Which of the following is true in case of reversible adiabatic expansion?

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Solution

The correct option is B

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

$Δ V = q + ω$

$\Rightarrow Δ V = ω [q = 0]$

If $ω = - v e \Rightarrow Δ V = - v e$ [by the system]

If $ω = + v e \Rightarrow Δ V = + v e$ [on the system]

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

As we know that,

$C_{v} = (\frac{d V}{d T}) v$

$\Rightarrow d V = C_{v} . d T$

and for finite change $\Rightarrow Δ V = C_{v} Δ T$

Therefore, $ω = Δ V = C_{v} Δ T$

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

Reversible Adiabatic Expansion

We know that,

$ω = - P Δ V$

& $ω = C_{v} Δ T$ (as we just saw)

$C_{v} Δ V = - P Δ V$

For very small change in reversible process,

$C_{v} Δ T = - P d V$

$\Rightarrow C_{v} Δ T = - \frac{R T}{v} d v$ (for one mole of gas) (As we know, $P v = n R T$ )

$\Rightarrow C_{v} . \frac{d T}{T} = - R . \frac{d V}{V}$

Integrating from $T_{1}$ to $T_{2}$ and $V_{1}$ to $V_{2}$

$C_{v} T_{2} \int T_{1} \frac{d T}{T} = - R v_{2} \int v_{1} \frac{d v}{v}$

$\Rightarrow C_{v} l o g_{e} \frac{T_{2}}{T_{1}} = - R l o g_{e} \frac{v_{2}}{v_{1}} = R l o g_{e} \frac{v_{1}}{v_{2}}$

$\Rightarrow l o g \frac{T_{2}}{T_{1}} = - \frac{R}{c_{v}} l o g \frac{v_{2}}{v_{1}} = \frac{R}{c_{v}} l o g \frac{v_{1}}{v_{2}}$

Now, as we know

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

$\Rightarrow \frac{c_{p}}{c_{v}} - 1 = \frac{R}{c_{v}}$

$\Rightarrow (γ - 1) = \frac{R}{c_{v}} [∵ \frac{c_{p}}{c_{v}} = γ]$

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

$L o g \frac{T_{2}}{T_{1}} = (γ - 1) l o g \frac{v_{1}}{v_{2}}$

= $l o g (\frac{v_{1}}{v_{2}})$ ...............(5)

or $\frac{T_{2}}{T_{1}} = {(\frac{v_{1}}{v_{2}})}^{γ - 1}$ .............(6)

$\frac{T_{1}}{T_{2}} = {(\frac{v_{2}}{v_{1}})}^{γ - 1}$ .............(7)

$\Rightarrow \frac{P_{1} V_{1}}{P_{2} V_{2}} = (\frac{v_{2}}{v_{1}}) γ - 1 \Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{v_{2}}{v_{1}})}^{γ}$

$\Rightarrow P_{1} v_{1}^{γ} = P_{2} v_{2}^{γ}$

$\Rightarrow P v^{γ} = c o n s t a n t$

Now, we know that

$\frac{V_{1}}{V_{2}} = \frac{P_{2} T_{1}}{P_{1} T_{2}}$ ................(8)

Substituting (8) in (4)

${(\frac{P_{2} T_{1}}{P_{1} T_{2}})}^{γ - 1} = \frac{T_{2}}{T_{1}}$

$\Rightarrow {(\frac{P_{2}}{P_{1}})}^{γ - 1} = {(\frac{T_{2}}{T_{1}})}^{γ}$

$\Rightarrow {(\frac{T_{1}}{T_{2}})}^{γ} = {(\frac{P_{1}}{P_{2}})}^{1 - γ}$

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