg 278.7456K 

 1.000m 3
2
p800
800x10 Pa
At 800 hPa, the parcels have the same pressure (mechanical equilibrium), but the
isothermal parcel has a larger volume because it has a higher temperature. Heat has been
added to the isothermal parcel, while the adiabatic parcel is allowed to cool as it rises.
(b) For the parcel that rises isothermally, derive an expression for the time rate of change
of the specific volume as a function of the vertical p-velocity , where
dp

.
dt
T is constant, so
dv d  RT 
RT
  
 

dt dt  p 
p2
2. (a) Show that d  c p d ln  for a reversible processes.
Take the ln of the definition of potential temperature, and then the differential:
R
 p  cp
R
R
  T 0 
 ln   ln T 
ln p 0 
ln p  d ln T  d ln   d ln p
cp
cp
 p 
Derive and expression for dlnT from the first law:
dT
dq
vdp
dq RTdp
dq
c p dT  vdp  dq 
 d ln T 




 d ln p
T
c p T c p T c p T c p pT c p T
Comparing the 2 equations above,
d ln   
dp
dq
dp
dq


 d ln  
 c p d ln   d
p cpT
p
cpT
(b) How did the specific entropy of Parcel A change in moving from the surface to 500
hPa in Question 1 above? How did the specific entropy of Parcel B change?
For parcel A,  doesn’t change. For parcel B,
d  c p d ln 
 sfc  Tsfc  278.7K
 p
800  T800  0
 p800
R
 cp
 1000 

 278.8K

 800 

287
1004
 297.2K
d  1004J / K  kg ln 800  ln 1000   64.53J / K  kg
(c) Were the changes in specific entropy found in (b) above consistent with Claussius’
inequality? Explain briefly.
Yes, since the change of entropy was zero (Parcel A) or positive (Parcel B).
3. Is the dependence of pressure on geometric height (depth) in the troposphere similar to
that in the ocean? Explain briefly.
No. Pressure decreases exponentially with height in the atmosphere, but increases
linearly with depth in the oceans.
4. Why is cp > cv?
Possibly Useful Formulas and Constants
acceleration due to gravity (g) = 9.81 m/s2
Stefan-Boltzman constant () = 5.67 x 10-8 W/m2-K4
solar “constant” (So) = 1370 W/m2
 q 
 q 
cp =   =1004 J/kg-K
cv =   = 717 J/kg-K
 T  v
 T  p
cp - cv = R
gas constants
Rd = dry air: 287 J/K-kg
Rv = water vapor: 461 J/K-kg
density (0˚ C, 1000 mb)
air=1.275 kg m-3
water=1.000 x 103kg m-3
sand = 2.65 x 103kg m-3
q  c p dT  vdp
q  cv dT  pdv
q  dh  vdp
R
 p  cp
  T 0 
 p 
q
 T 0
 
d ln  
q
cp T
 q 
d   
 T  rev
cp

cv
h = u+pv
R
 0.286
cp
Poisson’s Equations:
Tv  1  constant
Tp   constant
The Maxwell Relations:
 T 
 p 
    
 v  s
 s  v
 s 
 p 
   
 v T  T  v
 T 
 pv 
   

 p  s  s  p
 s 
 v 
    
 T  p
 p T
pv  constant