2. Thermodynamics

2.1. Basic problems

2.1.1.The costitution of air in first approach is as follos:

78 % N2

21 % O2

0,96 % Ar0,0388 % CO2.

These concentrations are given in volume%, so they mean the ratio of the number of different molecules to the whole number of particles.

Calculate the special gas constant of the air!

Solution:According to Dalton’s law, the moelcules and Argon atoms are making the so called partial pressure independently as it can be seen below:

pN2V=

mN2

MN2

¿RT

pO2V=

mO2

MO2

¿ RT

pArV=mAr

M Ar

⋅RT

pCO2V=

mCO2

MCO2

¿ RT

For the whole gas:

p⋅V=mM

⋅R⋅T=m⋅r⋅T

We know that the overall mass of the gas is the sum of the masses of the constitutig gases:

m=mN2+mO2

+mAr+mCO2

and,p=pN2

+ pO2+ pAr+ pCO2

thus:p⋅V=pN 2

⋅V + pO2¿V + pAr ¿V+ pCO2

¿V=m⋅r⋅T

that is

(mN2

M N2

+mO2

MO2

+mAr

M Ar

+mCO2

MCO2

)R⋅T=m⋅r⋅T

and so:

mN2

m⋅ RMN2

+mO2

m¿ RMO2

+mAr

m¿ RM Ar

+mCO2

MCO2

¿ RMCO2

=r

consequently if we introduce the concentration on mass basis of the partial gases with C than:

CN2⋅rN 2

+CO2¿ rO2

+C Ar¿ r Ar+CCO2¿ rCO2

=r

where,

CN2=mN2

m=N N2

¿ μN2

L⋅μL

=0 ,78⋅L⋅μN2

L⋅μL

=0 ,78⋅L⋅46 ,507⋅10−27 kgL⋅48 ,084⋅10−27kg

=0 ,78⋅46 ,50748 ,084

=0 ,754

, CO2

=mO2

m=0 ,21⋅53 ,123

48 ,084=0 ,232

, C Ar=

mAr

m=0 ,0096⋅66 ,333

48 ,084=0 ,0132

, és

CCO2=mCO2

m=0 ,000388⋅73 ,063

48 ,084=5,9⋅10−4

,

and,

rN2=

RMN 2

=8 ,314

JK

28⋅10−3 kg=296 ,93

Jkg⋅K

,

rO2= RMO2

= 8 ,314

32⋅10−3=259 ,81

Jkg⋅K

,

rAr=R

M Ar

= 8 ,314

40⋅10−3=207 ,85

Jkg⋅K , és

rCO 2= R

MCO2

= 8 ,314

44⋅10−3=188 ,95

Jkg⋅K

are the specific gas constant of the individual gases.

Thus the specific gas constant for the air is:

r=CN2⋅rN2

+CO2¿ rO2

+CAr ¿ r Ar+CCO2¿ rCO2

=0 ,754⋅286 ,93+0 ,232⋅259 ,81+

+0 ,0132⋅207 ,85+5,9⋅10−4⋅188 ,95=216 ,345+60 ,276+2 ,743+0 ,115=289 ,48Jkg⋅K

2.1.2. There is oxygen in a bombe with the volume of 40 l. The pressure gauge on the bombe indicates 150 bar. How much is the mass of the gas in the bombe if the temperature of the gas is 27°C?

Solution:

pV= mM

⋅RT

m= pVMRT

=151⋅105⋅40⋅10−3⋅32g8 ,314⋅300

=7749 g=7 ,749kg

2.1.3. A gas bottle has a filling volume of 60 l. It is containing 100 bar of N2 gas. How much is discharged if the remaining pressure 30 ? The initial temperature was 27°C, and the final one is 13°C.

Solution:

p1V=m1

M⋅RT 1→m1=

p1V

T 1

⋅MR

p2V=m2

M⋅RT 2→m2=

p2V

T 2

⋅MR

Δm=(p1

T1

−p2

T 2

)⋅V⋅MR

=(100⋅105 Pa300 K

−30⋅105Pa286

)⋅60⋅10−3 ℓ⋅28g

8 ,314JK

=4 ,616⋅103 g=4 ,616kg

2.1.4. We introduce 4000 J heat to a container of 4 l. The container has a gas in it with the pressure of 100 bar, the gas is CO2 (M=44g), the temperature is 0°C. How much will be the temperature and the pressure after the heat is tranferred to the vessel?

