During an adiabatic compression, $830 J$ of work is done on $2 m o l e s$ of a diatomic ideal gas to reduce its volume by $50$ %. The change in its temperature is nearly :
$(R = 8.3 J K^{- 1} m o l^{- 1})$

20 K

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14 K

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40 K

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33 K

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Solution

The correct option is A $20 K$
The work done in adiabatic process is given as
$W = \frac{n R Δ T}{γ - 1}$
or
$830 = \frac{2 \times 8.3 \times Δ T}{1.4 - 1}$
or
$Δ T = 20 K$

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