Temperature of the planet if gases such as carbon dioxide and methane are completely eliminated

 

Author; Rogelio Pérez C

Summary;

Science teaches us that there is a greenhouse effect caused by certain gases; this greenhouse effect is attributed almost 33°C of temperature on the planet. The main greenhouse gases are CO2 and Methane.

As the atmosphere is a gas and temperature is the measure of the average kinetic energy of atoms or molecules in systems, in the case of the study Atmospheric system, therefore based the laws for the temperature of gases or ideal gas law , this work presents the following conclusions; If we completely remove gases such as CO2 and methane from the atmosphere, the temperature on the planet only decreases by 0.2 degrees centigrade, in other words the temperature goes from 288k or 15°C average, to 287.8k or 14.8°C.

Introduction

On the planet we talk about global warming or climate change that affects everyone, the science says the cause is a greenhouse effect , originated by certain gases known as greenhouse gases, according to science without greenhouse gases the average temperature of the planet would be -18,But thanks to the greenhouse effect caused by these gases, the average temperature is 15°C. This is a contribution of 33°C of these gases to the total temperature of the atmosphere, with mathematics to find the pressure and temperature in the gases,, with the laws of ideal gases, we can find the pressure and temperature in the gases, so we can also know the temperature in the atmosphere, if we completely eliminate gases such as carbon dioxide and methane, that are the gases that are being eliminated in the atmosphere as the main solution to the crisis of the high temperatures that the planet is experiencing.

Theory;

The Greenhouse Effect. If the earth only absorbed radiation from the sun without giving an equal amount of heat back to space by some means, the planet would continue to warm up until the oceans boiled. We know the oceans are not boiling, and surface thermometers plus satellites have shown that the earth's temperature remains roughly constant from year to year (the interannual globally-averaged variability of about 0.2 C or the 0.5 C warming trend in the 20th century, notwithstanding). This near constancy requires that about as much radiant energy leaves the planet each year in some form as is coming in. In other words, a near-equilibrium or energy balance has been established. The components of this energy balance are crucial to the climate.

All bodies with temperature give off radiant energy. The earth gives off a total amount of radiant energy equivalent to that of a black body -- a fictional structure that represents an ideal radiator -- with a temperature of roughly -18 C (255 K). The mean global surface air temperature is about 14 C (287 K), some 32 C warmer than the earth's black body temperature. The difference is due to the well-established greenhouse effect. U.S. Standard Atmosphere. [1]

Comparison of the 1962 US Standard Atmosphere graph of geometric altitude against air density, pressure, the speed of sound and temperature with approximate altitudes of various objects.[2]

The U.S. Standard Atmosphere is a static atmospheric model of how the pressure, temperature, density, and viscosity of the Earth's atmosphere change over a wide range of altitudes or elevations. The model, based on an existing international standard, was first published in 1958 by the U.S. Committee on Extension to the Standard Atmosphere, and was updated in 1962, 1966, and 1976. It is largely consistent in methodology with the International Standard Atmosphere, differing mainly in the assumed temperature distribution at higher altitudes.

Visualization of composition by volume of Earth's atmosphere, Water vapor is not included, as this is highly variable. Each tiny cube (such as the one representing krypton) has one millionth of the volume of the entire block. Data is from NASA Langley.

Methodology

The USSA mathematical model divides the atmosphere into layers with an assumed linear distribution of absolute temperature T against geopotential altitude h.[3] The other two values (pressure P and density ρ) are computed by simultaneously solving the equations resulting from:

The vertical pressure variation, which relates pressure, density and geopotential altitude (using a standard pressure of 101,325 Pascals (14.696 psi) at mean sea level as a boundary condition):

Dp/dh = -pg,and

The ideal gas law in molar form, which relates pressure, density, and temperature:

 P= p.(R specific)T at each geopotential altitude, where g is the standard acceleration of gravity, and (R specific) is the specific gas constant for dry air.

Air density must be calculated in order to solve for the pressure, and is used in calculating dynamic pressure for moving vehicles. Dynamic viscosity is an empirical function of temperature, and kinematic viscosity is calculated by dividing dynamic viscosity by the density.

Thus the standard consists of a tabulation of values at various altitudes, plus some formulas by which those values were derived.

To allow modeling conditions below mean sea level, the troposphere is actually extended to −2,000 feet (−610 m), where the temperature is 66.1 °F (18.9 °C), pressure is 15.79 pounds per square inch (108,900 Pa), and density is 0.08106 pounds per cubic foot (1.2985 kg/m3).

