Question

solve the above problem ​

Answer 1

Here,

$\displaystyle\longrightarrow\sf{\gamma=\dfrac {C_P}{C_V}}$

By Meyer's Relation,

• $\displaystyle\sf{C_P=C_V+R}$

Hence,

$\displaystyle\longrightarrow\sf{\gamma=\dfrac {C_V+R}{C_V}\quad\quad\dots (1)}$

But we know that, internal energy,

$\displaystyle\longrightarrow\sf{u=\dfrac {n}{2}\,RT}$

$\displaystyle\longrightarrow\sf{\dfrac {du}{dT}=\dfrac {d}{dT}\left [\dfrac {n}{2}\,RT\right]}$

$\displaystyle\longrightarrow\sf{C_V=\dfrac {n}{2}\,R}$

Hence (1) becomes,

$\displaystyle\longrightarrow\sf{\gamma=\dfrac {\dfrac {n}{2}\,R+R}{\dfrac {n}{2}\,R}}$

$\displaystyle\longrightarrow\sf{\gamma=\dfrac {\left (\dfrac {n}{2}+1\right)\,R}{\dfrac {n}{2}\,R}}$

$\displaystyle\longrightarrow\sf{\gamma=\dfrac {\dfrac {n}{2}+1}{\dfrac {n}{2}}}$

$\displaystyle\longrightarrow\sf{\gamma=\dfrac {n+2}{n}}$

$\displaystyle\longrightarrow\sf{\underline {\underline {\gamma=1+\dfrac {2}{n}}}}$

Hence (1) is the answer.