EAS 651/ASTRO 651 Advanced Atmospheric Physics
Problem Set #1
Due: Thursday, September 16, 2004
1. In general, there are an infinite number of paths through state space that a
thermodynamic system can take in going from state A to state B. If the amount of work,

W, done by the system (or on the system) is independent of the path, dr , than the force

that acted to change the state of the system, F , is said to be conservative.
B 
W   F  dr .
A

Mathematically, F will be conservative if it can be written as the gradient of a potential
function, f, because
B  B
B
 B f
W   F  d r   f  d r   dr   df f B  f A ,
A
A
A r
A
i.e., W depends only on the initial and final states of the system, independent of path.
Note that for a cyclic process
W   dW  0

when F is conservative.




(a) Show that if F  M( x, y, z) i  N( x, y, z) j  P( x , y, z)k is conservative, then
P N

,
y z
M P

,
z x
and
N M
.

x
y
(b) Show that if a vector field is conservative, then it is also irrotational.
2. A plume of heated air leaves the cooling tower of a power plant at 1000 hPa with
T=30ºC. Assuming that the air is dry, calculate the level to which the plume will ascend
if the environmental temperature varies with altitude (z in km) as
(a) T(z) = 20 - 8z
(b) T(z) = 20 + z.
(c) Which situation, part (a) or (b) above, is applicable to the troposphere and which to
the stratosphere? What can you deduce from this problem about the vertical motion (and
vertical stability) of parcels that start off 10C warmer than the environment in these 2
different regions of the atmosphere?
(Ref. Problem 2.1 in Salby’s Fundamentals of Atmospheric Physics)
A parcel will rise until its density matches that of its environment. The pressure of the
parcel will adjust instantaneously to the environment. Assuming an ideal gas, this means
that the temperature of the parcel T  Tsurface  9.8 K / km z km (adiabatic cooling)


will equal the air temperature when the densities are equal. (i.e., it only takes 2 state
variables to completely describe an ideal gas, and the problem constrains pressure and
density).
(a)
T  20  8z
T  30  9.8z
Set them equal and solve for z to find z=5 km.
(b)
T  20  z
T  30  9.8z
Set them equal and solve for z to find z=0.9 km.
(c) (a) is the troposphere and (b) is the middle/lower stratosphere. Both parcels are
buoyant at z=0, but the parcel in the stratosphere only rises ~1 km to stabilize, where
the tropospheric parcel rises 5 km. This suggests that vertical motion is deeper in the
troposphere as compared with the stratosphere.
3. Problem 2.6 in Salby: A large helium balloon ascends from sea level to 10 hPa. If the
balloon ascends rapidly enough so that heat transfer across its surface is negligible, what
relative increase in volume must be accommodated?
Adiabatic conditions are assumed so Poisson’s equations apply:
pv  constant
1
3
 p  
1000  5
piv i  p f v f  v f   i  vi  
v i  15.85vi
 10 
pf 
So a 15.85-fold increase in volume for the balloon must be accommodated.
Note: Use  = 5/3 for a monotomic gas (helium).


4. Problem 2.12 in Salby: Commercial aircraft cruise near 200 hPa where T = -60ºC.
How much specific heat must be added or removed to maintain a cabin temperature of
25ºC if the cabin air is adiabatically compressed to maintain cabin pressure of 800 hPa?
(a) Tp  constant; =0.286

p 
Tf  Ti  f   316.87 K
p i 
(b) Need to remove an amount of heat given by
dq p  C p T  1.91x10 4 J / kg.
5. Problem 2.15 in Salby: A scuba diver releases a bubble at T=30ºC at a depth of 32 ft,
where the pressure is 2 atm. Assuming that the bubble remains intact, heat transfer
across its surface is negligible, and condensational heating in the bubble’s interior can be
neglected, derive an expression for the bubble’s temperature as a function of elevation z
above its initial location. What will be the temperature of the air inside the bubble when
it reaches the surface?
For the ocean (water) environment,
p w
  w g
z
Since ocean water is essentially incompressible (  w  constant), this equation
integrates to
pw  p0   w gz .
As the bubble rises, the bubble pressure always equals the pressure of the ocean
water, i.e.,
p B (z)  p W (z);  B (z)  W (z); TB (z)  T W (z);
p B (z)  p 0   w gz
(1)
Since the bubble expands adiabatically,
TB(z)pB (z)  TB (0)p B(z  0)  .
Using (1) to replace p B (z) in (2), and setting pB(z  0)  p0 ,

  gz 
TB(z)  TB(0)1 w  .
p 0 

(2)
6. Calculate changes in internal energy, enthalpy, and entropy for the following reversible
processes in a unit mass of dry air:
(a) isobaric cooling (at ~ surface temperature and pressure) from 26 to 18C at
p = 1000 mb
(b) isobaric cooling (at ~ upper-tropospheric temperature and pressure) from -55
to -65C at p = 200 mb
(c) isothermal expansion from v = 700 cm3/g to 750 cm3/g at T = 250 K.
7. Consider a Carnot cycle, which uses a unit mass of dry air as the working substance,
consisting of the following four steps:
I. adiabatic compression from 60 kPa and 0C to a temperature of 25C
II. isothermal expansion to p = 70 kPa
III. adiabatic expansion to T = 0C
IV. isotherml compression back to 60 kPa.
Draw this Carnot cycle on a p-v diagram. Calculate the work done and the engine's
efficiency.
Particulars for the drawing:
(a) adiabatic compression from p1 = 60 kPa to p2, with a temperature change of
T1 = 273 K to T2 = 298 K
Use the Poisson relation Tp   with k = 0.286 to find that p2 = 81 kPa. Also note
from the ideal gas law that v1 = 1.31 m3/kg and v2 = 1.05 m3/kg.
(b) isothermal expansion from p2 = 81 kPa to p3 = 70 kPa at T2 = T3 = 298 K.
Use the ideal gas law to find that v3 = 1.215 m3/kg.
(c) adiabatic expansion to T4 = 273 K. Again, Poisson’s relation Tp   can be
used to find that p4 = 51.5 kPa and v4 = 1.52 m3/kg.

(b) work done = Rd T1  T2 ln C
B
where
T1 = the warm temperature
T2 = the cold temperature
C = specific volume resulting from the isothermal expansion done in contact
with the warm reservoir (fromB)
B = specific volume resulting from the adiabatic compression done
Calculate B and C
Consider the first adiabatic leg. It starts at (pA,T2) and compresses to T1. Using the
ideal gas law and one of Poisson’s equations,
B 
T2 p A
Rd T1
pB

1
T  

1
 T1 p B  pB  
T 
 p A  81.5 kPa
 2 
Now consider the isothermal expansion at T1 to C - It starts at (pB,T1) and finishes with
(pc,T1), where pc = 70 kPa. . Using the ideal gas law,
C 
So,
Rd T1
.
pC
 C pB 81.5


 B pC
70

p
Therefore, WORK  Rd T1  T2 ln C  Rd T1  T2 ln B  1,091 J / kg
B
pC
T
(c)   1  2  0.08  8%
T1
8. Problem 3.3 in Salby. Consider a parcel crossing isentropic surfaces. If radiative
cooling is just large enough to keep the parcel on a fixed isobaric surface, what is the
relative change of the parcel’s temperature when its potential temperature has decreased
to 90% of its initial value?
For an isobaric process,
q dT
d ln  

 d ln T
cp T T
Tf  f

,
Ti
i
and the temperature must decrease to 90% of its initial value.
Therefore,