Real Gas

, Real Gases
[From Basic to Advance]

1. Van dar Waal’s Equation of State for Real Gases:

In reality no gas will obey ideal gas equation because the real gas molecules are not point
mass. These molecules have finite volume and the collision among them is not elastic. Also
their interactions are weak but not zero. For all these causes, the scientist Van dar Waal
corrected ideal gas equation of state by taking two corrections – volume correction and
pressure correction. After these two corrections Van dar Waal establish real gas equation of
state which is given by
𝐚 𝐧𝟐 𝐚
(𝐏 + 𝐕 𝟐 ) (𝐕 − 𝐛) = 𝐑𝐓 (For one mole real gas), (𝐏 + ) (𝐕 − 𝐧𝐛) = 𝐧𝐑𝐓 (For n mole real
𝐕𝟐
gas)

2. Behavior of Real Gases: Deviations from the Ideal Gas Equation: The Virial
Equation:

Basically at constant temperature, “PV-P” Graph for real gases not parallel to P axis. It will be
straight line with non zero slope or parabolic. For
real gas𝐏𝐕 ≠ 𝐑𝐓. Now for ideal gas at constant
temperature 𝐏𝐕 = 𝐑𝐓 = 𝐀 =Constant. But for
real gas 𝐏𝐕 ≠ 𝐀 ,
𝐏𝐕 = 𝐀 + 𝐁 𝐩 + 𝐂𝐩𝟐 + 𝐃𝐩𝟑 +…………….. where
A, B, C, ……………are called Virial coefficients and
by observation we get
𝐀 > 𝐁 > 𝐂 > 𝐃 >………………………..

Now from van dar Waal equation of state
𝐚 𝐑𝐓 𝐚
(𝐏 + 𝟐 ) (𝐕 − 𝐛) = 𝐑𝐓 =) 𝐏 = − 𝟐
𝐕 𝐕−𝐛 𝐕

𝐑𝐓𝐕 𝐚 𝐑𝐓 𝐚 𝐛 𝐚 𝐛 𝐛𝟐 𝐛𝟑 𝐚
𝐏𝐕 = 𝐕−𝐛 − 𝐕 = 𝐛 − 𝐕 = 𝐑𝐓(𝟏 − 𝐕)−𝟏 − 𝐕 =) 𝐏𝐕 = 𝐑𝐓(𝟏 + 𝐕 + 𝐕 𝟐 + 𝐕 𝟑 +…………) – 𝐕
(𝟏− )
𝐕
𝟏 𝐑𝐓𝐛𝟐 𝟏 𝐤 𝟏 𝐏
=) 𝐏𝐕 = 𝐑𝐓 + (𝐑𝐓𝐛 − 𝐚) 𝐕 + +……………. ∵ 𝐏 ∝ 𝐕 =) 𝐏 = 𝐕 . =) 𝐕 = 𝐤
𝐕𝟐
𝟏 𝐑𝐓𝐛𝟐
And 𝐏𝐕 = 𝐑𝐓 + 𝐤 (𝐑𝐓𝐛 − 𝐚)𝐏 + 𝐏 𝟐 +……= 𝐀 + 𝐁𝐏 + 𝐂𝐏 𝟐 +…………
𝐤𝟐
𝟏
Now 𝐀 = 𝐑𝐓, 𝐁 = 𝐤 (𝐑𝐓𝐛 − 𝐚),…………..
𝟏 𝟏 𝐚
Now at Boyel’s temperature 𝐓 = 𝐓𝐁 , 𝐁 = 𝐤 (𝐑𝐓𝐁 𝐛 − 𝐚) or 𝐁 = 𝐤 (𝐑𝐛. 𝐛𝐑 − 𝐚) = 𝟎 . Thus at
Boyel’s temperature the second Virial coefficient will be zero.