Simply one equation can be used to distinguish between

**from real gas and this is,***ideal gas*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*

- The
cannot be liquefied since it has no inter-molecular attraction and so that molecule will not condense.*Ideal gas* - The coefficient of thermal expansion(ɑ) depends on temperature(T) of the gas and does not depends on the nature of the gas.
- The coefficient of compressibility(β) similarly depends on the Pressure(P) of the gas and will be the same for all gases.
- When P is plotted against V, at constant temperature(T) a rectangular hyperbola curve is obtained as demanded by Boyle's Low,
- 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.
- When an
*ideal gas*passes through a porous plug from higher pressure to lower pressure within the insulated enclosure, there will be no change of temperature of the gas. This confirms that the*ideal gas*has no inter-molecular attraction.

PV = Constant |

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

Graphical Representation of Boyle's Law |

*Problem*- Show that the coefficient of thermal expansion of

*ideal gas*depends on the temperature of the gas.

*Answer*
Coefficient of thermal expansion(α) is defined as, α = (1/V)[dV/dT]

Ideal gas Equation for 1 mole gas is,PV = RT

Hence [dV/dT]

Thus α = (1/V) × (R/P) = (R/PV) = 1/T

This means all the gases have the same coefficient of thermal expansion.

_{P}Ideal gas Equation for 1 mole gas is,PV = RT

Hence [dV/dT]

_{P}= R/PThus α = (1/V) × (R/P) = (R/PV) = 1/T

This means all the gases have the same coefficient of thermal expansion.

*Problem*- Show that the coefficient of compressibility of an

**depends on the pressure of the gas.**

*ideal gas**Answer*

Coefficient of Compressibility(β) is defined as, β = - (1/V)[dV/dP]

Ideal gas equation for 1-mole gas is, PV = RT

Hence [dV/dP]

Thus β = (1/V) × (RT/P

This means the coefficient of Compressibility depends on the Pressure(P) of the gas.

_{T}Ideal gas equation for 1-mole gas is, PV = RT

Hence [dV/dP]

_{T}= - (RT/P^{2})Thus β = (1/V) × (RT/P

^{2}) = (RT/P^{2}V) = (RT/PV) × (1/P) = (1/P)This means the coefficient of Compressibility depends on the Pressure(P) of the gas.

###
*Real gas*

*Real gas*

- This gas could be liquefied since it has an intermolecular attraction which helps to coalesce the gas molecules.
- The coefficient of thermal expansion (ɑ) is found to vary from gas to gas that is α depends on the nature of the gas.
- The coefficient of compressibility (β) also is found to depend on the nature of the gas.
- When P is plotted against V, a rectangular hyperbola is obtained only at high temperature (above the critical temperature).
- When PV is plotted against P for
, following plots, called Amagat Curve are obtained.*real gas* - When
pass through porous plug from higher pressure to lower pressure within the 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.*real gases*

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 the liquid and gas can be indistinguishable.

- It shows that for most gases, the value of Z 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 the increase of pressure(P) from the very beginning.

- For CO₂, there is a large dip in the beginning, In fact, for gases which are easily liquefied, Z dips sharply below the ideal line in the low-pressure region.

###
*Boyle temperature*

- At some intermediate temperature TB called Boyle temperature, the initial slope is zero.

- At the Boyle temperature, the Z versus P line of an

**is tangent to that of a**

*ideal gas***when P approaches zero. The latter rises above the ideal gas line only very slowly.**

*real gas*- Thus, at the Boyle temperature, the

**behaves ideally over a wide range of pressure, because the effect of size of molecules and intermolecular forces roughly compensate each other. Boyle Temperature(T**

*real gas*_{B}) is given by,

[d(PV)/dP]_{T} when P → 0 |

- The Boyle temperature of Some gases are given below:

Gases | T_{B} |

Hydrogen (H_{2}) | -156^{0}C |

Helium (He) | -249^{0}C |

Nitrogen (N_{2}) | 59^{0}C |

Methane (CH_{4}) | 224^{0}C |

Ammonia (NH_{3}) | 587^{0}C |

- Thus we can see that for H₂ and He, the temperature of 0⁰C is above their respective Boyle temperature and so they have Z values greater than unity.

- The other gases at 0⁰C are below their respective Boyle temperature and so hay has Z values less than unity in the low-pressure region.

###
*Compressibility factor*

*Compressibility factor*

- An important single parameter, called Compressibility factor (Z) is used to measure the extent of deviation of the

**from ideal behavior. It is defined as,**

*real gases*Z = PV/RT |

- 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 a measure of the extent of non-ideality of the gas.

- When Zく1, the gas is more compressible then

**and When Z 〉1, the gas is a less compressible then ideal gas.**

*ideal gas*

*Problem*- At 273 K and under pressure of 100 atm the compressibility factor of O

_{2}is 0.97. Calculate the mass of O

_{2}necessary to fill a gas cylinder of 108.5 lit capacity under the given conditions.

*Answer*- Mass of O

_{2}= 1600 gm = 1.6 Kg