GEF2200 spring 2018: Solutions thermodynam-ics 1

A.1.TWhat is the difference between R and R∗?

R∗ is the universal gas constant, with value 8.3143JK−1mol−1. R is the gas constant for a specificgas, given by

R =R∗

M(1)

where M is the molecular weight of the gas (usually given in g/mol). In other words, R takesinto account the weight of the gas in question so that mass can be used in the equation of state.For the equation of state:

pV = nR∗T =m

MR∗T = mRT (2)

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

A.2.T

What is apparent molecular weight, and why do we use it?

Apparent molecular weight is the average molecular weight for a mixture of gases. We introduceit to calculate a gas constant R = R∗/Md for the mixture , where Md is the apparent molecularweight of i different gases given by Equation (3.10):

Md =

∑mi∑ miMi

=

∑mi

n=

∑niMi

n

=∑ ni

nMi (3)

In meteorology the most common apparent molecular weight is the one of air.

WH06 3.19Determine the apparent molecular weight of the Venusian atmosphere, assuming that it consistsof 95% CO2 and 5% N2 by volume. What is the gas constant for 1 kg of such an atmosphere?(Atomic weights of C, O and N are 12, 16 and 14 respectively.)

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

vN2 =VN2

V(4)

vCO2 =VCO2

V(5)

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Assuming ideal gas, we have total volume V = VN2 + VCO2 at a given temperature (T ) andpressure (p). This means volume concentration is equal to molecular fraction:

vi =ViV

=niR

?T/p

nR?T/p=nin

(6)

The appearent molecular mass is the molecular mass 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.TShow that the gas constant for moist air is greater than for dry air.

We have thatR =

R∗

M(10)

So proving that the gas constant is greater for moist than dry air is the same as proving thatthe apparent molecular weight of moist air, Mw, is smaller than that for dry air, Md.For dry air the apparent molecular weight is:

Md =

∑mi∑ miMi

=md

nd(11)

Adding mass mv of water vapor with molecular weight Mv gives

Mw =

∑mi +mv∑ miMi

+ mvMv

=md +mvmdMd

+ mvMv

= Mdmd +mv

md +mvMdMv

(12)

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Since Mv < Md, we see that Mw < Md (lighter than dry air), which means that Rw > R.If, on the other hand, we had added a gas x with Mx > Md, then Mnew > Md (heavier than dryair) and Rnew < R.

A.5.T

Why do we introduce virtual temperature?

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

RT = RdTv (13)

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

Virtual temperature is the temperature dry air would have to have the same density at the samepressure as the moist package. Since moist air has lower apparent molecular weight than dry air,this means that viritual temperature must always be higher than absolute temperature.A.6.TShould we use virtual temperature when the gas in question is water vapor?

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

WH06 3.20

If water vapor comprises 1% of the volume of the air (i.e. if it accounts for 1% of the moleculesin air), what is the virtual temperature correction?

The virtual temperature depends on the temperature and ratio between the partial pressure ofwater vapor over the total pressure e/p. Since we assume that the gasses can be approximatedas ideal gases, the volume fraction is equal to the molefraction and thus equal to the e/p by theequation of state pV = nR∗T : e

p = eVpV = nvR∗T

ndR∗T = nvnd.

Finally, we can now compute Tv as a function of T , by using e/p = 0.01.

Tv =T

1− ep(1− ε)

Tv =T1

1− 0.01(1− 0.622)

TvT

= 1.00379

(14)

If we take T = 288K as an example, then Tv = T · 1.00379 = 289.09K.A.7.TThe pressure of water vapor in a sample of air at 20◦C taken at sea level is 18hPa. What is the

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mole fraction of water vapor?Hint: Make use of Dalton’s law and equation (3.6).

