A certain quantity of ideal gas takes up $60 J$ of heat in the process AB and $360 J$ in the process AC. What is the number of degrees of freedom of the gas. (Answer upto two digit after decimal point)

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Solution

$C_{P} = \frac{R γ}{γ - 1}$ , $C_{V} = \frac{R}{γ - 1}$ & $γ = 1 + \frac{2}{f}$

$Δ Q_{A B} = n C_{P} Δ T = \frac{γ}{γ - 1} n R Δ T = \frac{γ}{γ - 1} [3 P_{0} V_{0} - P_{0} V_{0}] Δ Q_{A B} = 2 P V_{0} \times \frac{γ}{γ - 1} . . . . . . . . (1)$

$Δ Q_{A C} = Δ U + Δ W$
Where, $Δ W$ is the area under the curve $A C$ & $Δ U = n C_{V} Δ T$

$Δ Q_{A C} = \frac{n R}{γ - 1} Δ T + \frac{1}{2} (4 V_{0} - V_{0}) (4 P_{0} + P_{0})$

$Δ Q_{A C} = \frac{[16 P_{0} V_{0} - P_{0} V_{0}]}{γ - 1} + \frac{15 P_{0} V_{0}}{2} . . . . . . (2)$

From equation (1)
$60 = 2 P_{0} V_{0} \times \frac{γ}{γ - 1} . . . . . . (3)$

From equation (2)
$360 = 15 P_{0} V_{0} [\frac{γ - 1 + 2}{2 (γ - 1)}] . . . . . . . (4)$

Dividing $(4)$ by $(3)$ :
$\frac{360}{60} = \frac{15}{4} \frac{(γ + 1)}{γ}$
$24 γ = 15 γ + 15$
$γ = \frac{15}{9} = 1 + \frac{2}{f} \Rightarrow f = 3$

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