
In terms of mechanical units, ${C_p} - {C_v} = \cdots $ where ${C_p}$ and ${C_v}$ are principal specific heats.
A) $2R$
B) $\dfrac{R}{M}$
C) $\dfrac{R}{J}$
D) $\dfrac{R}{{M}}$
Answer
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Hint: ${C_p}$ is molar specific heat at constant pressure and ${C_v}$ is molar specific heat at constant volume. Use mayor’s relation for specific heat and specific heat capacity. One is for 1 mole and the other is for 1gm.
Complete step by step solution:
The relation given by Robert Mayer between ${C_p}$ and ${C_v}$ is –
${C_p} - {C_v} = R$$ \cdots eq(1)$
Here in $ \cdots eq(1)$ the relation is in between molar specific heats but we have to find out the relation between principal specific heats in terms of mechanical unit
So the relation mechanical unit (Joule) is –
${C_p} - {C_v} = \dfrac{R}{J}$$ \cdots eq(2)$
Where $J$ is used for unit conversion of $R$ from calories to joule, because ${C_p}$ and ${C_v}$ are given in joule, in mechanical units and $R$ was in calories.
Hence, option $\left( C \right)$ is the correct choice.
Note: If one of ${C_p}$ or ${C_v}$ are given in joule and other one in question is expecting to find out then we are going to use $ \cdots eq(2)$ because here $R$ is joule and we will get answer in joule. And for calorie we have to use $ \cdots eq(1)$.
Complete step by step solution:
The relation given by Robert Mayer between ${C_p}$ and ${C_v}$ is –
${C_p} - {C_v} = R$$ \cdots eq(1)$
Here in $ \cdots eq(1)$ the relation is in between molar specific heats but we have to find out the relation between principal specific heats in terms of mechanical unit
So the relation mechanical unit (Joule) is –
${C_p} - {C_v} = \dfrac{R}{J}$$ \cdots eq(2)$
Where $J$ is used for unit conversion of $R$ from calories to joule, because ${C_p}$ and ${C_v}$ are given in joule, in mechanical units and $R$ was in calories.
Hence, option $\left( C \right)$ is the correct choice.
Note: If one of ${C_p}$ or ${C_v}$ are given in joule and other one in question is expecting to find out then we are going to use $ \cdots eq(2)$ because here $R$ is joule and we will get answer in joule. And for calorie we have to use $ \cdots eq(1)$.
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