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 ideal gas and the gas which does not obey this equation under all conditions of temperature and pressure is called real gases. A number of points can be discussed to compare the two types of gases.

__Ideal Gas:__

__Ideal Gas:__

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

**at each temperature is called one isotherm and at a different temperature, we have different isotherms. Two isotherms will never intersect.**__hyperbola Curve__Graphical Representation of Boyle's Low |

- Problem 1:

- Show that the

**of Ideal Gas depends on the temperature of the gas.**

__coefficient of thermal expansion__- Answer:

- Coefficient of

**is defined as, α = (1/V)[dV/dT]**

__thermal expansion(α)___{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__- Problem 2:

- Show that the coefficient of

**of Ideal Gas depends on the Pressure of the gas.**

__Compressibility__- Answer:

- Coefficient of

**is defined as, β = - (1/V)[dV/dP]**

__Compressibility(β)___{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 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
the gas molecules.__coalesce__ - The coefficient of
is found to vary from gas to gas that is α depends on the nature of the gas.__thermal expansion (ɑ)__ - The coefficient of
also is found to depend on the nature of the gas.__compressibility (β)__ - 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 real gas, following plots, called Amagat Curve are obtained. - When real gases 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.

But a temperature below the

**, the gas is liquefied after certain pressure depends on temperature. The point**__critical temperature(C)__**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__(T_{B})__:__

__Boyle Temperature__(T

_{B})

__:__

- At some intermediate temperature

**TB**called Boyle temperature, the initial slope is zero.

- At the Boyle temperature, the Z versus P line of an ideal gas is tangent to that of a real gas when P approaches zero. The latter rises above the ideal gas line only very slowly.

- Thus, at the Boyle temperature, the real gas behaves ideally over a wide range of pressure, because the effect of size of molecules and inter-molecular forces roughly compensate each other. Boyle Temperature(

**T**) is given by,

_{B}[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(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 |

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

- Problem 3:

- At

**273 K**and under pressure of

**100 atm**the compressibility factor of

**O**is

_{2}**0.97**. Calculate the

**mass of O**necessary to fill a gas cylinder of

_{2}**108.5 lit**capacity under the given conditions.

- Answer:

- Mass of O

_{2}= 1600 gm = 1.6 Kg

- See Question Answers Physical Chemistry

**.**

__Problem 8__