A gas performs Q work when it expand at constant pressure. During this process heat absorbed by the gas is 4Q. The average number of degrees of freedom for the gas is:

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Solution

The correct option is B 6
$G i v e n :$ A gas performs $Q$ work when it expand at constant pressure. During this process heat absorbed by the gas is $4 Q .$
$S o l u t i o n :$ In a isobaric process,
$U = 4 Q - Q = n C_{v} d T = 3 Q$
Work = $n R d T$ So, U/Work $= C_{v} / R$
$3 = C_{v} / R$
$R = C_{p} - C_{v}$
$3 = \frac{C_{v}}{C_{p} - C_{v}}$
$\frac{C_{p}}{C v} = 4 / 3 = γ$
$γ = 1 + \frac{2}{f}$
where f is degree of freedom $4 / 3 = 1 + \frac{2}{f}$
$f = 6$
$S o, t h e$ $c o r r e c t$ $o p t i o n : B$

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