Question

For a perfect gas, the ratio of volume coefficient of expansion to pressure coefficient is:
(A) Equal to one
(B) Less than one
(C) More than one
(D) An imaginary quantity

Answer 1

Hint Volume coefficient of expansion is a coefficient for Joule expansion which is an expansion with no change in internal energy. Meanwhile, pressure coefficient of expansion is a coefficient for Joule-Kelvin expansion which is an expansion with no change in enthalpy.

Formula Used: The formulae used in this solution are given as,
 $ {\beta _V} = \dfrac{1}{V}{\left. {\dfrac{{\partial V}}{{\partial T}}} \right|_P} $
 $ {\beta _P} = \dfrac{1}{P}{\left. {\dfrac{{\partial P}}{{\partial T}}} \right|_V} $
 $ PV = nRT $
Here, $ {\beta _V} $ is the volume coefficient of expansion, $ {\beta _P} $ is the pressure coefficient of expansion, $ T $ is the temperature, $ V $ is the volume, $ P $ is the pressure, $ n $ is the number of moles of the gas and $ R $ is the perfect gas constant.

Complete step by step answer
We already know that Volume coefficient of expansion is given as,
 $ {\beta _V} = \dfrac{1}{V}{\left. {\dfrac{{\partial V}}{{\partial T}}} \right|_P} $
Note that the partial differentiation is done keeping pressure constant.
Now since perfect gas equation is,
 $ PV = nRT $
In terms of Volume, this can be written as,
 $ V = \dfrac{{nRT}}{P} $
So, Putting this value of $ V $ in $ {\left. {\dfrac{{\partial V}}{{\partial T}}} \right|_P} $ we get,
 $
  {\left. {\dfrac{{\partial V}}{{\partial T}}} \right|_P} = {\left. {\dfrac{{\partial (\dfrac{{nRT}}{P})}}{{\partial T}}} \right|_P} \\
   \Rightarrow {\left. {\dfrac{{\partial V}}{{\partial T}}} \right|_P} = \dfrac{{nR}}{P}\left( {\dfrac{{\partial T}}{{\partial T}}} \right) \\
 $
This gives,
 $ {\left. {\dfrac{{\partial V}}{{\partial T}}} \right|_P} = \dfrac{{nR}}{P} $
Putting this value in the expression for $ {\beta _V} $ we get,
 $ {\beta _V} = \dfrac{1}{V}{\left. {\dfrac{{\partial V}}{{\partial T}}} \right|_P} = \dfrac{1}{V}\left( {\dfrac{{nR}}{P}} \right) $
Now since,
 $ \dfrac{{nR}}{P} = \left( {\dfrac{V}{T}} \right) $
So, we can write,
 $
  {\beta _V} = \dfrac{1}{V}\left( {\dfrac{V}{T}} \right) \\
   \Rightarrow {\beta _V} = \dfrac{1}{T} \\
 $
Now, Similarly, Pressure Coefficient can be written as,
 $ {\beta _P} = \dfrac{1}{P}{\left. {\dfrac{{\partial P}}{{\partial T}}} \right|_V} $
This on similar simplifications leads to,
 $ {\beta _P} = \dfrac{1}{T} $
Thus the ratio of volume of coefficient to Pressure coefficient is given as,
 $
  ratio = \dfrac{{{\beta _V}}}{{{\beta _P}}} \\
    \\
 $
So, putting in the required values we get,
 $
  ratio = \left( {\dfrac{{\left( {\dfrac{1}{T}} \right)}}{{\left( {\dfrac{1}{T}} \right)}}} \right) \\
   \Rightarrow ratio = 1 \\
 $
$ \therefore $ Option (A) is the correct option out of the given options.

Additional Information
We should keep in mind that the concept used above is for the process of thermal expansion. The concept and the formulas would differ, if the expansion was Joule-Thomson Expansion.

Note
Here it is important to note that to calculate pressure coefficient, volume is kept constant, and while calculating volume coefficient, pressure is kept constant. Do not confuse one with the other.