3 Ideal Gas Examples: Under Which Circumstance:


In this article “Ideal gas examples” and ideal gas examples related facts will be discuss. Ideal gas examples are based on law of ideal gas. But in practical life ideal gas is not present in the universe.

3+ Ideal Gas Examples are listed below,

Example 1:-

Calculate the amount of density for the Nitrogen gas at the pressure of 256 Torr and 25 degree centigrade temperature.

Solution: – Given data are,

P = 256 Torr = 256 Torr \times \frac{1 atm}{760 Torr} = 0.3368 atm

V =?

R = [latex]0.0820574 L*atm*mol^-^1*K^-^1[/latex]

T = (25 + 273) K = 298 K

n =?

Now we apply the formula for ideal gas,

PV = nRT   ………. eqn (1)

So, we also can write the density is,

[latex]\rho = \frac{m}{v}[/latex] ………. eqn (2)

Where,

[latex]\rho[/latex] = Density of the ideal gas

m = Mass of the ideal gas

v = Volume of the ideal gas

Now, [latex] m = M \times n [/latex] ………. eqn (3)

Where,

m = Mass

M = Molar mass

n = Moles

From eqn (2) and eqn (3) we get,

[latex]\rho = \frac{m}{v}[/latex] …… (4)

Arranging the eqn (2) and eqn (3) we get,

[latex]\rho = \frac{M\times n}{V}[/latex] ……eqn(5)

[latex]\frac{\rho }{M} = \frac{n}{V}[/latex] ……eqn(6)

Now applying the equation of ideal gas,

PV = nRT

[latex]\frac{n}{V} = \frac{\rho }{M}[/latex] ……eqn(7)

[latex]\frac{n}{V} = \frac{P}{RT}[/latex] ……eqn(8)

From eqn (6) and eqn (8) we get,

[latex]\frac{\rho }{M} = \frac{P}{RT}[/latex] ……eqn(9)

Isolate density,

[latex]\rho = \frac{PM}{RT}[/latex]……eqn(10)

[latex]\rho = \frac{(0.3368 atm)(2 \times 14.01gram/mol)}{(0.08206 L*atm*mol^-^1*K^-^1 )(298 K)}[/latex]

[latex]\rho[/latex] = 0.3859 gram / mol

The amount of density for the Nitrogen gas at the pressure of 256 Torr and 25 degree centigrade temperature is 0.3859 gram / mol.

Ideal gas examples
Image – Nitrogen : Example of ideal gas;
Image Credit – Wikimedia Commons

Example 2:-

A container which is filled with the Neon gas. The amount Neon in container is 5.00 Litre that time the temperature is 26 degree Centigrade at 750 mm Hg. A carbon dioxide vapour is now added to the container. The quantity of carbon dioxide added to the container is 0.627 gram.

Now determine these factors,

Partial pressure for Neon in atm.

Partial pressure for carbon dioxide in atm.

Total pressure present in the container.

Solution: – Given data are,

P = 750 mm Hg -> 1.01 atm

V = 5.00 Litre

T = (26 + 273) K= 299 K

[latex]n_n_e[/latex] =?

[latex]n_c_o_2[/latex] =?

For Carbon dioxide the number of mole is,

[latex]n_c_o_2 = 0.627 gram CO_2 = \frac{1 mol}{44 gram} = 0.01425 mol CO_2[/latex]

Now for Neon the number of mole is,

[latex]n_N_e = \frac{PV}{RT}[/latex]

[latex]n_N_e = \frac{1.01 atm\times 5.00 Litre}{(0.0820574 L*atm*mol^-^1*K^-^1)(298)}[/latex]

[latex]n_N_e[/latex] = 0.206 mol Ne

Before adding the carbon dioxide to the container we can get only pressure for neon. So the partial pressure for neon is definitely the amount of pressure is already discussed in question.

Now for the carbon dioxide,

Using the equation of ideal gas equation we can write,

[latex]\frac{P_N_e\times V}{n_N_e \times RT} = \frac{P_C_O_2\times V}{n_C_O_2 \times RT}[/latex]

For the both Carbon dioxide and Neon Temperature, volume and gas constant remain same.

