$1 m o l e$ of a non linear triatomic ideal gas is expanded adiabatically at $300 K$ from $16 a t m$ to $1 a t m$ . Calculate workdone under the following conditions -
$(i)$ Free expansion
$(i i)$ Reversible expansion
$(i i i)$ Irreversible expansion

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Solution

(i)Free expansion
No work is done during free expansion as their is
no external pressure
$∴ w = 0$
(ii) Reversible expansion
as $P_{1}^{1 - r} T_{1}^{r} = P_{2}^{1 - r} T_{2}^{r}$
$T_{2} = T_{1} {(\frac{P_{1}}{P_{2}})}^{\frac{1 - v}{v}}$
as for triatomic ideal gas, $C_{v} = 3 R$ & $v = 4 / 3$
$∴ T_{2} = (300) [\frac{16}{1}]^{\frac{1 - 4 / 3}{4 / 3}} = 300 \times (16)^{- 1 / 4} = 150 k$
$w = c_{v} (T_{2} - T_{1}) = 3 R (150 - 300) = - 450 R$
(iii) Irreversible expansion
$- P_{e x t} (v_{2} - v_{1}) = C_{v} (T_{2} - T_{1})$
as $n = 1$ & $P_{e x} = 1$
$∴ - 1 [\frac{R T_{2}}{P_{2}} - \frac{R T_{1}}{P_{1}}] = 3 R [T_{2} - T_{1}]$
$\frac{- 1}{16} [16 T_{2} - T_{1}] = 3 [T_{2} - T_{1}]$
On solving, $T_{2} = 229.69 k$
Using $= C_{v} (T_{2} - T_{1}) = 3 R (229.69 - 300) = - 210.93 R$

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