GEF2200 vår 2017 Løsningsforslag sett 1

A.1.T

R∗ is the universal gas constant, with value8.3143JK−1mol−1. R is the gas constant fora speci�c gas, given by

R =R∗

M(1)

where M is the molecular weight of the gas(usually given in g/mol). For the equation ofstate:

pV = nR∗T =m

MR∗T = mRT (2)

It is important to note that we usually usemass m in units of [kg], which requires thatthe units of R is changed accordingly (if givenin [J/gK], it must be multiplied by 1000 [g/kg]).

A.2.T

Apparent molecular weight is the averagemolecular weight for a mixture of gases. Weintroduce it to calculate a gas constant R =R∗/Md for the mixture , whereMd is the appar-ent molecular weight of i di�erent gases givenby Equation (3.10):

Md =

∑mi∑ miMi

=

∑mi

n=

∑niMi

n

=∑ ni

nMi (3)

In meteorology the most common apparentmolecular weight is the one of air.

WH06 3.19

Concentrations by volume (See exercise A.8.T):

vN2 =VN2

V(4)

vCO2 =VCO2

V(5)

Assuming ideal gas, we have total volume V =VN2 + VCO2 at a given temperature (T ) and

pressure (p). This means volume concentrationis equal to molecular fraction:

vi =ViV

=niR

?T/p

nR?T/p=nin

(6)

The appearent molecular mass is the molecularmass a mixture of n moles of gass and a totalmass m appears to have:

Mve =m

n=

∑mi

n=

∑niMi

n

=∑ ni

nMi =

∑viMi (7)

Inserting values for CO2 and N2:

Mve = 0.95× (12 + 2× 16) g/mol

+0.05× (2× 14) g/mol

= 43.2g/mol (8)

This is quite higher than for our atmosphere,and gives the gas constant

Rve =R?

Mve=

8.3143JK−1mol−1

43.2gmol−1

= 0.1925Jg−1K−1

= 192.5Jkg−1K−1 (9)

which is lower than for the Earth's atmosphere.

A.4.T

We have that

R =R∗

M(10)

To �nd if R increases when adding water vapor(or any gas) to dry air, will be the same as �nd-ing that the mixture becomes lighter (molecularmass decreases).Looking at dry air:

Md =

∑mi∑ miMi

=md

nd(11)

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Adding mass mx with molecular weight Mx

gives

Mnew =

∑mi +mx∑ miMi

+ mxMx

=md +mxmdMd

+ mxMx

= Mdmd +mx

md +mxMdMx

(12)

If Mx < Md, we see that Mnew < Md (lighterthan dry air), which means that Rnew > R.If Mx > Md, then Mnew > Md (heavier thandry air) and Rnew < R.

A.5.T

Because R is dependent on humidity, we movethis dependency from R to T :

RT = RdTv (13)

so that we can use the gas constant for dry airin calculations.

A.6.T

When we look at a single gas we use the gasconstant for that gas. For water vapor we useRv = 461JK−1kg−1. We therefore do not usevirtual temperature.

WH06 3.20

Given 1% of water vapor in the air, what is Tv−T? The mole fraction of water vapor is equal tothe pressure fraction exerted by the water vapor(see A.7.T). Thus e/p = 0.01. Using a typicaltemperature T = 288 K, equation (3.16) gives:

Tv − T =T

1− ep(1− ε)

− T

=288K

1− 0.01 · (1− 0.622)− 288K

= 1.1K

A.7.T

At sea level the pressure is about 1013 hPa.The pressure fraction exerted by water vapor is

therefore

e

p=

18hPa

1013hPa= 0.0178 (14)

The ideal gas law as stated in (3.6) is

pV = nR∗T (15)

The partial pressure exerted by water vapor isthe pressure it would exert had it alone occu-pied the same volume as the total gas mixture.We may therefore use (3.6) again on water va-por alone:

eV = nvR∗T (16)

where nv is the number of moles of water vaporcontained in the volume V . Combining the twoequations, we �nd that

nvn

=e

p(17)

(From this deduction, we see that the mole frac-tion of any gas component in a mixture equalsthe pressure fraction exerted by that compo-nent.)So the mole fraction of water vapor in this caseis 1.78 %.

A.9.T

The hydrostatic equation gives the relationshipbetween gravity and pressure for an air mass atrest, a relationship between a change in pres-sure dp and a change in height dz (see Figure1). Since the pressure at a given height is due tothe weight of the above air, we expect the pres-sure to decrease with height, meaning that a de-crease in pressure equals an increase in height.