Common forms

The most frequently introduced forms are:

PV=nRT=nkBNAT

Where:

P is the pressure of the gas,

V is the volume of the gas,

n is the amount of substance of gas (also known as number of moles),

R is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant,

KB is the Boltzmann constant

 NA is the Avogadro constant

T is the absolute temperature of the gas.

In SI units, p is measured in pascals, V is measured in cubic metres, n is measured in moles, and T in kelvins (the Kelvin scale is a shifted Celsius scale, where 0.00 K = −273.15 °C, the lowest possible temperature). R has the value 8.314 J/(K·mol) ≈ 2 cal/(K·mol), or 0.0821 l·atm/(mol·K).

Molar form

How much gas is present could be specified by giving the mass instead of the chemical amount of gas. Therefore, an alternative form of the ideal gas law may be useful. The chemical amount (n) (in moles) is equal to total mass of the gas (m) (in kilograms) divided by the molar mass (M) (in kilograms per mole): [4]

n=m/M

By replacing n with m/M and subsequently introducing density ρ = m/V, we get:

PV =m/M.RT

P=m/V.RT/M

P=p. R/M.T

Defining the specific gas constant Rspecific(r) as the ratio R/M,

P=p.R specific.T

This form of the ideal gas law is very useful because it links pressure, density, and temperature in a unique formula independent of the quantity of the considered gas. Alternatively, the law may be written in terms of the specific volume v, the reciprocal of density, as Pv=R.specific.T.

 

1962 version

The basic assumptions made for the 1962 version were:[5]

Air is a clean, dry, perfect gas mixture (cp/cv = 1.40)

Molecular weight to 90 km of 28.9644 (C-12 scale)

Principal sea-level constituents are assumed to be:

N2 – 78.084%

O2 – 20.9476%

Ar – 0.934%

CO2 – 0.0314%

Ne – 0.001818%

He – 0.000524%

CH4 – 0.0002%.

Assigned mean conditions at sea level are as follows:

P = 14.696 psi = 2116.22 psf = 101325 Pa = 29.92 inHg = 0.1013250 MN/m2

T = 59 °F = 518.67 °R = 15 °C = 288.15 K

ρ = 0.0764734 lb/(cu ft) = 1.225 0 kg/m3

g = 32.174 1 ft/s2 = 9.80665 m/s2

R = 1545.31 ft⋅lb/(lbmol⋅°R) = 8.31432 J/(mol⋅K).

Development

I will calculate the average molecular weight of the atmosphere under normal conditions, with greenhouse gases such as carbon dioxide and methane, multiplying the mole numbers of each gas in the atmosphere by their molecular weights.

 

Columna1	Columna2	Columna3	Columna4
GAS	No. of Moles in 1 Mole Air	Molecular Weight	Mass/grams
Nitrogen (N2)	0,78084	28	21,86352
Oxygen (02)	0,20951	32	6,70432
Argon (Ar)	0,00934	40	0,3736
CO2	0,000415	44	0,01826
Neon (NE)	0,00001818	20,17	0,000366691
Helium (HE)	0,00000524	4	0,00002096
Methane (CH4)	0,00000187	16,04	2,99948E-05
Total weight in 1 mole of air			28,960117645

 

Now I will find the average density of the atmosphere under normal conditions. But first I have to find the volume occupied by a mole of gas in the atmosphere at 15°C.

V = nRT/P = (1mole)(0,0821 L atm /mole K)(288 K)/1 atm=23,6448L

V=23, 6448L/1000=0.0236448m³        

M=28.96011 grams /mole=0.02896011Kg/mol.

Calculating molar air density = mass/ volume

p=M/V

Then;

p=M/V=0.02896011 kg/mole/0.0236448m³/mole=1,224798588 kg/m³.

Density of the atmosphere =1, 224798588 kg/m³.

Calculating atmospheric pressure

PV =m/M.RT

P=m/V.RT/M

P=p. R/M.T

Where ;

P= pressure?

p= Density = 1.224798588 kg/m³

R. specific = is the gas constant specific to dry air. = 8.31432 J/(mol⋅K).

M= Molar mass =28.96011 grams /mole=0.02896011Kg/mole.

T=Atmosphere temperature. 288k o 15°C.

Then;

P=p. R/M.T

P=1.224798588kgm³.(8.31432 J/mole⋅K)/ 0.02896011Kg/mole.).288k

P=1.224798588kgm³.(287.0845449J.kg.k).288K

P=101266.77Jm³

Calculating the atmospheric temperatura

P=p.R/M.T

p.R/M.T=P

T=P/p.R/M

Where;

P= pressure 101266.77Jm³

p= Density = 1.224798588 kgm³

R. specific = is the gas constant specific to dry air. = 8.31432 J/(mole⋅K).

M= Masa Molar =28.96011 grams/mole=0.02896011Kg/mole.

