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Question: \({C_P}\;and\,\;{C_v}\)are specific heats at constant pressure and constant volume respectively. It ...

CP  and   Cv{C_P}\;and\,\;{C_v}are specific heats at constant pressure and constant volume respectively. It is observed that CP − CV = a{C_P}\, - \,{C_V}\, = \,a for hydrogen gas CP − CV = b{C_P}\, - \,{C_V}\, = \,bfor nitrogen gas.The correct relation between a and b is
A. a = 28ba\, = \,28b
B. a = 114ba\, = \,\dfrac{1}{{14}}b
C. a = ba\, = \,b
D. a = 14ba\, = \,14b

Explanation

Solution

Concept of conductivity and the factors on which it depends. Good conductivity refers to a good conductor.

Complete step by step answer:
By Mayer’s formula, for 1 g mole of a gas
CP−CV=R{C_P} - {C_V} = R
Where CP{C_P} is the specific heat at constant pressure
CV{C_V} is the specific heat at constant volume and R is the gas constant.
Now, when gram moles are given, this formula reduces to
CP−CV=RM{C_P} - {C_V} = \dfrac{R}{M}
Where
M is the molar mass
For Hydrogen gas, M=2×1=2M = 2 \times 1 = 2
So, CP−CV=R2{C_P} - {C_V} = \dfrac{R}{2} ….(1)
But according to question, CP − CV = a{C_P}\, - \,{C_V}\, = \,a …..(2)
a=R2 R=2a  a = \dfrac{R}{2} \\\ R = 2a \\\ ….(3)
For Nitrogen gas, Molar mass, M=2×14=28M = 2 \times 14 = 28
So, Mayer’s formula becomes
CP−CV=R28{C_P} - {C_V} = \dfrac{R}{{28}} ….(4)
But CP − CV = b{C_P}\, - \,{C_V}\, = \,b ….(5) (given)
From (4) and (5),
b=R28b = \dfrac{R}{{28}} ….(6)
Putting equation (3) in (6), we get
b=a14b = \dfrac{a}{{14}}
⇒      a=14 b\Rightarrow \;\;\;a = 14\,b
So, D is the correct option.

Note: CP{C_P} is greater than CV{C_V} because when a gas is heated at constant volume, no external work is done and so the heat supplied is consumed only in increasing the internal energy of a gas. But if the gas is heated at constant pressure, the gas expands against the external pressure so does some external work, in this case the heat is used up in increasing the internal energy of the gas and in doing some external work. Since the internal energy depends only on temperature the internal energy of a mass of a gas will increase by the same amount whether the pressure or volume remains constant. But since external work is additionally done for constant pressure than at constant volume to produce the same rise in temperature of the gas.