So,

[latex]\frac{P}{n_N_e} = \frac{P}{n_C_O_2}[/latex]

[latex]\frac{1.01 atm}{0.206 mol Ne} = \frac{P_C_O_2}{0.01425 mol CO_2}[/latex]

[latex]P_C_O_2 = 0.698 atm[/latex]

Total pressure,

[latex]P_t_o_t_a_l = P_N_e + P_C_O_2[/latex]

[latex]P_t_o_t_a_l = 1.01 atm + 0.698 atm[/latex]

[latex]P_t_o_t_a_l[/latex] = 1.708 atm

Partial pressure for Neon is 1.01 atm.

Partial pressure for carbon dioxide 0.698 atm.

Total pressure present in the container is 1.708 atm.

Example 3:-

Determine the amount of volume.

In a glass container carbon dioxide gas is present. The temperature of the carbon dioxide gas is 29 degree centigrade, pressure is 0.85 atm and the mass of the carbon dioxide gas is 29 gram.

Solution: – Given data are,

P = 0.85 atm

m = 29 gram

T = (273 + 29) K = 302 K

The mathematical form of the ideal gas is,

PV = nRT ……..eqn (1)

Where,

P = Pressure for the ideal gas

V = Volume for the ideal gas

n = Molar number for the ideal gas

R = Universal gas constant for the ideal gas

T = Temperature for the ideal gas

If in a matter M denoted as molar mass and mass of a matter denoted as m then the total number of moles for that particular matter can be expressed s,

[latex]n = \frac{m}{M}[/latex] ……..eqn (2)

Combine the ……..eqn (1) and ……..eqn (2) we get,

[latex]PV = \frac{mRT}{M}[/latex] ……..eqn (3)

We know the value of molar mass for carbon dioxide is,

M = 44.01 gram/ mol

From eqn (3) we can write,

[latex]V = \frac{mRT}{M} = \frac{29 gram \times 0.0820574 L*atm*mol^-^1*K^-^1\times 302}{44.01 gram/mol\times 0.85 atm}[/latex]

V = 19.21 Litre

In a glass container carbon dioxide gas is present. The temperature of the carbon dioxide gas is 29 degree centigrade, pressure is 0.85 atm and the mass of the carbon dioxide gas is 29 gram. Then the amount of volume is 19.21 Litre.

Image – Carbon dioxide;
Image Credit – Wikipedia

Real gas vs. Ideal gas:

Ideal gases follow the law of gas in a particular constant condition but real gas does not follow the law of gas in a particular constant condition. In practical life ideal gas is not exist but real gas is exist.

The major points are derive about the difference between real gas and ideal gas,

ParameterIdeal gasReal gas
DefinitionThe gas which are follow the law of gas at particular condition of constant pressure and temperatureThe gas which are not follow the law of gas at particular condition of constant pressure and temperature
Movement of particlesThe particle present in the ideal gas is free to move and the particle does not attend in interparticle interaction.The particle present in the real gas is not free to move and compete with each other, the particle attend in interparticle interaction.
Volume occupiedNegligibleNot negligible
PressureHigh pressure is presentLower pressure than the ideal gas pressure
Force presentIntermolecular force of attraction is not presentIntermolecular force of attraction is present
FormulaThe formula which ideal gas is follow,
PV = nRT
Where,
P = Pressure
V = Volume
n = Amount of substance R = Ideal gas constant
T = Temperature
The formula which real gas is follow,
[latex]{(P + an^2/V^2)(V – nb)} = nRT[/latex]
Where,
P = Pressure
a = Parameter which need to determine empirically for individual gas
V = Volume
b = Parameter which need to determine empirically for individual gas
n = Amount of substance
R = Ideal gas constant
T = Temperature  
AvailabilityNot existExist

Read more about Isothermal process : Its’s all Important facts with 13 FAQs

Frequent asked questions:-

Question: – Derive the limitations of ideal gas.