We need to consider the forces acting upon thelayer, and to do so, we look at the edges of thelayer (height z and z+dz. Gravity works down-wards, and the force on the top of the layer isFg(z + dz). Furthermore, the layer acts on theair below with a force Fg(z). Since the layer isat rest, we have from Newton's thrid law that

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Figure 1: The forces on a layer of air.

the layer experience a similar force in the op-posite direction (−Fg(z)).For a body with mass m, the gravity is mg =%V g, where V is the volume. Given that thelayer has a thickness dz and area A, and there-fore a volume dV = Adz, we have (assumingpositive axis upwards):

Fg = Fg(z + dz)− Fg(z)= −%gA(z + dz − z)= −%gAdz (18)

Forces acting upwards: Pressure (force perarea). The upward force at the layer top isopposed by a similar force acting on the layer(−Fp(p+ dp)).

Fp = Fp(p)− Fp(p+ dp)

= [p− (p+ dp)]A = −Adp (19)

Where Fp(p) > Fp(p + dp), that is, dp isnegative. Fp is then value of the force actingupwards.

To �nd the equilibrium between gravity andpressure forces, we set Fp + Fg = 0, and cancel

A.

−%gdz − dp = 0 (20)

It is very common to do such calculations per

area, leaving out A from the start.

When using vector calculation (k-vector):

Fgk = −%gA(z + dz)k− (−%gAz)k= −%gAdzk (21)

Fpk = [p− (p+ dp)]Ak = −Adpk (22)

In any case, when the sum of forces is zero, weget

dp

dz= −%g (23)

A.10.T

The geopotential at a given location is the en-ergy required to lift 1 kg from sea level and upto that location.

φ(z) =

∫ z

0gdz′ (24)

Dividing by the average surface acceleration ofgravity, we get the geopotential height.

Z =1

g0

∫ z

0gdz′ (25)

Using the hydrostatic equation, the equation ofstate and the de�nition of virtual temperature:

dp

dz= −%g = − p

RTg = − p

RdTvg (26)

Using the de�nition of geopotential height

gdz = g0dZ (27)

we havedp = − p

RdTvg0dZ (28)

Which we integrate from p1, Z1 to p2, Z2:

dZ = −RdTvg0

dp

p∫ Z2

Z1

dZ =

∫ p2

p1

−RdTvg0

dp

p(29)

= −Rdg0

∫ p2

p1

Tvdp

p(30)

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By introducing the weighted mean virtual tem-perature

Tv =

∫ p2p1Tv

dpp∫ p2

p1dpp

(31)

and inserting into Equation (30), we may write

Z2 − Z1 = −Rdg0Tv

∫ p2

p1

dp

p(32)

= −Rdg0Tv ln

(p2p1

)(33)

This means that the thickness is linearly pro-portional to the mean temperature of the layer.A horizontal gradient in this temperaturewill then mean a horizontal gradient in layerthickness. This is an important feature of theatmosphere, especially when it comes to stormsystems and low/high pressures.

WH06 3.27

Meteorological station 50m below sea surface.The temperature at the surface is Tvs = 15◦C= 288K, and the mean temperature in the layerbetween 1000 and 500hPa is Tv = 0◦C = 273K.

Zs = −50mps = 1020hPa

g0 = 9.81ms−2

R = 287Jkg−1K−1

To �nd the height of the 500hPa level, we usethe hypsometric equation. You have to as-sume that the temperature from the surfaceup to 1000hPa is constant. (If the layer be-tween 1000hPa and 500hPa was isothermal, thewould be physically correct to calculate a meantemperature for the layer between 1020hPa and1000hPa, to avoid temperature jumps.) Assum-ing the mean temperature in the layer betweenps and p1000=1000hPa to be Tvs , we integrate

the hypsometric equation∫ Z500

Zs

dZ = −Rdg0

∫ p500

ps

Tvdp

p(34)

Z500 − Zs = −Rdg0

[∫ p1000

ps

Tvsdp

p

+

∫ p500

p1000

Tvdp

p

](35)

Z500 − Zs = −Rdg0

[Tvs ln

p1000ps

+ Tv lnp500p1000

](36)

Inserting the values:

Z500 = 5.7km (37)

WH06 3.26

Ignoring the virtual temperature correction, wemay use the hypsometric equation (3.29) on thelocation at the storm centre

Z200hPa − Zsea level =RdTAg0

ln940hPa

200hPa(38)

and on the surrounding region:

Z200hPa − Zsea level =RdTBg0

ln1010hPa

200hPa(39)

TA is the (unknown) mean temperature be-tween the sea surface and 200 hPa at the stormcentre, and TB = −3◦ C = 270.2 K is the meantemperature in the surrounding region.Since the 200 hPa surface is perfectly �at, andthe ocean surface is also �at, the left-hand sidesof the two above equations are the same. There-fore:

RdTAg0

ln940hPa

200hPa=RdTBg0

ln1010hPa

200hPa(40)

TA = TBln 1010hPa

200hPa

ln 940hPa200hPa

(41)

= 270.2Kln 1010hPa

200hPa

ln 940hPa200hPa

(42)

= 282.7K = 9.5◦C (43)

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The air column above the storm centre has inaverage 12.5◦C higher temperature than thesurroundings.

