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Chapter 8: Problem 1145

If \(r\) denotes the ratio of adiabatic of two specific heats of a gas. Then what is the ratio of slope of an adiabatic and isothermal \(\mathrm{P} \rightarrow \mathrm{V}\) curves at their point of intersection ? (A) \((1 / \gamma)\) (B) \(\gamma-1\) (C) \(\gamma\) (D) \(\gamma+1\)

Short Answer

Expert verified
The ratio of the slopes of adiabatic and isothermal P-V curves at their point of intersection is \(\gamma\).

Step by step solution

01

Write down the equations of adiabatic and isothermal P-V curves

For an adiabatic process, we have the equation: \(PV^{\gamma}=K_{1}\), where K1 is a constant For an isothermal process, we have the equation: \(PV = K_{2}\), where K2 is a constant
02

Differentiate the equations with respect to V

Now we need to find the slope of both curves at any point, so we will differentiate them with respect to V: For adiabatic: \(\frac{dP}{dV} = -\frac{\gamma K_{1}}{V^{\gamma+1}}\) For isothermal: \(\frac{dP}{dV}=-\frac{K_{2}}{V^{2}}\)
03

Calculate the ratio of slopes

Now, let's find the ratio of slopes of adiabatic and isothermal processes: \(\frac{dP_{ad}/dV}{dP_{iso}/dV} = \frac{-\frac{\gamma K_{1}}{V^{\gamma+1}}}{-\frac{K_{2}}{V^{2}}}\)
04

Simplify the ratio

Simplifying the ratio, we get: \(\frac{dP_{ad}/dV}{dP_{iso}/dV}=\frac{\gamma K_{1}V^{2}}{K_{2}V^{\gamma+1}}\) At the point of intersection of both curves, \(K_{1} = K_{2}\) and \(PV^{\gamma}=PV\) which gives us: \(V^{\gamma-1}=1\) This implies that: \(V^{\gamma+1}=V^{2}\) Now substituting this value in the ratio: \(\frac{dP_{ad}/dV}{dP_{iso}/dV}=\frac{\gamma K_{1}V^{\gamma+1}}{K_{2}V^{\gamma+1}}\) \(\Rightarrow \frac{dP_{ad}/dV}{dP_{iso}/dV}=\frac{\gamma K_{1}}{K_{2}}\) Since, at the point of intersection, \(K_{1}=K_{2}\), our final answer becomes: \(\frac{dP_{ad}/dV}{dP_{iso}/dV}=\gamma\) The correct option is (C) \(\gamma\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Ratio
The specific heat ratio, often denoted as \( \gamma \), is a crucial factor in the study of thermodynamics as it relates to gases. It is defined as the ratio of the specific heat at constant pressure \( C_p \) to the specific heat at constant volume \( C_v \). Understanding \( \gamma \) is essential in analyzing processes such as adiabatic and isothermal, as it tells us how a gas behaves when heat is added or removed.
  • In an adiabatic process, no heat is transferred to or from the system, which means that the specific heat ratio directly influences how the pressure and volume change.
  • The value of \( \gamma \) is greater than 1 for gases, with typical values ranging for air around 1.4, indicating that gases expand more at constant pressure compared to constant volume.
Slope of Curves
In thermodynamics, the slope of a curve tells us about the rate of change of one variable with respect to another. For P-V (pressure-volume) curves, understanding the slope is critical for both adiabatic and isothermal processes.
  • In an adiabatic process, the slope is derived from the equation \( PV^{\gamma} = K_1 \). Differentiating gives the slope \( \frac{dP}{dV} = -\frac{\gamma K_1}{V^{\gamma+1}} \), showing dependence on both \( \gamma \) and the volume \( V \).
  • For an isothermal process described by \( PV = K_2 \), the slope is simpler, \( \frac{dP}{dV} = -\frac{K_2}{V^2} \), indicating a direct relation to \( V \).
By comparing these slopes at a point, one can glean insights into the behavior of the gas under different conditions.
Differentiation of Equations
Differentiation is used to find slopes of curves representing physical processes, such as the pressure-volume relationships in gas laws like the adiabatic and isothermal equations. Differentiating these equations provides the rate at which pressure changes with respect to volume.
  • For the adiabatic equation \( PV^{\gamma} = K_1 \), differentiation with respect to \( V \) yields \( \frac{dP}{dV} = -\frac{\gamma K_1}{V^{\gamma+1}} \).
  • For the isothermal equation, \( PV = K_2 \), differentiation gives \( \frac{dP}{dV} = -\frac{K_2}{V^2} \).
Understanding these derivatives is important since it helps in calculating the precise behavior of gases under these conditions, allowing for the comparison of slope ratios in this process geometry.
Point of Intersection
The point of intersection between adiabatic and isothermal curves is a critical concept when comparing processes involving gases. At this intersection, both curves satisfy their respective equations equally for a given pressure and volume.
  • At the intersection, \( PV^{\gamma} = K_1 = K_2 = PV \).
  • This point conditionally implies \( V^{\gamma-1} = 1 \), simplifying to \( V^{\gamma+1} = V^2 \), which allows for simplification in comparing slopes.
Understanding the intersection helps in both calculating slope ratios and making physics assessments about gas behavior under these conditions.
P-V Curves
P-V curves, or pressure-volume curves, are graphical representations of the relationships between pressure (P) and volume (V) for a gas under different processes. These curves are foundational to understanding thermodynamic processes like adiabatic and isothermal.
  • An adiabatic P-V curve does not transfer heat and is steeper due to the energy conservation equation \( PV^{\gamma} = K_1 \).
  • An isothermal P-V curve represents constant temperature processes with the equation \( PV = K_2 \), indicating a hyperbolic curve behavior.
Through analyzing these curves, we observe how gases react to changes in pressure or volume while maintaining certain energy conservation conditions, offering insights into more complex thermodynamic systems.

