The volume of 1 mole of an ideal gas with the adiabatic exponent $γ$ is changed according to the relation $V = \frac{b}{T}$ , where b = constant. The amount of heat absorbed by the gas in the process if the temperature is increased by $∆ T$ will be :

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$\frac{R}{γ - 1} ∆ T$

$\frac{R ∆ T}{γ - 1}$

$(\frac{1 - γ}{γ + 1}) R ∆ T$

$(\frac{2 - γ}{γ - 1}) R ∆ T$

answer is B.

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Detailed Solution

$\begin{array}{l} V = \frac{b}{T}, Δ Q = ? P V = n R T \\ V = \frac{b}{\frac{P V}{n R}} \Rightarrow P V^{2} = c o n s t . \end{array}$

It is a Polytropic Process with n = 2

$\begin{array}{l} C = C_{V} + \frac{R}{1 - n} = \frac{R}{γ - 1} + \frac{R}{1 - 2} = (\frac{2 - γ}{γ - 1}) R \\ Q = nCΔT = (\frac{2 - γ}{γ - 1}) R ΔT \end{array}$

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