We saw how the gas laws depict the behaviour of a gas under certain conditions of temperature and pressure. can we somehow combine these laws into one equation? The answer is yes and the equation that determines the behaviour of an ideal gas is the Ideal gas equation. Let’s learn more!

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## Ideal Gas Equation

The ideal gas equation is a combination of all the individual gas laws. Boyle’s Law, Charle’s law, Gay-Lussac’s Law and Avogadro’s Laws are the basis of the Ideal gas equation.

After the analysis of theÂ experiments in various gas laws, we can understand the idea behind the ideal behaviour of gases. The ideal gases perfectly obey the Ideal Gas Laws. The law of Ideal gases states that theÂ volume ofÂ aÂ specifiedÂ amount of gas isÂ inversely proportional to pressure and directly proportional to volume and temperature. Before deriving the Ideal Gas Law lets revise what various Gas Laws say.

**Browse more Topics under States Of Matter**

- Behaviour of Real Gases â€“ Deviations From Ideal Behaviour
- Gas Laws
- Intermolecular Forces
- Kinetic Molecular Theory of Gases
- Liquefaction of Gases
- The Gaseous State
- The Liquid State

### Boyle’s Law

It states that at a constant temperature, the volume of a gas is indirectly proportional to pressure. This implies that with an increase in pressure the volume of the gas will decrease. Mathematically Boyle’s law is summarized as VÂ \(\propto\)Â 1/P

### Charle’s Law

According to the Charle’s Law, at constant pressure, the volume of a gas is directly proportional to temperature. This means that with increasing temperature, the volume of the gas increases. Mathematically Charle’s Law can be written as VÂ \(\propto\)Â T

### Avogadro’s Law

Avogadro’s law states that at standard temperature and pressure conditions, the volume of a gas is directly proportional to the number of molecules of the gas. Therefore, mathematically we write it as VÂ \(\propto\)Â n

### The Equation

Now putting all the three laws together, we find that at standard conditions, Volume of a gas can be represented as follows:

VÂ \(\propto\) Â T Ã— n Ã— 1/P

V = R Ã— T Ã— n Ã— (1/P)

R, here is a proportionality constant. Therefore we can write, P V = n R T or R = PV/nT. Also called as the Gas Constant, R is same for all gases. This is therefore also called Universal Gas Constant. From theÂ analysis of the three Gas laws, we get the Ideal Gas Equation:

PV=nRT

Therefore if the number of moles (n) is 1 we get pV= RT. Now taking the value of m/M for n we write: pV=(m/M) RT ; (n=m/M)

Since pV= mRT /MÂ = (mRT)/VM

Therefore, p = (dRT)/MÂ Â (Since, d =m/V)

Hence we have: d = (pM) /RT

M= dRT/P. This equation brings us to the conclusion that under unchanged conditions of temperature, and pressure the density of a gas is directly proportional to the molar mass of the gas.

## Ideal Gas Behaviour

Many properties of Ideal gases are very closely similar to real gases. Some of these properties specific to Ideal gases are:

- Ideal gas constitutes a large number of molecules that are identical to each other.
- These molecules move in random motion and obey Newton’s Laws of Motion.
- The volume occupied by the gas is larger as compared to that of the volume occupied by the molecules.

It is only during a collision that molecules experience external forces. These collisions are elastic in nature and end in very short intervals of time.

## Value of R

R is the gas constant or proportionality constant in the Ideal Gas Equation. Mathematically, if you need to find the value of any variable, then you can do so if you have the other values. In the Ideal gas law equation pV = nRT, we can write R =Â pV/ nT. Now at Standard Temperature Pressure conditions i.e. when T = 273.15 K and p = 1 bar the volume is 22.710 L mol^{-1 }and number of moles(n) is 1, then R = (10^{5 }Pa)(22.71Ã—10^{-3 }m^{3}) / (1 mol)(273.15K)

= 8.314 Pa m^{3}K^{-1} mol^{-1}

= 8.314 JK^{-1} mol^{-1Â }

## Combined Gas Law

Now if the physical conditions of temperature, pressure and volume show variation then the initial values shall be T_{1}, p_{1Â }and V_{1} while the final values areÂ T2, p_{2Â }and V2Â then, we take, p_{1}V_{1}/T_{1Â }= nR and p_{2}V_{2}/T_{2Â }= nR. Therefore we have,

p_{1}V_{1}/T_{1Â Â }=Â p_{2}V_{2}/T_{2}

This equation is known as the Combined Gas Law.

## SolvedÂ Example For You

Q: At 45Â° C and 900mm of Hg pressure a gas occupies 500ml volume. What will be its pressure at a height where the temperature is 15Âº C and volume is 750mL.

Solution: Here we use the equation of Combined Gas Law, p_{1}V_{1}/T_{1Â Â }=Â p_{2}V_{2}/T_{2}

Therefore we have: p_{2Â }=Â p_{1}V_{1Â }T_{2}/T_{1Â }V_{2}

Hence, p_{1Â }= 900 mmHg, T_{1Â }= 45 +273= 318 K, V_{1Â }= 500 mL, T_{2}= 15+273 = 288 and V_{2Â }= 750 mL

p_{2}Â = (900 mm Hg) Ã— (500mL) Ã— 288 / 318 K Ã— 750 mL = 543.96 mm Hg

Therefore the height will be = 543.96 mm

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