Solution:

The mass of the gas in the container is:

m= p⋅V⋅MR⋅T

=100⋅105 Pa⋅4⋅10−3 m3⋅44 g

8 ,314JK

⋅273 K=775 g=0 ,775kg

Take the gas to be ideal, so the specific heat constant at constant volume is

cv=f2⋅RM

=52⋅

8 ,314JK

44⋅10−3kg=472 ,4

JkgK .

The measured vale is: 632 J/kgK, so we will use that.Similarly if we take the gas to be ideal, so the specific heat constant at constant volume is

c p=f +2

2⋅RM

=72⋅

8 ,314JK

44⋅10−3kg=661

JkgK ,

The measured value is 821 J/kgK, so we use that.

The 1st Law of Termodinamycs states:

ΔU=Q+W ,and we can take the volume of the container constant, therefore W=0, which means that all of the heat will change the internal energy.

The change of the internal energy:

ΔU=cv⋅m⋅ΔT=632J

kgK⋅0 ,775 kg⋅ΔT=4000 J

.

From this

ΔT=8,2K ,

so the final temperature in the bottle is: 8,2 oC.The pressure can be determined from the unified gas law:

p1

T1

=p2

T 2

p2=p1⋅T 2

T 1

=100⋅105Pa⋅381 ,2 K273K

=139 ,6⋅105 Pa

2.1.5. The volume of a cylinder of a compressor at initial stage is 0,7 liters. The pressure of the gas in it is 1 bar. A temperature is 27°C. Let the process of the compression be adiabatic with the ratio of specific heats of 1.4. The compression rate is 16. How much are the volume, the pressure and the temperature of the gas after compression? Calculate how much work has to be done for that process.

Solution:

The determination of final pressure can be made using the form below:

p1⋅V 1κ=p2⋅V 2

κ

p2=p1⋅(V 1

V 2)κ

=1bar⋅161,4=48 ,5bar.

The determination of final temperature can be made using the form below:

T 1⋅V 1κ−1=T 2⋅V 2

κ−1

T 2=T 1⋅(V 1

V 2)κ−1

=300 K⋅160,4=909 ,4K.

The work done on the gas:

W=−∫V 1

V 2

p⋅dV.

Th following equation is valid for any point of the process:

p⋅V κ=p1⋅V 1κ

,

thus:

p=p1⋅V 1

κ

V κ.

Substituting it into the integral we get:

W=−∫V 1

V 2

p⋅dV =∫V1

V2

p1 ¿V 1

κ

V κ ¿dV=−p1¿V 1κ ¿∫

V1

V2dVV κ =−p1 ¿V 1

κ ¿1−κ+1

⋅[V −κ+1 ]VV2 =

=1κ−1

⋅p1⋅V 1κ⋅(V 2

1−κ−V 11−κ )=1

κ−1⋅(p1⋅V 1

κ⋅V 21−κ−p1⋅V 1 )=

p1

κ−1⋅(V 1

κ⋅V 21−κ−V 1 )=

¿p1⋅V 1

κ−1⋅((V 1

V 2)κ−1

−1)=p1⋅V 1

κ−1⋅(ε κ−1−1 )=105⋅0,7⋅10−3

1,4−1⋅(160,4−1 )=355 ,5 J

.

2.1.6. How much is the polytrophic power factor in the following process?

Solution:

The inclination of the strate line is:

5

1

1

5

p[bar]

V[l]

m=5bar−1bar5 ℓ−1 ℓ

=1barℓ

=1⋅107 Pa

m3.

Thus

p=1⋅107⋅V ,therefore

p⋅V−1=107,

and from this the polytrophical powet factor is n = -1.On the other hand the definition of the polythrophic power factor is:

n=cn−c p

cn−cv ,so the polythropic specific heat is:

cn=c p+cv

2 .