![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
Determine the value of $ {C_p},{C_v} $ and $ \gamma $ for a monatomic, diatomic and polyatomic gas.
Answer
393k+ views
Hint: In a thermodynamic system, $ {C_p} $ is the amount of heat energy absorbed or released by a unit mass of a given substance with the change in temperature at a constant pressure. $ {C_v} $ is the amount of heat energy transferred between a system and its surroundings at a constant volume. $ \gamma $ is the ratio of $ {C_p} $ and $ {C_v} $
Complete answer:
The degree of freedom of a molecule will affect the value of all the specific heat constants.
For any molecule the degrees of freedom is given by the equation:
$ f = 3N - m $
Where f is the degree of freedom
N is the number of particles in the given system/molecule
m is the number of constraints.
So for the given type of molecules we can say that:
Now we can say that the specific heat constants are related to the degree of freedom by the following equations:
$ \Rightarrow {C_p} = R(1 + \dfrac{1}{2}f) $
$ \Rightarrow {C_v} = \dfrac{1}{2}f \times R $
$ \Rightarrow \gamma = 1 + \dfrac{2}{f} $
Therefore we can substitute the values of f in the equations and obtain the value of all the constants for monoatomic, diatomic and polyatomic gas.
Note:
For gas with molecules having higher numbers of atomicity, the degree of freedom would vary even more because there are a large number of arrangements that are possible. This will mean that there will be more degrees of freedom with each type of molecule. Thus for such types of gases a known degree of freedom can be used to figure out the specific molar heats and their ratios, which will always be decreasing with increasing atomicity of the molecule.
Complete answer:
The degree of freedom of a molecule will affect the value of all the specific heat constants.
For any molecule the degrees of freedom is given by the equation:
$ f = 3N - m $
Where f is the degree of freedom
N is the number of particles in the given system/molecule
m is the number of constraints.
So for the given type of molecules we can say that:
Type of molecule | Degrees of freedom |
Monoatomic molecule | 3 (N=1, m=0) |
Diatomic molecule | 5 (N=2, m=1) |
Polyatomic molecule | >6 |
Now we can say that the specific heat constants are related to the degree of freedom by the following equations:
$ \Rightarrow {C_p} = R(1 + \dfrac{1}{2}f) $
$ \Rightarrow {C_v} = \dfrac{1}{2}f \times R $
$ \Rightarrow \gamma = 1 + \dfrac{2}{f} $
Therefore we can substitute the values of f in the equations and obtain the value of all the constants for monoatomic, diatomic and polyatomic gas.
Type of Gas molecule | Degree of freedom f | $ {C_p} $ | $ {C_v} $ | $ \gamma $ |
Monoatomic | 3 | $ \dfrac{5}{2}R $ | $ \dfrac{3}{2}R $ | $ \dfrac{5}{3} $ |
Diatomic | 5 | $ \dfrac{7}{2}R $ | $ \dfrac{5}{2}R $ | $ \dfrac{7}{5} $ |
Triatomic (A case of polyatomic) | 7 (where N=3 and m=2) | $ \dfrac{9}{2}R $ | $ \dfrac{7}{2}R $ | $ \dfrac{9}{7} $ |
Note:
For gas with molecules having higher numbers of atomicity, the degree of freedom would vary even more because there are a large number of arrangements that are possible. This will mean that there will be more degrees of freedom with each type of molecule. Thus for such types of gases a known degree of freedom can be used to figure out the specific molar heats and their ratios, which will always be decreasing with increasing atomicity of the molecule.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Find the value of x if the mode of the following data class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How many valence electrons does nitrogen have class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Soap bubble appears coloured due to the phenomenon class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How is the brain protected from injury and shock class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)