For an ideal gas, in an isobaric process, the ratio of heat supplied ' Q ' to the work done ' w ' by the system is ( γ = ratio of specific heat at constant pressure to that at constant volume) (a) (1)/(γ) (b) (1)/(γ - 1) (c) (γ)/(γ - 1) (d) (γ - 1)/(γ)

Answer: (c). Isobaric: Q = nCpΔ T, W = nRΔ T. Q/W = Cp/R = (Cp)/(Cp - Cv) = (γ)/(γ - 1) So the answer is (c) (γ)/(γ - 1).

For an ideal gas, in an isobaric process, the ratio of heat supplied… | MHT CET 2024 (9 May, Shift 1)

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Thermodynamic Processes

For an ideal gas, in an isobaric process, the ratio of heat supplied ' $Q$ ' to the work done ' $w$ ' by the system is ( $\gamma =$ ratio of specific heat at constant pressure to that at constant volume)

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Answer: (c)

Isobaric: $Q = nC_p\Delta T$ , $W = nR\Delta T$ .
$\frac{Q}{W} = \frac{C_p}{R} = \frac{C_p}{C_p - C_v} = \frac{\gamma}{\gamma - 1}$
So the answer is (c) $\frac{\gamma}{\gamma - 1}$ .

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