Question

Derive an expression for the ratio of Cp/Cv for a diatomic gas.

Answer 1

Answer: The ratio of the $\dfrac{C_{p}}{C_{v}}$ for diatomic gas is $\dfrac{C_{p}}{C_{v}}= \dfrac{7}{5}$.

Explanation:

Using specific heat,

$C_{v} = \dfrac{dQ}{(ndt)}$

Using first law of thermodynamics

$dQ = dU + dW$

We know,

dQ = dU

At constant volume

We know that,

$U = fn \dfrac{RT}{2}$

On differentiate

$\dfrac{dU}{dT} = fn\dfrac{R}{2}$

Where, f = degree of freedom

Now, the molar specific heat at constant

$C_{v} = \dfrac{dU}{n dT}$.....(I)

Now put the value of dU in equation (I)

$C_{v} = \dfrac{fR}{2}$...(II)

Using mayer's law

$C_{p}-C_{v} = R$

$C_{p} =R+C_{v}$

Now, put the value of $C_{v}$

$C_{p} = R+\dfrac{fR}{2}$

$C_{p} = (\dfrac{f}{2}+1)R$...(III)

Now, the ratio of $C_{p}$ and $C_{v}$

$\dfrac{C_{p}}{C_{v}}=\dfrac{(\dfrac{f}{2}+1)R}{\dfrac{fR}{2}}$

For diatomic gas f = 5

$\dfrac{C_{p}}{C_{v}}= \dfrac{\dfrac{5}{2}+1}{\dfrac{5}{2}}$

$\dfrac{C_{p}}{C_{v}}= \dfrac{7}{5}$

Hence, the ratio of the $\dfrac{C_{p}}{C_{v}}$ for diatomic gas is $\dfrac{C_{p}}{C_{v}}= \dfrac{7}{5}$.

Answer 2

"Derivation for the ratio $\frac { { C }_{ p } }{ { C }_{ v } }$

Using specific heat,

${ C }_{ v }\quad =\quad \frac { dQ }{ (ndt) }$

Using first law of thermodynamics

$dQ\quad =\quad dU+dW$

We know,

dQ = dU

At constant volume

We know that,

$U\quad =\quad { f }_{ n }\frac { RT }{ 2 }$

On differentiate

$\frac { dU }{ dT } \quad =\quad { f }_{ n }\frac { R }{ 2 }$

Where, f = degree of freedom

Now, the molar specific heat at constant

${ C }_{ v }\quad =\quad \frac { dU }{ ndT }.....(I)$

Now put the value of dU in equation (I)

${ C }_{ v }\quad =\quad \frac { fR }{ 2 }...(II)$

Using mayer's law

${ C }_{ p }-{ C }_{ v }\quad =\quad R$

${ C }_{ p }\quad =\quad R+{ C }_{ v }$

Now, put the value of Cv

${ C }_{ p }\quad =\quad \frac { R+fR }{ 2 }$

${ C }_{ p }\quad =\quad (\frac { f }{ 2 } +1)R..(III)$

Now, the ratio of  and

$\frac { { C }_{ p } }{ { C }_{ v } } \quad =\quad \frac { (\frac { f }{ 2 } +1)R }{ \frac { fR }{ 2}}$

For diatomic gas f = 5

$\frac { { C }_{ p } }{ { C }_{ v } } \quad =\quad \frac { (\frac { 5 }{ 2 } +1) }{ \frac { 5 }{ 2}}$

$\frac { { C }_{ p } }{ { C }_{ v } } \quad =\quad \frac { 7 }{ 5 }$

Hence, the ratio of the Cp/Cv for diatomic gas is $\frac { { C }_{ p } }{ { C }_{ v } } \quad =\quad \frac { 7 }{ 5 }$"