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Most popular questions from this chapter

Work done by \(0.1\) mole of a gas at \(27^{\circ} \mathrm{C}\) to double its volume at constant Pressure is \(\quad\) Cal. \(R=2\left\\{(\mathrm{Cal}) /\left(\mathrm{mol}^{\circ} \mathrm{K}\right)\right\\}\) (A) 600 (B) 546 (C) 60 (D) 54

A Carnot engine takes \(3 \times 10^{6}\) cal of heat from a reservoir at \(627^{\circ} \mathrm{C}\), and gives to a sink at \(27^{\circ} \mathrm{C}\). The work done by the engine is (A) \(4.2 \times 10^{6} \mathrm{~J}\) (B) \(16.8 \times 10^{6} \mathrm{~J}\) (C) \(8.4 \times 10^{6} \mathrm{~J}\) (D) Zero

In a container of negligible heat capacity, \(200 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) and \(100 \mathrm{~g}\) steam at \(100^{\circ} \mathrm{C}\) are added to \(200 \mathrm{~g}\) of water that has temperature \(55^{\circ} \mathrm{C}\). Assume no heat is lost to the surroundings and the pressure in the container is constant \(1 \mathrm{~atm} .\) Amount of the Sm left in the system, is equal to (A) \(16.7 \mathrm{~g}\) (B) \(8.4 \mathrm{~g}\) (C) \(12 \mathrm{~g}\) (D) \(0 \mathrm{~g}\) Copyright () StemEZ.com. All rights reserved.

In Column I different Process is given match corresponding option of column \(\mathrm{I}_{1}\) Column - I Columm - II (A) adiabatic process (p) \(\Delta \mathrm{p}=0\) (B) Isobaric process (q) \(\Delta \mathrm{u}=0\) (C) Isochroic process (r) \(\Delta Q=0\) (D) Isothermal process (s) \(\Delta \mathrm{W}=0\) (A) A-p, B-s, C-r, D-q (B) (B) A-s, B-q, C-p, D-r (C) \(\mathrm{A}-\mathrm{r}, \mathrm{B}-\mathrm{p}, \mathrm{C}-\mathrm{s}, \mathrm{D}-\mathrm{q}\) (D) \(\mathrm{A}-\mathrm{q}, \mathrm{B}-\mathrm{r}, \mathrm{C}-\mathrm{q}, \mathrm{D}-\mathrm{p}\)

Work done per mole in an isothermal ? change is (A) RT log \(_{e}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (B) \(\mathrm{RT} \log _{10}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (C) RT \(\log _{10}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) (D) RT \(\log _{\mathrm{e}}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\)
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