Solution: – The limitations of ideal gas is listed below,

  • Ideal gas could not work in high density, low temperature and high pressure
  • Ideal gas not applicable for heavy gases
  • Ideal gas not applicable strong intermolecular forces.
Image – Ideal gas;
Image Credit – Wikipedia

Read more about Gauge Pressure : It’s Important Properties with 30 FAQs

Question: – Write down the assumptions about the ideal gas.

Solution: – Actually in our surrounding ideal gas is no present. The law of ideal gas is a simple equation by which we can understand the relation between pressures, volume and temperature for gases.

The assumptions about the ideal gas is listed below,

  • The gas particles of ideal gas have negligible volume.
  • The size of the gas particles of ideal gas is equal and they don’t have intermolecular force.
  • The gas particles of ideal gas have follows the law of motion of Newton’s.
  • There is no loss of energy.
  • The gas particles of ideal gas have elastic collision.

Question: – Derive the different form equation for ideal gas.

Solution: – Ideal gas formula actually combination of Boyle’s law, Avogadro’s law, Charle’s law and Gay Lussac’s law.

The different form equation for ideal gas is briefly summarize below,

Common form of ideal gas:

[latex]PV = nRT = nk_bN_AT = Nk_BT[/latex]

Where,

P = Pressure for the ideal gas measured in Pascal

V = Volume for the ideal gas measured in cubic meter

 n = The total of ideal gas which is measured in moles measured in mole

R = Gas constant for the ideal gas which value is [latex]8.314 J/K.mol = 0.0820574 L*atm*mol^-^1*K^-^1[/latex]

T = Temperature for the ideal gas measured in Kelvin

N = The total number of the ideal gas molecules

[latex]k_b[/latex] = Boltzmann constant for the ideal gas

[latex]N_A[/latex] = Avogadro constant

Molar form of ideal gas:

Pv = Rspecific T

P = Pressure for the ideal gas

v = Specific volume for the ideal gas

Rspecific = Specific gas constant for the ideal gas

T = Temperature for the ideal gas

Statistical form of ideal gas:

[latex]P = \frac{k_b}{\mu m_\mu } \rho T[/latex]

Where,

P = Pressure for the ideal gas

[latex]k_b[/latex] = Boltzmann constant for the ideal gas

[latex]\mu[/latex] = Average partial mass for the ideal gas

[latex]m_\mu[/latex] = Atomic mass constant for the ideal gas

[latex]\rho[/latex] = Density for the ideal gas

T = Temperature for the ideal gas

Combined gas law:-

[latex]\frac{PV}{T} = k[/latex]

P = Pressure

V = Volume

T = Temperature

k = Constant

When same matter in present two different condition that time we can write,

[latex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/latex]

Question: –Derive the law of Boyle.

Solution: – Boyle’s law is a gas law. The gas law of Boyle’s derive that the pressure exerted by a gaseous substance(of a given mass, kept at a constant temperature) is inversely proportional to the volume occupied by it.

Image – Boyle’s law;
Image Credit – Wikimedia

In other words, the pressure and volume of a gas are indirectly proportional to each other to the temperature and the quantity of gas are kept constant. 

The gas law of Boyle’s can be expressed mathematically as follows:

[latex]P_1V_1 = P_2V_2[/latex]

Where,

[latex]P_1[/latex] = The initial pressure exerted by the gaseous substance

[latex]V_1[/latex] = The initial volume occupied by the gaseous substance

[latex]P_2[/latex] = The final pressure exerted by the gaseous substance

[latex]V_2[/latex] = The final volume occupied by the gaseous substance

This expression can be obtained from the pressure-volume relationship suggested by Boyle’s law. For a fixed amount of gas kept at a constant temperature, PV = k. Therefore,

[latex]P_1V_1[/latex] = k (initial pressure \times initial volume)[/latex]

[latex]P_2V_2[/latex] = k (final pressure \times final volume)[/latex]

∴ P1V1 = P2V2

As per Boyle’s law, any change in the volume occupied by a gas (at constant quantity and temperature) will result in a change in the pressure exerted by it.

Indrani Banerjee

Hi..I am Indrani Banerjee. I completed my bachelor's degree in mechanical engineering. I am a enthusiastic person and I am a person who is positive about every aspect of life. I like to read Books and listening to music.

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