Q: The average degrees of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is
(a) 75 J
(b) 100 J
(c) 150 J
(d) 125 J
Ans: (b)
Sol: f = 6
$\gamma = 1 + \frac{2}{\gamma} $
$\gamma = 1 + \frac{2}{6} = \frac{4}{3} $
$C_p = \frac{\gamma R}{\gamma -1} $
$\large C_p = \frac{\frac{4}{3} R}{\frac{4}{3} -1} $
Cp = 4 R
∆W = n R ∆T
∆Qp = n Cp ∆T
$\large \frac{\Delta W}{\Delta Q_p} = \frac{R}{C_p} = \frac{1}{4}$
∆Qp = 4 ∆W = 4 × 25 = 100 J