We use that the pressure at sea level is approximately 1013hPa.The ideal gas law as stated in (3.6) is

pV = nR∗T (15)

The partial pressure exerted by water vapor is the pressure it would exert had it alone occupiedthe same volume as the total gas mixture. We may therefore use (3.6) again on water vaporalone:

eV = nvR∗T (16)

where nv is the number of moles of water vapor contained in the volume V . Combining the twoequations, we find that

nvn

=e

p=

18hPa1013hPa

= 0.0178 (17)

(From this deduction, we see that the mole fraction of any gas component in a mixture equalsthe pressure fraction exerted by that component.)So the mole fraction of water vapor in this case is 1.78 %.

A.9.TDerive the hydrostatic equation. Why must the atmospheric pressure decrease with height?

The hydrostatic equation gives the relationship between gravity and pressure for an air mass atrest, a relationship between a change in pressure dp and a change in height dz (see Figure 1).Since the pressure at a given height is due to the weight of the above air, we expect the pressureto decrease with height, meaning that a decrease in pressure equals an increase in height.

We need to consider the forces acting upon the layer, and to do so, we look at the edges of thelayer (height z and z + dz. Gravity works downwards, and the force on the top of the layer isFg(z+ dz). Furthermore, the layer acts on the air below with a force Fg(z). Since the layer is atrest, we have from Newton’s thrid law that the layer experience a similar force in the oppositedirection (−Fg(z)).For a body with mass m, the gravity is mg = %V g, where V is the volume. Given that the layerhas a thickness dz and area A, and therefore a volume dV = Adz, we have (assuming positiveaxis upwards):

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

Forces acting upwards: Pressure (force per area). The upward force at the layer top is opposedby a similar force acting on the layer (−Fp(p+ dp)).

Fp = Fp(p)− Fp(p+ dp)

= [p− (p+ dp)]A = −Adp (19)

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Figure 1: The forces on a layer of air.

Where Fp(p) > Fp(p+ dp), that is, dp is negative. Fp is then value of the force acting upwards.

To find the equilibrium between gravity and pressure 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, we get

dp

dz= −%g (23)

A.10.TWhat is the geopotential? Use this quantity together with the hydrostatic equation and the idealgas law to derive the hypsometric equation.

The geopotential at a given location is the energy required to lift 1 kg from sea level and up tothat location.

φ(z) =

∫ z

0gdz′ (24)

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

Z =1

g0

∫ z

0gdz′ (25)

Using the hydrostatic equation, the equation of state and the definition of virtual temperature:

dp

dz= −%g = − p

RTg = − p

RdTvg (26)

Using the definition 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)

By introducing the weighted mean virtual temperature

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 proportional to the mean temperature of the layer. Ahorizontal gradient in this temperature will then mean a horizontal gradient in layer thickness.This is an important feature of the atmosphere, especially when it comes to storm systems andlow/high pressures.

WH06 3.27

A meteorological station is located 50m below sea level. If the surface pressure at this station is1020 hPa, the virtual temperature at the surface is 15◦C, and the mean virtual temperature for

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the 1000- to 500-hPa is 0◦C, compute the height of the 500-hPa pressure level above sea level atthis station.

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

Zs = −50mps = 1020hPag0 = 9.81ms−2

R = 287Jkg−1K−1

To find the height of the 500hPa level, we use the hypsometric equation. You have to assumethat the temperature from the surface up to 1000hPa is constant. (If the layer between 1000hPaand 500hPa was isothermal, the would be more correct to calculate a mean temperature for thelayer between 1020hPa and 1000hPa, to avoid temperature jumps.) We will use the hypsometricequation and split the calculation in two: first we calculate the difference in geopotential heightbetween 1000hPa and 500hPa and then between the surface and 1000hPa:The hypsometric equation:

Z2 − Z1 = −Rdg0Tv ln

(p2p1

)(34)

Z500-Z1000:

Z500hPa − Z1000hPa = −Rdg0Tv ln

( 500hPa1000hPa

)(35)

= −287Jkg−1K−1

9.81ms−2· 273.15K · ln

( 500hPa1000hPa

)(36)

= 5539.1m (37)

Z1000-Zs:Now we first calculate the difference in geopotential height between 1020hPa and 1000hPa, andthen subtract the 50 meters to account for the fact that the station is below sea level:

Z1000hPa − Z1020hPa = −Rdg0Tv ln

(1000hPa1020hPa

)(38)

