Which are true for reversible adiabatic process :

w = 2.303 R T log \frac{V_{2}}{V_{1}}

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w = \frac{n R}{(γ - 1)} (T_{2} - T_{1})

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w = 2.303 R T log \frac{V_{1}}{V_{2}}

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Δ U = w

n C_{v} Δ T

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Solution

The correct options are
B $w = \frac{n R}{(γ - 1)} (T_{2} - T_{1})$
D $Δ U = w$ = $n C_{v} Δ T$
We know, from first law, $Δ U = q + w$ .
For adibatic process, $q = 0$ .
$w = Δ U$
Also $Δ U$ = $n C_{v} d T$
(using $C_{v} = \frac{R}{γ - 1}$ )
$Δ U = w = \frac{n R}{(γ - 1)} (T_{2} - T_{1})$ .
These are the derived formula for $W_{r e v}$ in adiabatic process.

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\begin{matrix} C o l u m n - I (I d e a l G a s) & Column - II (Related equation) (A) Reversible isothermal process & (P) W = - 2.303 n R T l o g (V_{2} / V_{1}) (B) Reversible adiabatic process & (Q) W = n C_{v, m} (T_{2} - T_{1}) (C) Irreversible adiabatic process & (R) P V = n R T (D) Irreversible isothermal process & (S) W = - \int_{v_{i}}^{v_{f}} P_{e x t .} d V \end{matrix}

\begin{matrix} L i s t I & L i s t I I A . Reversible isothermal expansion & (p) w = - R P_{e x t} \frac{[T_{2} P_{1} - T_{1} P_{2}]}{P_{1} P_{2}} B . Irreversible isothermal expansion & (q) w = \frac{R}{γ - 1} [T_{2} - T_{1}] C . Reversible adiabatic expansion & (r) w = - P_{e x t} (V_{2} - V_{1}) D . Irreversible adiabatic expansion & (s) w = - 2.303 n R T l o g_{10} V_{2} / V_{1} \end{matrix}

w_{1}, w_{2}, w_{3}, w_{4}

\frac{2.303 \times R T}{F} = 0.06]

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