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Chapter 8: Problem 1170

The Volume of air increases by \(5 \%\) in an adiabatic expansion. The percentage decrease in its Pressure will be (A) \(5 \%\) (B) \(6 \%\) (C) \(7 \%\) (D) \(8 \%\)

Short Answer

Expert verified
The percentage decrease in pressure during an adiabatic expansion when the volume of air increases by 5% is approximately 7% (C).

Step by step solution

01

Express the volume increase in terms of initial and final volumes

If the volume of air increases by 5%, then the final volume can be expressed as: \(V_2 = V_1 \times (1 + 0.05) = 1.05 V_1\)
02

Substitute the expression for final volume into the adiabatic process equation

We can now substitute the expression for \(V_2\) into the adiabatic process equation and solve for the final pressure: \(P_1 V_1^{\gamma} = P_2 (1.05 V_1)^{\gamma}\)
03

Solve for the final pressure

Divide both sides of the equation by \(V_1^{\gamma}\) and use the adiabatic index value for air, \(\gamma = 1.4\): \(P_1 = P_2 (1.05)^{1.4}\) To find the final pressure, divide both sides of the equation by \((1.05)^{1.4}\): \(P_2 = \frac{P_1}{(1.05)^{1.4}}\)
04

Calculate the percentage decrease in pressure

The percentage decrease in pressure can be calculated as: Percentage decrease = \(\frac{P_1 - P_2}{P_1} \times 100\% = \frac{P_1 - \frac{P_1}{(1.05)^{1.4}}}{P_1} \times 100\% \) Simplify the expression: Percentage decrease = \(\frac{(1 - \frac{1}{(1.05)^{1.4}})P_1}{P_1} \times 100\% \) The \(P_1\) terms cancel out: Percentage decrease = \((1 - \frac{1}{(1.05)^{1.4}}) \times 100\% \) Calculate the percentage decrease: Percentage decrease ≈ \((1 - \frac{1}{1.0726}) \times 100\% \approx 7.26\% \)
05

Compare the result to the given options and identify the correct answer

The calculated percentage decrease in pressure is approximately 7.26%, which is closest to 7% among the given options. Thus, the correct answer is: (C) 7%

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Most popular questions from this chapter

When a System is taken from State \(i\) to State \(f\) along the path iaf, it is found that \(\mathrm{Q}=70 \mathrm{cal}\) and \(\mathrm{w}=30 \mathrm{cal}\), along the path ibf. \(\mathrm{Q}=52\) cal. \(\mathrm{W}\) along the path ibf is (A) 6 cal (B) \(12 \mathrm{cal}\) (C) \(24 \mathrm{cal}\) (D) 8 cal

For an adiabatic expansion of a perfect gas, the value of $\\{(\Delta \mathrm{P}) / \mathrm{P}\\}$ is equal to (A) \(-\sqrt{\gamma}\\{(\Delta \mathrm{r}) / \mathrm{v}\\}\) (B) \(-\\{(\Delta \mathrm{v}) / \mathrm{v}\\}\) (C) \(-\gamma^{2}\\{(\Delta \mathrm{v}) / \mathrm{v}\\}\) (D) \(-\gamma\\{(\Delta \mathrm{v}) / \mathrm{v}\\}\)

The temperature of sink of Carnot engine is \(27^{\circ} \mathrm{C}\). Efficiency of engine is \(25 \%\) Then find the temperature of source. (A) \(227^{\circ} \mathrm{C}\) (B) \(327^{\circ} \mathrm{C}\) (C) \(27^{\circ} \mathrm{C}\) (D) \(127^{\circ} \mathrm{C}\)

In an isochoric Process \(\mathrm{T}_{1}=27^{\circ} \mathrm{C}\) and \(\mathrm{T}_{2}=127^{\circ} \mathrm{C}\) then $\left(\mathrm{P}_{1} / \mathrm{P}_{2}\right)$ will be equal to (A) \((9 / 59)\) (B) \((2 / 3)\) (C) \((4 / 3)\) (D) \((3 / 4)\)

A Carnot engine having a efficiency of \(\mathrm{n}=(1 / 10)\) as heat engine is used as a refrigerators. if the work done on the system is \(10 \mathrm{~J}\). What is the amount of energy absorbed from the reservoir at lowest temperature ! (A) \(1 \mathrm{~J}\) (B) \(90 \mathrm{~J}\) (C) \(99 \mathrm{~J}\) (D) \(100 \mathrm{~J}\)
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