Consider a system consisting of one mole of a diatomic gas, contained in a piston. What is the temperature change of the gas, if $q = 50 J a n d W = - 100 J$ ?

- 50 K

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- 2.4 K

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- 5.8 K

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- 1.25 K

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Solution

The correct option is B $- 2.4 K$
According to first law of thermodynamics,
$△ U = q + W$
$= 50 - 100 = - 50 J$
we know that
$△ U = n C_{v} △ T$
$n = 1$
$C_{v} = \frac{5}{2} R$ for diatomic gas

$∴ △ T = \frac{△ U}{C_{V}} = \frac{- 50 J}{\frac{5}{2} R}$
putting values we get,

$= \frac{- 50}{\frac{5}{2} \times 8.314} = - 2.41 K$

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