A.11.T

dq is the heat added to a system.dw is the work done by the system.du is the change in internal energy.Di�erent kinds of work include expansion,surface expansion, extension and electricalwork. Expansion work is the important one inthermodynamics of the atmosphere, so we canassume dw = pdα.

WH06 3.18j

../FIGURES/tire-eps-converted-to.pdf

Figure 2: An air parcel expands when leavinga tire: T1 = T , p1 > p.

We follow an air parcel leaving the tire, to getout into the surroundings. The pressure insidethe tire is higher than the pressure of the sur-roundings (p1 > p), but the temperature is thesame (T1 = T ). The air parcel will then expandadiabatically, since it is entering air with lower

pressure. From the �rst law of thermodynam-ics, we have

dq = du+ dw = cvdT + pdα = 0 (44)

In the expansion, the speci�c volume (volumeper mass) increases (dα > 0), so that the tem-perature change is negative:

dT = − p

cvdα < 0 (45)

Since the air expands, the work done on thesurroundings is positive, equals to dw = pdα.The decrease in temperature means that theinternal energy (du = cvdT ) also decreases.This is what happens when an air parcel risein the atmosphere; it expands and cools. Thisis also important for the heat transport fromthe equator to the poles.

A.12.T

An adiabatic process is a process where noheat exchange between the system and thesurroundings is allowed. That means dq = 0.

3.18h

When a low-pressure is colder than its sur-roundings, the isobars are closer together inthe low-pressure than in the surroundings.Going upwards (towards lower pressure), theheight of a given isobar will get increasinglylower in the low-pressure centre compared toin the surroundings (see �gure 3.3b in WH06).

3.18k

In our study of thermodynamics, the only worktaken into account is the work done in expand-ing the volume of the system. We then have:

dw = pdα (46)

Here, p is the pressure at the 'edge' of oursystem, where the environment is 'pushed'away as the system expands by dα. To getzero work we can either let the volume remain

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constant (dα = 0) or let the system expandinto a vacuum (p = 0, there are no surround-ings to do work on).

WH06 3.32 (WH78 2.24)

We have isothermal compression of 2kg dry airat T = 15◦C, to one tenth of its initial volume.

Work done on the air parcel = − work doneby the air parcel. The only work done by theparcel is expansion work:

W =

∫ V2

V1

pdV (47)

We have the ideal gas equation

p =mRT

V(48)

Inserting and integrating gives, when remem-bering that T is constant,

W =

∫ V2

V1

pdV = mRT lnV2V1

(49)

= 2kg · 287Jkg−1K−1 · 288K ln1

10= −3.806 · 105J

This is the work done by the parcel, so thework done on the parcel when compressing itis −W = 3.806 · 105J.

WH06 3.33 (WH78 2.23)

� a �We have an ideal gas, undergoing an adiabatictransformation, and we will prove that pV γ =constant for this process, where γ = cp/cv.The �rst law of thermodynamics (TD1) is givenby Eqn. (WH06 3.41):

dq = cvdT + pdα (50)

where dq = 0 for an adiabatic process. Di�er-entiating the ideal gas equation and inserting

(50) for dT , we get

pα = RT (51)

αdp+ pdα = RdT (52)

= −R p

cvdα (53)

αdp = −(1 +

R

cv

)pdα (54)

Using Eqn. (WH06 3.45), we have that

1 +R

cv=

1

cv(cv +R) =

cpcv

= γ (55)

and we get

αdp = −(1 +

R

cv

)pdα

= −γpdα (56)

Re-arranging:

−dpp

= γdα

α(57)

Now er integrate from p0, α0 to p, α

−∫ p

p0

dp

p=

∫ α

α0

γdα

α(58)

lnp0p

= γ lnα

α0= ln

(α

α0

)γ(59)

p0p

=

(α

α0

)γ(60)

and because

α

α0=

1/%

1/%0=%0%

=m/V0m/V

=V

V0(61)

we have (V

V0

)γ=

p0p

pV γ = p0Vγ0 (62)

� b �For an isothermal transformation we have fromthe equation of state

p1V1 = p2V2 (63)

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So for an isothermal compression from pressurep1 =1000hPa and volume V1 =7.5cm3 to thevolume V2 =2.5cm

3, we �nd the pressure

p2 = p1V1V2

= 1000hPa7.5

2.5= 3000hPa

To �nd the pressure after adiabatic expansionuntil the original volume V3 = V1, we use theexpression from Equation (62)

p3Vγ3 = p2V

γ2

p3 = p2

(V2V1

)γ(64)

= 3000hPa

(2.5

7.5

)γ= 643hPa

The temperature after expansion is found usingthe equation of state

p3V3T3

=p2V2T2

(65)

