A gas at equilibrium has definite value of Pressure(

**P**), Volume(**V**), Temperature(**T**) and Composition(n). These are called state Variables and are determined experimentally. The state of the gas can be defined by these variables.**Boyle's(1662)**,**Charles's(1787)**and**Avogadro laws**gives the birth of an**equation of state for Ideal Gas**.Boyle's Low, V ∝ 1/PWhen n and T are constant for a gas. |

Charl's Low, V ∝ TWhen n and P are constant for a gas. |

Avogadro's Low, V ∝ nWhen P and T are constant for a gas. |

When all the variables are taken into account, The variation rule states that,

V ∝ (1/P) × T × n

or, V = R × (1/P) × T × n

∴ PV = nRT |

Ideal Gas Equation. |

Where

**R**is the__. This is called ideal gas equation of state for ideal gas.__**Universal Gas Constant**
This equation is found to hold most satisfactory when

**P**tense to zero. At ordinary temperature and pressure, the equation is found to deviated about**5%**.

__Value Of Universal Gas constant (R) at NTP:__
At NTP 1 mole gas at 1 atm Pressure occupied 22.4 lit of gas.

Thus, R = (PV)/(nT)

Putting the values above equation,

We have,

**R**= ( 1 atm × 22.4 lit)/(1 mol × 273 K)= 0.082 lit atm mol^{-1} K^{-1} |

__Value of R in C.G.S. and S.I. system:__**P**= 1 atm = 76 cm Hg

= 76 cm × 13.6 gm cm

^{-2}× 981 cm sec^{-2}
= 76 × 13.6 × 981 dyne cm

^{-2}
Thus,

**R**= (76 × 13.6× 981 dyne cm^{-2}×22.4 × 10^{3}cm^{3})/(1 mol × 273 K)= 8.314 × 10^{7} dyne cm^{2} mol^{-1} K^{-1} |

Again, Work (

**W**) = Force(**F**) × Displacement(**d**),
So, erg = dyne cm

^{2}.Thus, R = 8.314 × 10^{7} erg mol^{-1} K^{-1} |

We Know That, 1 J = 10

^{7}erg,
Thus the vale of

**R**in**S.I**. Unit,= 8.314 J mol^{-1} K^{-1} |

Again, 4.18 J = 1 Cal,

hence, R = 8.314 / 4.18 Cal mol

^{-1}K^{-1}
= 1.987 Cal mol

^{-1}K^{-1}≃ 2 Cal mol^{-1} K^{-1} |

__Physical Significance of Gas Constant R:__
The universal gas constant

**R = PV/nT**
Thus, it has the units of (Pressure × Volume)/(amount of gas × temperature).

Now the dimension of

**pressure**and**volume**are,
Pressure = (force/area)

= (force/ length

^{2})
= force × length

^{-2}
and Volume = length

^{3}
∴

**R**= (force×length^{-2}×length^{3})/(amount of gas×Kelvin)
= (force × length)/(amount of gas × kelvin)

= (Work or Energy)/(amount of gas × kelvin)

Thus, the dimensions of

**R**are energy per mole per kelvin and hence it represents the amount of work or energy) that can be obtained from one mole of a gas when its temperature is raised by one kelvin.
Determine the value of gas constant

**R**when pressure is expressed in**Torr**and**Volume**in**dm**.^{3}
61.54 Torr dm

For Solution See ^{3}mol^{-1}K^{-1}

__Problem 6__Properties of Gases |

Derive the value of

**R**when, (a)**pressure**is expressed in**atom**, and**volume**in**cm**and (b)^{3}**Pressure**in**dyne m**and^{-2}**volume****mm**.^{3}
(a) 82.05 atm cm

^{3}mol^{-1}K^{-1}
(b) 8.314 × 10

^{14}dyne m^{-2}mm^{3}mol^{-1}K^{-1}

__Problem 7__Properties of Gases |

__Determination of Molar mass from Ideal Gas Equation:__
The Ideal Gas Equation is,

**PV = nRT**

or, PV= (g/M)RT

Where

**g**=__weight of the gas__in gm and**M**=__Molar mass of the gas__.
Again,

**P = ( g/V) (RT/M)**
We know that, Density (

**d**) = Weight (**g**)/Volume (**V**).∴ P = dRT/M |

Find the Molar mass of

**ammonia**at**5 atm**pressure and**30**temperature (Density of ammonia =^{0}C**3.42 gm lit**).^{-1}
17 gm mol

For Solution See ^{-1}

__Problem 8__Properties of Gases |

What is the molecular weight of a gas,

**12.8 gms**of which occupy**10 liters**at a pressure of**750 mm**and at**27°C**?
31.91 gm mol

For Solution See ^{-1}

__Problem 9__Properties of Gases |

__Determination of Number of Molecule Present in Ideal Gas From Ideal Gas Equation:__
The Ideal gas equation for n mole gas is,

**PV = nRT**

Again, PV= (N/N

_{0}) RT
Where

**N**= Number of molecules present in the gas and**N**= Avogadro Number._{0}
Thus, P = (N/V) × (R/N

_{0}) × T∴ P = N′ k T |

Where

**N′**= number of molecules present per unit Volume and**k**= Boltzmann Constant =

**R**/

**N**

_{0}
= 1.38 × 10

^{-16}erg molecule^{-1}K^{-1}
Calculate the

**number of molecules present per ml**of an ideal gas maintained at pressure of**7.6 × 10**of Hg at 0°C.^{-3}mm
We have given that, V = 1ml = 10

^{-6}dm^{3}
P = 7.6 × 10

^{-3}mmHg
= (7.6 × 10

^{-3}mmHg) (101.235 kPa/760 mmHg)
= 1.01235 × 10

^{-3}**kPa**
Amount of the gas, n = PV/RT

= (1.01235 × 10

^{-3}kPa)(10^{-6}dm^{3})/(8.314 kPa dm^{3}mol^{-1}K^{-1})(273 K)**= 4.46 × 10**

^{-13}mol
Hence the number of molecules,

**N = n N**_{0}
= (4.46 × 10

^{-13}mol)(6.023 × 10^{23}mol^{-1})**= 2.68 × 10**^{-11}