= −287Jkg−1K−1

9.81ms−2· 288.15K · ln

(1000hPa1020hPa

)(39)

= 166.9m (40)

Now we subtract 50m to get the height from sea level, Zs, to 500 hPa:

(Z1000hPa − Zs) = 166.9m− 50m = 116.9m (41)

Finally we add together Z500hPa − Z1000hPa and Z1000hPa − Zs:

(Z500hPa − Zs) = 5539.1m+ 116.9m = 5656m (42)

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Inserting the values:Z500 = 5.7km (43)

WH06 3.26Ignoring the virtual temperature correction, we may use the hypsometric equation (3.29) on thelocation at the storm centre

Z200hPa − Zsea level =RdTAg0

ln940hPa200hPa

(44)

and on the surrounding region:

Z200hPa − Zsea level =RdTBg0

ln1010hPa200hPa

(45)

TA is the (unknown) mean temperature between the sea surface and 200 hPa at the storm centre,and TB = −3◦ C = 270.2 K is the mean temperature in the surrounding region.Since the 200 hPa surface is perfectly flat, and the ocean surface is also flat, the left-hand sidesof the two above equations are the same. Therefore:

RdTAg0

ln940hPa200hPa

=RdTBg0

ln1010hPa200hPa

(46)

TA = TBln 1010hPa

200hPa

ln 940hPa200hPa

(47)

= 270.2Kln 1010hPa

200hPa

ln 940hPa200hPa

(48)

= 282.7K = 9.5◦C (49)

The air column above the storm centre has in average 12.5◦C higher temperature than the sur-roundings.

A.11.TThe first law of thermodynamics states that dq − dw = du.What is dq, dw and du?

dq is the heat added to a system.dw is the work done by the system.du is the change in internal energy.Different kinds of work include expansion, surface expansion, extension and electrical work. Ex-pansion work is the important one in thermodynamics of the atmosphere, so we can assumedw = pdα.

WH06 3.18jExplain why air released from a tire is cooler than its surroundingsWe follow an air parcel leaving the tire, to get out into the surroundings. The pressure insidethe tire is higher than the pressure of the surroundings (p1 > p), but the temperature is the

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

same (T1 = T ). The air parcel will then expand adiabatically, since it is entering air with lowerpressure. From the first law of thermodynamics, we have

dq = du+ dw = cvdT + pdα = 0 (50)

In the expansion, the specific volume (volume per mass) increases (dα > 0), so that the temper-ature change is negative:

dT = − p

cvdα < 0 (51)

Since the air expands, the work done on the surroundings is positive, equals to dw = pdα. Thedecrease in temperature means that the internal energy (du = cvdT ) also decreases.This is what happens when an air parcel rise in the atmosphere; it expands and cools. This isalso important for the heat transport from the equator to the poles.

A.12.TDefine adiabatic process

An adiabatic process is a process where no heat exchange between the system and the surround-ings is allowed. That means dq = 0.

3.18h Explain why: If a low pressure system is colder than its surroundings, the amplitude ofthe depression in the geopotential height field increases with height

9

When a low-pressure is colder than its surroundings, the isobars are closer together in the low-pressure than in the surroundings. Going upwards (towards lower pressure), the height of a givenisobar will get increasingly lower in the low-pressure centre compared to in the surroundings (seefigure 3.3b in WH06).

3.18k Explain why: Under what conditions can an ideal gas undergo a change of state withoutdoing external work?In our study of thermodynamics, the only work taken into account is the work done in expandingthe volume of the system. We then have:

dw = pdα (52)

Here, p is the pressure at the ’edge’ of our system, where the environment is ’pushed’ awayas the system expands by dα. To get zero work we can either let the volume remain constant(dα = 0) or let the system expand into a vacuum (p = 0, there are no surroundings to do work on).

WH06 3.32 (WH78 2.24)

Calculate the work done in compressing isothermally 2 kg of dry air to one tenth of its volumeat 15 ◦C.We have isothermal compression of 2kg dry air at T = 15◦C, to one tenth of its initial volume.

