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# If for hydrogen ${C_P} - {C_V} = m$ and for nitrogen ${C_P} - {C_V} = n$ , where ${C_P}$​ and ${C_V}$ refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between m and n is (molecular weight of hydrogen=2 and molecular weight of nitrogen=14)(A) $n = 14m$(B) $n = 7m$(C) $m = 7n$(D) $m = 14n$

Last updated date: 12th Sep 2024
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Hint In this question, the specific heats per unit mass is given and we know the relation between molar specific heats of gases which is the amount of heat required by a mole of gas to raise its temperature by a degree . So, we can use the relation between mass and mole to get the result.
Formula Used:
${C_P} - {C_V} = R$

We know the relation ${C_P} - {C_V} = R$ where ${C_P}$ is the specific heat per unit mole at constant pressure and ${C_V}$ is the specific heat per unit mole at constant volume. It is for 1 mole of ideal gas. But in the question, specific heats per unit mass is mentioned.
If the molecular weight of a gas is ‘M’ then, 1 mole of gas = M gram that is ‘n’ moles of gas $= n \times M$ grams
So, for ‘M’ grams, the equation will be ${C_P} - {C_V} = \dfrac{R}{M}$
From the question for hydrogen, we get $2({C_P} - {C_V}) = R$
and for nitrogen, $14({C_P} - {C_V}) = R$
for hydrogen, $2m = R$ and for nitrogen $14n = R$ . As we can see, the right side of both the equations are same so equating them as equal,
$2m = 14n \Rightarrow m = 7n$
The specific heat of gas is at constant pressure is greater than specific heat of gas is at constant volume. The difference between the two is utilized as the heat used in doing work against the external pressure applied to raise 1 mole of gas temperature by ${1^0}C$ at constant pressure. There is a difference in notation of specific heat and molar specific heat. If it is written in small like ${c_P}$ or ${c_V}$ , then it is specific heats at constant pressure and volume respectively and if it is written in capital like ${C_P}$ or ${C_V}$ , then it is molar specific heat at constant pressure and volume respectively.