For an isothermal expansion of a Perfect gas, the value of $(\Delta \mathrm{P} / \mathrm{P})$ is equal to (A) \(-\gamma^{(1 / 2)}\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) (B) \(-\gamma\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) (C) \(-\gamma^{2}\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) \((\mathrm{D})-(\Delta \mathrm{V} / \mathrm{V})\)

Short Answer

Expert verified
The short answer for the given problem is: (D) \( -(\Delta V / V) \).

Step by step solution

01

Recap Boyle's Law and γ (Gamma)

Boyle's Law for an ideal gas states that at constant temperature, the pressure of an ideal gas is inversely proportional to its volume. Mathematically, it can be written as \(PV = Constant\) or \(P_1V_1 = P_2V_2\). Gamma (γ) in thermodynamics is a specific heat ratio. For a perfect gas, γ = Cp/Cv, where Cp is the Specific heat at constant pressure and Cv is the Specific heat at constant volume.
02

Express the change in terms of ratio

Let's denote the change in Pressure as ΔP and the change in Volume as ΔV. The percentages in change of the pressure and volume can be given as \( (\Delta P / P) \) and \( (\Delta V / V) \) respectively.
03

Derive the required expression

Starting from Boyle's Law, we can write, \( P_1V_1 = P_2V_2 \). This can be rewritten as, \( P = \frac{P_2V_2}{V} \). Now, change in Pressure, \( ΔP = P_2 - P = P_2(1 - \frac{V}{V_2}) \). Hence, \( \frac{ΔP}{P} = \frac{P_2(1 - \frac{V}{V_2})}{P} = 1 - \frac{V}{V_2} = 1 - \frac{1}{(1 + \frac{ΔV}{V})} \). Considering the fact for small changes, \( (1 + x)^n \approx 1 + nx \), where \( |x| \ll 1 \), we can write, \( \frac{ΔP}{P} = -\frac{ΔV}{V} \). It indicates that the change in pressure is directly proportional to the change in volume, but they are in opposite direction. Hence, the correct answer is (D) \( -(\Delta V / V) \).

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