Which of the following is true in case of reversible ad

SOLUTION

Adiabatic Expansion or Compression Process

As we know, in adiabatic process, heat change is zero.

$q=0$ [No heat is allowed to enter or leave the system]

Now, as we know

$\Delta V=q+\omega$

$\Rightarrow \Delta V=\omega [q=0]$

If $\omega =-ve\Rightarrow \Delta V=-ve$ [by the system]

If $\omega =+ve\Rightarrow \Delta V=+ve$[on the system]

Now, Let's try to find out the work done in adiabatic expansion.

As we know that,

$C_v=\left(\frac{dV}{dT}\right)v$

  $\Rightarrow dV=C_v.dT$

and for finite change $\Rightarrow \Delta V=C_v\Delta T$

Therefore, $\omega =\Delta V=C_v\Delta T$

Here, the value of $\Delta T$ depends on the process whether it is reversible or irreversible.

Reversible Adiabatic Expansion

We know that,

$\omega =-P\Delta V$

& $\omega =C_v\Delta T$     (as we just saw)

$C_v\Delta V=-P\Delta V$

For very small change in reversible process,

$C_v\Delta T=-PdV$

$\Rightarrow C_v\Delta T=-\frac{RT}{v}dv$(for one mole of gas) (As we know, $Pv=nRT$)

$\Rightarrow C_v.\frac{dT}{T}=-R.\frac{dV}{V}$

Integrating from $T_1$ to $T_2$ and $V_1$ to $V_2$

$C_v\underset{T_1}{\overset{T_2}{\int }}\frac{dT}{T}=-R\underset{v_1}{\overset{v_2}{\int }}\frac{dv}{v}$

$\Rightarrow C_{v}lo{g}_e\frac{T_2}{T_1}=-Rlo{g}_e\frac{v_2}{v_1}=Rlo{g}_e\frac{v_1}{v_2}$

$\Rightarrow log\frac{T_2}{T_1}=-\frac{R}{c_v}log\frac{v_2}{v_1}=\frac{R}{c_v}log\frac{v_1}{v_2}$

Now, as we know

$C_p-C_v=R$

$\Rightarrow \frac{c_p}{c_v}-1=\frac{R}{c_v}$

$\Rightarrow (\gamma -1)=\frac{R}{c_v}[\because \frac{c_p}{c_v}=\gamma ]$

Now, put value of $\frac{R}{C_v}$ in eq. (4),

$Log\frac{T_2}{T_1}=(\gamma -1)log\frac{v_1}{v_2}$

= $log\left(\frac{v_1}{v_2}\right)$       ...............(5)

or $\frac{T_2}{T_1}={\left(\frac{v_1}{v_2}\right)}^{\gamma -1}$ .............(6)

$\frac{T_1}{T_2}={\left(\frac{v_2}{v_1}\right)}^{\gamma -1}$ .............(7)

$\Rightarrow \frac{P_1{V}_1}{P_2{V}_2}=\left(\frac{v_2}{v_1}\right)\gamma -1\Rightarrow \frac{P_2}{P_1}={\left(\frac{v_2}{v_1}\right)}^{\gamma }$

$\Rightarrow P_1{v}_1^{\gamma }=P_2{v}_2^{\gamma }$

$\Rightarrow P{v}^{\gamma }=constant$

Now, we know that

$\frac{V_1}{V_2}=\frac{P_2{T}_1}{P_1{T}_2}$................(8)

Substituting (8) in (4)

${\left(\frac{P_2{T}_1}{P_1{T}_2}\right)}^{\gamma -1}=\frac{T_2}{T_1}$

$\Rightarrow {\left(\frac{P_2}{P_1}\right)}^{\gamma -1}={\left(\frac{T_2}{T_1}\right)}^{\gamma }$

$\Rightarrow {\left(\frac{T_1}{T_2}\right)}^{\gamma }={\left(\frac{P_1}{P_2}\right)}^{1-\gamma }$