The molar specific heat at constant pressure for a certain gas is given by \(C_{p}=a+b T+c T^{2},\) where \(a=33.6 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\), \(b=2.93 \times 10^{-3} \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}^{2},\) and \(c=2.13 \times 10^{-5} \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}^{3} .\) Find the entropy change when 2 moles of this gas are heated from \(20^{\circ} \mathrm{C}\) to \(200^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Substitute the values of \(a\), \(b\), \(c\), \(T_{i}\), and \(T_{f}\) into the final integral equation and calculate the entropy change \(\Delta S\). And then, multiply the number of moles, which is \(2\) moles, to obtain the total entropy change.

Step by step solution

01

Define Variables and Constants

First, understand the symbols and values given in the equation \(C_{p}=a+b T+c T^{2}\). Here, \(a=33.6 \, \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\), \(b=2.93 \times 10^{-3} \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}^{2}\), and \(c=2.13 \times 10^{-5} \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}^{3}\) are constants. The substance, which is being heated from \(20^{\circ} \mathrm{C}\) to \(200^{\circ} \mathrm{C}\), is equal to \(2\) moles.
02

Convert the Temperature to Kelvin

Note that temperature in the formula must be in Kelvin. Thus, convert given temperatures from Celsius to Kelvin by adding \(273.15\). Hence, the initial temperature \(T_{i}\) is \(20^{\circ}C + 273.15 = 293.15K\) and the final temperature \(T_{f}\) is \(200^{\circ}C + 273.15 = 473.15K\).
03

Calculate the Entropy Change

Next, use the thermodynamic formula for entropy change \(\Delta S\) at constant pressure, which is \(\Delta S = \int_{T_{i}}^{T_{f}} \frac{C_{p}}{T}\,dT\). Here, \(C_{p}\) is a function of \(T\), so by substituting \(C_{p}\) into the equation, it becomes \(\Delta S = \int_{T_{i}}^{T_{f}} \frac{a+b T+c T^{2}}{T}\,dT\). Separate and integrate each term separately, the result will be \(\Delta S = \int_{T_{i}}^{T_{f}} a\,dT + \int_{T_{i}}^{T_{f}} b T\,dT + \int_{T_{i}}^{T_{f}} c T^{2}\,dT\).
04

Evaluate the Integrals

After integrating, substitute the limits (\(T_{i}\) and \(T_{f}\)) to find the final value. The result should be multiplied by the number of moles, which is \(2\) in this case, to obtain the total entropy change.

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Key Concepts

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

Molar Specific Heat
Molar specific heat is an essential concept in thermodynamics that refers to the amount of heat required to raise the temperature of one mole of a substance by one degree Kelvin (or Celsius). It's usually denoted by the symbol C, with subscripts p or v to indicate whether the measurement is taken at constant pressure or constant volume, respectively.

The given formula Cp = a + bT + cT2 represents the variation of molar specific heat with temperature for the subject gas. The constants a, b, and c characterize the gas's unique thermal properties. Understanding how these constants function in the formula can help us better comprehend how the molar specific heat of a gas changes as its temperature increases.

This variable nature of Cp has significant implications when calculating thermal processes like the one in our exercise, where we determine entropy changes.
Thermodynamics
Thermodynamics is the study of heat, energy, and work, and how they interact with each other. It is a fundamental theory for various branches of science and engineering, explaining how thermal energy is converted to other energy forms and vice versa.

In our exercise, we focus on one of the central concepts of thermodynamics: entropy. Entropy is a measure of the disorder or randomness of a system and is significant because it describes the feasibility of a thermodynamic process. The Second Law of Thermodynamics tells us that entropy within a closed system tends to increase over time.

The finding of entropy change involves the calculation of heat exchange and understanding how it affects the disorder of the system. In this case, the heat exchange is characterized by the molar specific heat that changes with temperature.
Integral Calculus in Physics
Integral calculus is a mathematical tool frequently used in physics to calculate quantities that accumulate over time or space. It allows us to solve problems involving rates of change when the rates are expressed as functions.

In our entropy change calculation, integral calculus is indispensable because it helps us determine the total change when a variable specific heat is considered. As the function Cp varies with temperature, integrating it with respect to temperature allows us to accumulate the infinitesimal changes over the range from the initial to the final temperatures.

Applying Integral Calculus

By integrating the given function of Cp over the temperature limits, we are essentially summing up all the little bits of heat absorbed by the gas as its temperature raises incrementally, which then allows us to calculate the overall entropy change.
Kelvin Temperature Conversion
The Kelvin scale is an absolute thermodynamic temperature scale used widely in the physical sciences. Unlike Celsius or Fahrenheit, Kelvin begins at absolute zero, - the point where no more thermal energy can be extracted from a substance, which theoretically means no molecular motion.

In practical scenarios such as our exercise, converting Celsius to Kelvin is straightforward: we add 273.15 to the Celsius temperature. This conversion is critical because all thermodynamic calculations must be carried out in Kelvin to maintain consistency with the formulas' derivation, which assumes an absolute scale.

Importance in Calculations

When calculating thermodynamic properties like entropy, using the Kelvin scale ensures that we don't run into negative temperatures, which would be nonsensical in the context of the laws of thermodynamics. This also avoids the potential for mathematical errors that could arise if using a relative temperature scale like Celsius or Fahrenheit.

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Most popular questions from this chapter

You're the environmental protection officer for a \(35 \%\) efficient nuclear power plant that produces 750 MW of electric power, situated on a river whose minimum flow rate is \(110 \mathrm{m}^{3} / \mathrm{s}\). State environmental regulations limit the rise in river temperature from your plant's cooling system to \(5^{\circ} \mathrm{C}\). Can you achieve this standard if you use river water for all your cooling, or will you need to install cooling towers that transfer some of your waste heat to the atmosphere?

Why doesn't the evolution of human civilization violate the second law of thermodynamics?

Problem 74 of Chapter 16 provided an approximate expression for the specific heat of copper at low absolute temperatures: \(c=31(T / 343 \mathrm{K})^{3} \mathrm{J} / \mathrm{kg} \cdot \mathrm{K} .\) Use this to find the entropy change when \(40 \mathrm{g}\) of copper are cooled from \(25 \mathrm{K}\) to \(10 \mathrm{K}\). Why is the change negative?

A heat pump extracts energy from groundwater at \(10^{\circ} \mathrm{C}\) and transfers it to water at \(70^{\circ} \mathrm{C}\) to heat a building. Find (a) its COP and (b) its electric power consumption if it supplies heat at the rate of \(20 \mathrm{kW}\). (c) Compare the pump's hourly operating cost with that of an oil furnace if electricity costs \(15.5 \notin / \mathrm{kWh}\) and oil costs \(\$ 2.60 /\) gallon and releases about \(30 \mathrm{kWh} / \mathrm{gal}\) when burned.

A refrigerator maintains an interior temperature of \(4^{\circ} \mathrm{C}\) while its exhaust temperature is \(30^{\circ} \mathrm{C} .\) The refrigerator's insulation is imperfect, and heat leaks in at the rate of 340 W. Assuming the refrigerator is reversible, at what rate must it consume electrical energy to maintain a constant \(4^{\circ} \mathrm{C}\) interior?

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