T=  temperature ?

Then;

T=P/p.R/M

T=101266.77Jm³.(1.224798588 kgm³.8.31432 J/mole⋅K)/ 0.02896011Kg/mole).

T=101266.77Jm³.(10.182975 kgm³J/mole⋅K)/ 0.02896011Kg/mole).

T=101266.77Jm³.(10.182975 kgm³J/mole⋅K)/ 0.02896011Kg/mole).

T=101266.77Jm³.(351.620 m³J⋅K)

T=288⋅K

The temperature of the atmosphere under normal conditions is =288k

Columna1	Columna2	Columna3	Columna4
GAS	No. of Moles in 1 Mole Air	Molecular Weight	Mass/grams
Nitrogen (N2)	0,78084	28	21,86352
Oxygen (02)	0,20951	32	6,70432
Argon (Ar)	0,00934	40	0,3736
Neon (NE)	0,00001818	20,17	0,000366691
Helium (HE)	0,00000524	4	0,00002096
Total weight in 1 mole of air			28,941827651

 

Now i will find the average density of the atmosphere without greenhouse gases such as carbon dioxide and methane. But first i have to find the volume occupied by a mole of gas in the atmosphere at 15°C.

V = nRT/P = (1mole)(0,0821 L atm /mole K)(288 K)/1 atm=23,6448L

V=23, 6448L/1000=0.0236448m³        

M=28.941827 grams/mol=0.028941827Kg/mole.

Calculating molar air density = mass/volume

p=M/V

Then;

p=M/V=0.028941827 kg/mole/0.0236448m³/mole=1,224025kg/m³.

Density of the atmosphere without greenhouse gases as carbon dioxide and methane is =1, 224025 kg/m³.

Calculating atmospheric pressure.

PV =m/M.RT

P=m/V.RT/M

P=p. R/M.T

Where;

P= pressure ?

p= Density = 1,224025 kg/m³.

R. specific = is the gas constant specific to dry air. = 8.31432 J/(mole⋅K).

M= 28.96011 grams/mole=0.02896011Kg/mole.

T= Atmosphere temperature.288k o 15°C.

Then;

P=p. R/M.T

P=1.224025kgm³.(8.31432 J/mole⋅K)/ 0.02896011Kg/mole.).288k

P=1.224025kgm³.(287.0845449J.kg.k).288K

P=101202.81Jm³

Calculating atmosphere temperature

P=p.R/M.T

p.R/M.T=P

T=P/p.R/M

Where;

P= pressure 101202.81Jm³

p= Density = 1,224025 kg/m³.

R. specific = is the gas constant specific to dry air. = 8.31432 J/mole⋅K).

M= 28.941827 grams/mole=0.028941827Kg/mole.

T=  temperature?

Then;

T=P/p.R/M

T=101202.81Jm³. (1.224025 kgm³.8.31432 J/mole⋅K)/ 0.028941827Kg/mole)

T=101202.81Jm³.(10.176543 kgm³J/mole⋅K)/ 0.028941827Kg/mole).

T=101202.81Jm³.(10.176543 kgm³J/mole⋅K))/ 0.028941827Kg/mole).

T=101202.81Jm³.(351.620 m³J⋅K)

T=287.8K

The temperature of the atmosphere without greenhouse gases such as carbon dioxide and methane is 287.8k

Conclusión,

It can be concluded that if we remove gases such as carbon dioxide and methane from the atmosphere, the temperature of the atmosphere would be almost the same as what we currently have.

It can also be concluded that gases known as greenhouse gases do not contribute at 33oC to the average temperature of the planet.

Finally we can say that the strategy of removing carbon dioxide and methane from the planet's atmosphere is not the best tool for combating heat climates.

 

Bibliography    

1.   https://stephenschneider.stanford.edu/Mediarology/Statement_of_Stephen_Schneider_Climate_Change.html

https://en.wikipedia.org/wiki/U.S._Standard_Atmosphere.

2-Geometric altitude vs. temperature, pressure, density, and the speed of sound derived from the 1962 U.S. Standard Atmosphere.

3- Gyatt, Graham (2006-01-14): "The Standard Atmosphere". A mathematical model of the 1976 U.S. Standard Atmosphere.

4- Moran; Shapiro (2000). Fundamentals of Engineering Thermodynamics (4th ed.). Wiley. ISBN 0-471-31713-6.

5- U.S. Standard Atmosphere, 1962, U.S. Government Printing Office, Washington, D.C., 1962

https://en.wikipedia.org/wiki/Ideal_gas_law