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Chapter 19: Problem 49

An ideal gas undergoes a process that takes it from pressure \(p_{1}\) and volume \(V_{1}\) to \(p_{2}\) and \(V_{2},\) such that \(p_{1} V_{1}^{\gamma}=p_{2} V_{2}^{\gamma},\) where \(\gamma\) is the specific heat ratio. Find the entropy change if the process consists of constant-pressure and constant-volume segments. Why does your result make sense?

Short Answer

Expert verified
The entropy change for the ideal gas undergoing a process consisting of constant-pressure and constant-volume segments is given by \(\Delta S = nR \frac{\ln T_{2}}{1-\gamma} -\gamma nR \frac{\ln T_{1}}{1-\gamma}\). The result makes sense as it aligns with the Second Law of Thermodynamics, which states that the total entropy of a system should increase for an irreversible process.

Step by step solution

01

Identify formulas

We should start by identifying the proper formulas for entropy change in a constant volume and a constant pressure process: \(\Delta S_{V} = C_{v}\ln \frac{T_{2}}{T_{1}}\) and \(\Delta S_{P} = C_{p}\ln \frac{T_{2}}{T_{1}}\) respectively. \(C_v\) and \(C_p\) are the specific heat capacity at constant volume and constant pressure, respectively. Using \(C_{p}=C_{v}+\kappa\), \(nR=\kappa\), and \(C_{p}/C_{v}=\gamma\), we can isolate \(C_v = nR / (\gamma - 1)\) and \(C_p = \gamma nR / (\gamma - 1)\). We will apply these formulas to calculate the entropy change in the next steps.
02

Calculate Entropy Change

Let's calculate entropy changes for the constant-volume and constant-pressure processes separately. For the constant-volume process, \(\Delta S_{V} = C_{v}\ln \frac{T_{2}}{T_{1}} = nR \frac{\ln T_{2}}{(\gamma - 1)} - nR \frac{\ln T_{1}}{(\gamma - 1)}\). And for the constant-pressure process, \(\Delta S_{P} = C_{p}\ln \frac{T_{2}}{T_{1}} = \gamma nR \frac{\ln T_{2}}{(\gamma - 1)} - \gamma nR \frac{\ln T_{1}}{(\gamma - 1)}\). Notice that because \(T_2\) and \(T_1\) are the final and initial temperatures of the gas, respectively, we have used ln \(\frac{T_{2}}{T_{1}}\) - which indicates the change in the natural logarithm of temperature.
03

Summarize the entropy change

The total entropy change is the sum of the entropy change in the constant-volume and constant-pressure processes. Therefore, \(\Delta S = \Delta S_{V} + \Delta S_{P} = nR \frac{\ln T_{2}}{(\gamma - 1)} - nR \frac{\ln T_{1}}{(\gamma - 1)} + \gamma nR \frac{\ln T_{2}}{(\gamma - 1)} - \gamma nR \frac{\ln T_{1}}{(\gamma - 1)} = nR \frac{\ln T_{2}}{1-\gamma} -\gamma nR \frac{\ln T_{1}}{1-\gamma}\).
04

Interpreting the result

The values obtained for the entropy change will only be positive for an isothermal process and for a reversible process the entropy change will be zero. Since entropy is a measure of disorder or randomness, a positive change in entropy (\(\Delta S>0\)) means the disorder of the system has increased during this process. This rise in entropy is in line with the Second Law of Thermodynamics, which states that processes occur in the direction of increasing total entropy. In this case, because \(\gamma\) is always > 1 for any gas, \(\Delta S\) will always be greater than zero due to the constant-volume and constant-pressure segments.

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Most popular questions from this chapter

The temperature of \(n\) moles of ideal gas is changed from \(T_{1}\) to \(T_{2}\) at constant volume. Show that the corresponding entropy change is \(\Delta S=n C_{V} \ln \left(T_{2} / T_{1}\right)\)

A power plant extracts energy from steam at \(280^{\circ} \mathrm{C}\) and delivers 880 MW of electric power. It discharges waste heat to a river at \(30^{\circ} \mathrm{C} .\) The plant's overall efficiency is \(29 \% .\) (a) How does this efficiency compare with the maximum possible at these temperatures? (b) Find the rate of waste-heat discharge to the river. (c) How many houses, each requiring \(23 \mathrm{kW}\) of heating power, could be heated with the waste heat from this plant?

You operate a store that's heated by an oil furnace supplying \(30 \mathrm{kWh}\) of heat from each gallon of oil. You're considering switching to a heat-pump system. Oil costs \(\$ 1.75 /\) gallon, and electricity costs \(16.5 \mathrm{c} / \mathrm{kWh}\). What's the minimum heat-pump COP that will reduce your heating costs?

A Carnot engine absorbs \(900 \mathrm{J}\) of heat each cycle and provides \(350 \mathrm{J}\) of work. (a) What's its efficiency? (b) How much heat is rejected each cycle? (c) If the engine rejects heat at \(10^{\circ} \mathrm{C}\), what's its maximum temperature?

Refrigerators remain among the greatest consumers of electrical energy in most homes, although mandated efficiency standards have decreased their energy consumption by some \(80 \%\) in the past four decades. In the course of a day, one kitchen refrigerator removes \(30 \mathrm{MJ}\) of energy from its contents, in the process consuming \(10 \mathrm{MJ}\) of electrical energy. The electricity comes from a \(40 \%\) efficient coal-fired power plant. The fuel energy consumed at the power plant to run this refrigerator for the day is a. \(12 \mathrm{MJ}\) b. \(25 \mathrm{MJ}\) c. \(40 \mathrm{MJ}\) d. \(75 \mathrm{MJ}\)
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