T3 =p3V3p2V2

T2 (66)

where we can insert the pressure p3 from Equa-tion (64) (or the value calculated). We alsonote that T2 = T1, since the �rst transforma-tion was an isothermal transformation (and stillV3 = V1):

T3 = T2V3p2V2

p2

(V2V3

)γ(67)

= T1

(V2V1

)−1(V2V1

)γ= T1

(V2V1

)γ−1= 290K

(2.5

7.5

)γ−1= 186.8K

Comment on the physical side:

The result that pV γ is constant for an adiabaticprocess means that the volume changes di�er-ently from an isothermal process (where pV is

constant). In the case of expansion from onepressure to another (p), the volum of the adia-batic system (Va) after transformation is givenby

pV γa = Ca

Va =

(Cap

)1/γ

(68)

where Ca is the constant value. The corre-sponding volume change for the isothermal sys-tem (Vi) is given by

pVi = Ci

Vi =Cip

(69)

where Ci is the constant value.Because γ = cp/cv > 1, isothermal expansiongives a faster/larger increase in volume thandoes the adiabatic expansion (Equation (68)vs (69)). The physical reason is that theadiabatic system cools during the expansion,whereas the isothermal system is kept atconstant temperature by adding heat to thesystem. The increased temperature increasesthe volume or the pressure.

In this exercise, the volume was the samefor the two processes, but the pressure wasnot: The adiabatic expansion resulted in alower pressure (643hPa vs 1000hPa), and alower temperature (187K vs 290K). Clearly,adding heat to keep the process isothermalmeans heating the air by 103K, which indeedshould increase the pressure (remember thatthe volume is constant).

In the case of compression, Equation (68) and(69) show that the decrease in volume is smallerfor the isothermal process than for the adia-batic process. The argument follows the samelines as for expansion; since an adiabatic com-pression means that the temperature increases,whereas heat is removed from the isothermal

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system to keep it at constant temperature, theisothermal volume (or pressure) will be smaller.

A.13.T

Speci�c heat: The amount of heat you haveto supply to a body of unit mass to raise itstemperature by 1 degree.

dq

dT= c (70)

So that if dT = 1, we have dq = c. c dependson whether the system is allowed to changeits volume because in that case, work will bedone either by or on the system. This will alsocontribute to changing its temperature.

To �nd the relationship cv + R = cp, we startwith constant volume:

cv =

(dq

dT

)αconst

=

(du

dT

)αconst

(71)

and because of Joule's law for an ideal gas, duis only dependent on temperature

cv =du

dT(72)

Then we have from the TD1, and by rewritingpdα

dq = cvdT + pdα (73)

= cvdT + d(pα)− αdp (74)

= cvdT + d(RT )− αdp (75)

= (cv +R)dT − αdp (76)

So we see that for constant pressure, we havedp = 0 and(

dq

dT

)pconst

= cp = cv +R (77)

E.1.T

� a �When the air is contained in a closed and rigid

box, its volume is constant. We therefore usethe speci�c heat at constant volume:

dq

dT= cv (78)

⇒ q = cv(T2 − T1) (79)

Since we have 2 kg of air, the heat that mustbe added is:

Q = mq (80)

= mcv(T2 − T1) (81)

= 2kg · 717J/Kkg · 10K (82)

= 14.3kJ (83)

� b �Now pressure is constant. We do the same asin (a), just using cp = 1004J/Kkg instead ofcv. We then get Q = 20.1kJ.

� c �CANCELLED� d �For an adiabatic process we have

T2T1

=(p2p1

)R/cp(84)

(Eq. 3.54 in WH06). We solve for the pressure:

p2 = p1

(T2T1

)cp/R(85)

= 1000hPa ·(283.2K273.2K

) 1004287.0

(86)

= 1134hPa (87)

We �nd the speci�c volume from the ideal gasequation:

α2 =RdT2p2

(88)

=287.0J/Kkg · 283.2K

1134 · 102J/m3 (89)

= 0.717m3/kg (90)

WH78 2.22

To �nd the height of a pressure level, given an

8

isotherm atmosphere and surface pressure p0 =1000hPa, we use the hypsometric equation, andassume T = Tv = 240K.

Z1 − Z0 =RdTvg0

lnp0p1

(91)

and for Z0 = 0 at p0 we get

Z1 =287 J

kgK× 240K

9.81ms−2lnp0p1

(92)

= 7.021km lnp0p1

(93)

For pressure levels p1 = 100, 10 and 1hPa, weget

Z100 = 16.2km (94)

Z10 = 32.3km (95)

Z1 = 48.5km (96)

This is the geopotential height, which di�erslightly from the geometrical height. Theassumption of isothermal atmosphere is alsonot very correct, but ok for this exercise.

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