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__Comparison Between Ideal and Real Gases____:__

__Comparison Between Ideal and Real Gases__

__:__

Simply one equation can be used to distinguish between ideal gas from real gas and this is,

**PV = nRT**

The gas which obeys this equation under all conditions of temperature and pressure is called

**and the gas which does not obey this equation under all conditions of temperature and pressure is called**__ideal gas__**. A number of points can be discussed to compare the two types of gases.**__real gases__###
__Ideal Gas :__

__Ideal Gas :__

(i) The Ideal Gas can not be liquefied since it has no inter-molecular attraction and so that molecule will not condense.

(ii) The coefficient of

**thermal expansion(𝛂)**depends on temperature(**T**) and does not depends on the nature of the gas.
Show that coefficient of thermal expansion of Ideal Gas depends on the temperature of the gas.

Coefficient of thermal expansion(𝛂) is defined as 𝛂 = (1/V)(dV/dT)P.

Ideal Gas Equation for 1 mole gas is PV = RT.

Hence (dV/dT)p = R/P.

Thus 𝛂 = (1/V) × (R/P) = (R/PV) = 1/T.

__This means all the gases have the same coefficient of thermal expansion.__

(iii) The coefficient of

**compressibility(β)**similarly depends on the Pressure(P) of the gas and will be same for all gases.
Show that coefficient of Compressibility of Ideal Gas depends on the Pressure of the gas.

Coefficient of Compressibility(β) is defined as β = - (1/V)(dV/dP)T.

Ideal Gas Equation for 1 mole gas is PV = RT.

Hence (dV/dP)T = - (RT/P²).

Thus β = (1/V) × (RT/P²) = (RT/P²V) = (RT/PV) × (1/P) = (1/P).

__This means__

**coefficient of Compressibility**depends on the Pressure(P) of the gas.
(iv) When P is plotted against V, at constant temperature(T) a rectangular hyperbola curve is obtained as demanded by Boyle's Low PV = Constant.

P vs V Plot |

The hyperbola Curve at each temperature is called one isotherm and at different tempera-ture we have different isotherms. Two isotherms will never intersect.

(v) When

**PV**is plotted against**P**, at constant**T**a straight line parallel to**P**-axis is obtained. At different temperature there will be different parallel lines.PV vs P Plot |

(vi) When an ideal gas passes through a porous plug from higher pressure to lower pressure within insulated enclosure, there will be no change of temperature of the gas . This confirms that the ideal gas has no inter-molecular attraction.

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__Real Gas :__

__Real Gas :__

(i) This gas could be

__liquefied__since__it has inter molecular attraction__which helps to coalesce the gas molecules.
(ii) The coefficient of

**thermal expansion (α)**is found to vary from gas to gas that is**α**depends on the nature of the gas.
(iii) The

**coefficient of compressibility (β)**also is found to depend on the nature of the gas.
(iv) When

**P**is plotted against**V**, a rectangular hyperbola is obtained only at high temperature (above the critical temperature).P vs V Plot for Real Gas |

But a temperature below the

**critical temperature(C)**, the gas is liquefied after certain pressure depends on temperature. The point**C**is the critical point where he liquid and gas can be indistinguishable.
(v) When

**PV**is plotted against**P**for__real gas__, following plots, called__Amagat Curve__are obtained.###
__Amagat Curve:__

__Amagat Curve:__

Amagat Curves for different gasesat a given Temperate( 0°C) |

It shows that for most gases, value of

**PV**decreases,__attain minimum and then increases with the increase of__.**P**
Only Hydrogen(

**H₂**) and Helium(**He**) baffle this trend and__the curve rises with increase of Pressure(__.**P**) from the very beginningAmagat Curves for a gas (CO₂)at different temperature |

This shows that for

At **CO₂**gas__the depth of the minimum shifts__towards the**PV**-axis with the increase of temperature (**T**).**T₃**temperature

**PV**runs parallel to the

**P**-axis up to certain range of

**P**at low pressure region (

**P→0**).

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__Boyle Temperature(TB):__

__Boyle Temperature(TB):__

In the above Curve

**T₃**is called**Boyle Temperature(TB)**at which real gas obeys Boyle’s Law up to certain range of pressure at the low pressure region.
The minimum coincides with the

**PV**-axis. The mathematical condition for the calculation of**Boyle Temperature(TB)**is give by,**[d(PV)/dP]T = 0**when**P**→ 0.
The curves obtained for Hydrogen(

**H₂**) and Helium(**He**) at 0°C is above their**Boyle Temperature**and so with increase of**P**, values of**PV**increases from the beginning.###
__Compressibility Factor(Z):__

__Compressibility Factor(Z):__

An important single parameter, called

**Compressibility factor (Z)**is used to measure the extent of deviation of the real gases from ideal behavior.__It is defined as,__

**Z = PV/RT**where

**V**is the

__molar volume.__

When

**Z=1**, the gas is ideal or there is no deviation from ideal behavior.
When

**Z ≠ 1**, the gas is non-ideal and departure of the value of**Z**from unity is measure of the extent of non-ideality of the gas.
When

**Zく1**, the gas is more compressible then ideal gas and When**Z 〉1**, the gas is less compressible then ideal gas.
(vi) When real gases pass through porous plug from higher pressure to lower pressure within insulated enclosure, there occurs a change of temperature. This is due to the fact that real gases have inter-molecular attraction and when the gas expands, the molecules have to spend energy to overcome inter-molecular attraction and so the temperature of the gas drops down.

At 273 K and under pressure of 100 atm the compressibility factor of O₂ is 0.97. Calculate the mass of O₂ necessary to fill a gas cylinder of 108.5 lit capacity under the given conditions.

T = 273 K, Z = 0.97 and P = 100 atm.

From the definition of compressibility factor Z = PV/RT where V is the molar Volume.

Thus the molar volume of O₂ is

Vm = ZRT/P = (0.97) × (0.082 lit atm mol⁻¹ K⁻¹) × (273 K)/(100 atm) = 2.17 lit mol⁻¹

The mass of this molar volume will be equal to the molar mass of oxygen, that is 2.17 lit of oxygen equal to 32 gm.

Thus the mass of oxygen required to fill the gas cylinder 108.5 lit under the given condition

= {(32 gm)/(2.17 lit)} × (108.5 lit)

**= 1600 gm = 1.6 Kg**