One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{o} C$ . If the work done during the process is $3 k J$ , the final temperature will be equal to: $(C_{V} = 20 J K^{- 1})$

150 K

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100 K

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26, 85 K

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295 K

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Solution

The correct option is A $150 K$
From first law of thermodynamics:
$Δ U = q + w$

For an adiabatic process $q = 0$

$∴ Δ U = w$

As, $d U = C_{V} . d T$

Therefore, $w = C_{V} . d T$ = $C_{V} (T_{1} - T_{2})$ (in expansion $T_{1} > T_{2}$ )

Given that $w = 3000 J$ , $T_{1} = 27^{\circ} C = 300 K$ and $C_{V} = 20 J / K$

Upon substitution we get:

$3000 = 20 (300 - T_{2})$

$T_{2} = 300 - \frac{3000}{20} = 300 - 150 = 150 K$
Therefore option A is correct.

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