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Chapter 8: Problem 1143

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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Most popular questions from this chapter

Carnot engine working between \(300 \mathrm{~K}\) and \(600 \mathrm{~K}\) has work output of \(800 \mathrm{~J}\) per cycle. What is amount of heat energy supplied to the engine from source per cycle (A) \(1600\\{\mathrm{~J} /\) (cycle)\\} (B) \(2000\\{\mathrm{~J} /(\mathrm{cycle})\\}\) (C) \(1000\\{\mathrm{~J} /(\) cycle \()\\}\) (D) \(1800\\{\mathrm{~J} /(\) cycle \()\\}\)

One mole of an ideal gas $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma$ at absolute temperature \(\mathrm{T}_{1}\) is adiabatically compressed from an initial pressure \(\mathrm{P}_{1}\) to a final pressure \(\mathrm{P}_{2}\) The resulting temperature \(\mathrm{T}_{2}\) of the gas is given by. (A) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{\gamma /(\gamma-1)\\}}$ (B) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{(\gamma-1) / \gamma\\}}$ (C) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\gamma}$ (D) $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\mathrm{p}_{2} / \mathrm{p}_{1}\right)^{\gamma-1}$

The isothermal Bulk modulus of an ideal gas at pressure \(\mathrm{P}\) is (A) \(\mathrm{vP}\) (B) \(\mathrm{P}\) (C) \((\mathrm{p} / 2)\) (D) \((\mathrm{p} / \mathrm{v})\)

Efficiency of a Carnot engine is \(50 \%\), when temperature of outlet is $500 \mathrm{~K}\(. in order to increase efficiency up to \)60 \%$ keeping temperature of intake the same what is temperature of out let. (A) \(200 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(600 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)

One \(\mathrm{kg}\) of adiatomic gas is at a pressure of $5 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$ The density of the gas is \(\left\\{(5 \mathrm{~kg}) / \mathrm{m}^{3}\right\\}\) what is the energy of the gas due to its thermal motion ? (A) \(2.5 \times 10^{5} \mathrm{~J}\) (B) \(3.5 \times 10^{5} \mathrm{~J}\) (C) \(4.5 \times 10^{5} \mathrm{~J}\) (D) \(1.5 \times 10^{5} \mathrm{~J}\)
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