Work done on the air parcel = − work done by the air parcel. Remember that dW = pdV . Theonly work done by the parcel is expansion work:

W =

∫ V2

V1

pdV (53)

We have the ideal gas equation

p =mRT

V(54)

Inserting and integrating gives, when remembering that T is constant,

W =

∫ V2

V1

pdV = mRT lnV2V1

(55)

= 2kg · 287Jkg−1K−1 · 288K ln1

10= −3.806 · 105J

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

WH06 3.33 (WH78 2.23)– a –Prove that when an ideal gas undergoes an adiabatic transformation pV γ = constant, where γ

10

is the ratio of the specific heat at constant pressure (cp) to the specific heat at constant volume(cp). [Hint:By combining 3.3 and 3.41 show that for an adiabatic transformation of a unit massof gas cv(pdα+ αdp) +Rpdα = 0.

We have an ideal gas, undergoing an adiabatic transformation, and we will prove that pV γ =constant for this process, where γ = cp/cv.The first law of thermodynamics (TD1) is given by Eqn. (WH06 3.41):

dq = cvdT + pdα (56)

where dq = 0 for an adiabatic process. Differentiating the ideal gas equation and inserting (56)for dT , we get

pα = RT (57)αdp+ pdα = RdT (58)

= −R p

cvdα (59)

αdp = −(1 +

R

cv

)pdα (60)

Using Eqn. (WH06 3.45), we have that

1 +R

cv=

1

cv(cv +R) =

cpcv

= γ (61)

and we get

αdp = −(1 +

R

cv

)pdα

= −γpdα (62)

Re-arranging:

−dpp

= γdα

α(63)

Now er integrate from p0, α0 to p, α

−∫ p

p0

dp

p=

∫ α

α0

γdα

α(64)

lnp0p

= γ lnα

α0= ln

(α

α0

)γ(65)

p0p

=

(α

α0

)γ(66)

and because

α

α0=

1/%

1/%0=%0%

=m/V0m/V

=V

V0(67)

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we have (V

V0

)γ=

p0p

pV γ = p0Vγ0 (68)

– b –7.5cm3 of air at 17◦C and 100 hPa is compressed isothermally to 2.5 cm3. The ar is then allowedto expand adiabatically to its original volume. Calculate the final temperature and final rpessureof the gas.For an isothermal transformation we have from the equation of state

p1V1 = p2V2 (69)

So for an isothermal compression from pressure p1 =1000hPa and volume V1 =7.5cm3 to thevolume V2 =2.5cm3, we find the pressure

p2 = p1V1V2

= 1000hPa7.5

2.5= 3000hPa

To find the pressure after adiabatic expansion until the original volume V3 = V1, we use theexpression from Equation (68)

p3Vγ3 = p2V

γ2

p3 = p2

(V2V1

)γ(70)

= 3000hPa(2.5

7.5

)γ= 643hPa

The temperature after expansion is found using the equation of state

p3V3T3

=p2V2T2

(71)

T3 =p3V3p2V2

T2 (72)

where we can insert the pressure p3 from Equation (70) (or the value calculated). We also notethat T2 = T1, since the first transformation was an isothermal transformation (and still V3 = V1):

T3 = T2V3p2V2

p2

(V2V3

)γ(73)

= T1

(V2V1

)−1(V2V1

)γ= T1

(V2V1

)γ−1= 290K

(2.5

7.5

)γ−1= 186.8K

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Comment on the physical side:The result that pV γ is constant for an adiabatic process means that the volume changes differentlyfrom an isothermal process (where pV is constant). In the case of expansion from one pressureto another (p), the volum of the adiabatic system (Va) after transformation is given by

pV γa = Ca

Va =

(Cap

)1/γ

(74)

where Ca is the constant value. The corresponding volume change for the isothermal system (Vi)is given by

pVi = Ci

Vi =Cip

(75)

where Ci is the constant value.Because γ = cp/cv > 1, isothermal expansion gives a faster/larger increase in volume than doesthe adiabatic expansion (Equation (74) vs (75)). The physical reason is that the adiabatic systemcools during the expansion, whereas the isothermal system is kept at constant temperature byadding heat to the system. The increased temperature increases the volume or the pressure.

In this exercise, the volume was the same for the two processes, but the pressure was not: Theadiabatic expansion resulted in a lower pressure (643hPa vs 1000hPa), and a lower temperature(187K vs 290K). Clearly, adding heat to keep the process isothermal means heating the air by103K, which indeed should increase the pressure (remember that the volume is constant).

In the case of compression, Equation (74) and (75) show that the decrease in volume is smallerfor the isothermal process than for the adiabatic process. The argument follows the same linesas for expansion; since an adiabatic compression means that the temperature increases, whereasheat is removed from the isothermal system to keep it at constant temperature, the isothermalvolume (or pressure) will be smaller.

A.13.TDefine specific heat. Its value is dependent on whether the material is allowed to change its vol-ume. Why?Show that cp = cv +R for an ideal gas.

Specific heat: The amount of heat you have to supply to a body of unit mass to raise itstemperature by 1 degree.

dq

dT= c (76)

So that if dT = 1, we have dq = c. c depends on whether the system is allowed to changeits volume because in that case, work will be done either by or on the system. This will also

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contribute to changing its temperature.

To find the relationship cv +R = cp, we start with constant volume:

cv =

(dq

dT

)αconst

=

(du

dT

)αconst

(77)

and because of Joule’s law for an ideal gas, du is only dependent on temperature

cv =du

dT(78)

Then we have from the TD1, and by rewriting pdα

dq = cvdT + pdα (79)= cvdT + d(pα)− αdp (80)= cvdT + d(RT )− αdp (81)= (cv +R)dT − αdp (82)

So we see that for constant pressure, we have dp = 0 and(dq

dT

)pconst

= cp = cv +R (83)

E.1.T

The temperature of 2kg of dry air is raised from 0◦C to 10◦C. How much heat must be added if

a. the air is within a closed and rigid box?

b. the air remains at constant pressure?

c. If the air obtains the increase in temperature in an adiabatic process, what is the pressureand specific volume of the air after the raising of the temperature? Assume the initialpressure is 1000 hPa.

– a –When the air is contained in a closed and rigid box, its volume is constant. We therefore use thespecific heat at constant volume:

dq

dT= cv (84)

⇒ q = cv(T2 − T1) (85)

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

Q = mq (86)= mcv(T2 − T1) (87)= 2kg · 717J/Kkg · 10K (88)= 14.3kJ (89)

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– b –Now pressure is constant. We do the same as in (a), just using cp = 1004J/Kkg instead of cv.We then get Q = 20.1kJ.

– c –For an adiabatic process we have

T2T1

=(p2p1

)R/cp(90)

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

p2 = p1

(T2T1

)cp/R(91)

= 1000hPa ·(283.2K273.2K

) 1004287.0 (92)

= 1134hPa (93)

We find the specific volume from the ideal gas equation:

α2 =RdT2p2

(94)

=287.0J/Kkg · 283.2K

1134 · 102J/m3 (95)

= 0.717m3/kg (96)

WH78 2.22Assuming an isothermal atmosphere with a temperature of -33◦C and a surface pressure of1000hPa, estimate the levels at which pressure equals 100, 10 and 1hPa, respectively.

To find the height of a pressure level, given an isotherm atmosphere and surface pressure p0 =1000hPa, we use the hypsometric equation, and assume T = Tv = 240K.

Z1 − Z0 =RdTvg0

lnp0p1

(97)

and for Z0 = 0 at p0 we get

Z1 =287 J

kgK × 240K9.81ms−2

lnp0p1

(98)

= 7.021km lnp0p1

(99)

For pressure levels p1 = 100, 10 and 1hPa, we get

Z100 = 16.2km (100)Z10 = 32.3km (101)Z1 = 48.5km (102)

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This is the geopotential height, which differ slightly from the geometrical height. The assumptionof isothermal atmosphere is also not very correct, but ok for this exercise.

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