If one mole of an ideal gas \(\left(C_{p, m}=\frac{5}{2} R\right)\) is expanded isothermally at 300 until it's volume is tripled, then change in entropy of gas is : (a) zero (b) infinity (c) \(\frac{5}{2} R \ln 3\) (d) \(R \ln 3\)

Short Answer

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\(\frac{5}{2} R \ln 3\)

Step by step solution

01

Understanding the Isothermal Process

In an isothermal process, the temperature of the system remains constant. The change in entropy (\(\Delta S\)) for an isothermal expansion of an ideal gas can be calculated using the formula \(\Delta S = nC_{p, m} \ln\left(\frac{V_f}{V_i}\right)\), where \(n\) is the number of moles, \(C_{p, m}\) is the molar heat capacity at constant pressure, \(V_f\) is the final volume, and \(V_i\) is the initial volume.
02

Applying the Formula for Entropy Change

Since one mole of gas is given and its volume is tripled during the process, we apply the values to the entropy change formula: \(\Delta S = (1) \left(\frac{5}{2} R\right) \ln\left(\frac{3V_i}{V_i}\right)\). The initial volume \(V_i\) cancels out leaving the formula \(\Delta S = \frac{5}{2} R \ln 3\).
03

Simplifying and Finding the Answer

By simplifying the equation from Step 2, we get the change in entropy as \(\Delta S = \frac{5}{2} R \ln 3\), which is the solution to the problem. Therefore, the correct answer is option (c).

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation that relates the pressure, volume, temperature, and the number of moles of an ideal gas. Expressed in the formulation \(PV = nRT\), where \(P\) represents pressure, \(V\) is the volume, \(n\) stands for the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature in kelvins. This equation provides a clear picture of how gases would behave under different conditions, assuming that the gas particles do not interact and the volume of the individual particles is negligible compared to the volume the gas occupies.

In practical terms, when dealing with problems relating to the expansion or compression of a gas, referencing the ideal gas law helps to establish relationships between the changing parameters. For instance, during an isothermal expansion, while the temperature (\(T\)) and the amount of substance (\(n\)) remain constant, the product of pressure and volume must also remain constant. These relationships are crucial when performing entropy calculations in thermodynamics since they dictate how the volume change will affect entropy.
Entropy in Thermodynamics
Entropy, often denoted as \(S\), is a central concept in thermodynamics dealing with disorder and randomness in a system. It quantifies the number of possible configurations that a system can have, and in layman's terms, it measures the amount of chaos or disorder within a physical system. The second law of thermodynamics states that for an isolated system, entropy can only increase over time, indicating that processes naturally move towards a state of maximum disorder.

When a system experiences an isothermal process, as in the case with the expansion of an ideal gas, entropy changes because the volume accessible to the gas molecules increases. This leads to a larger number of possible microstates for the system, and thereby, increases its entropy. The calculation of entropy change for an ideal gas undergoing an isothermal process is simplified by the fact that temperature remains constant. Thus, the change in entropy can be directly linked to the volume change, as illustrated by the formula \(\Delta S = nC_{p, m} \ln(\frac{V_f}{V_i})\), where \(C_{p, m}\) is the molar heat capacity at constant pressure.
Molar Heat Capacity
Molar heat capacity is a physical property of substances that describes the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius at constant pressure (\(C_{p, m}\)) or volume (\(C_{v, m}\)). In the context of ideal gases, molar heat capacities are particularly simple because they are assumed to have constant values over a wide range of temperatures for any given gas.

The value of molar heat capacity influences how much a substance's temperature will change with a given amount of heat. In the step-by-step solution to the textbook exercise, the value of \(C_{p, m}\) for the ideal gas is given as \(\frac{5}{2}R\), which reflects the energy required for the gas to expand isothermally. Molar heat capacities are essential in calculating entropy changes because they enable the quantification of the energy exchanges that occur during thermodynamic processes, without temperature change in the case of isothermal processes. This demonstrates the practical use of molar heat capacity beyond merely the heating or cooling of substances—it's integral to our understanding of energy distribution in thermodynamic systems.

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Most popular questions from this chapter

The entropies of \(\mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{H}(\mathrm{g})\) are \(130.6\) and \(114.6 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) respectively at \(298 \mathrm{~K}\). Using the data given below calculate the bond energy of \(\mathrm{H}_{2}\) (in \(\mathrm{kJ} / \mathrm{mol}\) ): \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) ; \Delta G^{\circ}=406.6 \mathrm{~kJ}\) (a) \(377.2\) (b) \(436.0\) (c) \(425.5\) (d) \(430.5\)

The enthalpies of neutralization of a weak base \(A O H\) and a strong base \(B O H\) by \(H C l\) are \(-12250 \mathrm{cal} / \mathrm{mol}\) and \(-13000 \mathrm{cal} / \mathrm{mol}\) respectively. When one mole of \(\mathrm{HCl}\) is added to a solution containing 1 mole of \(A \mathrm{OH}\) and 1 mole of \(B O H\), the enthalpy change was \(-12500 \mathrm{cal} /\) mol. In what ratio is the acid distribution between \(A O H\) and \(B O H ?\) (a) \(2: 1\) (b) \(2: 3\) (c) \(1: 2\) (d) None of these

10 mole of ideal gas expand isothermally and reversibly from a pressure of \(10 \mathrm{~atm}\) to \(1 \mathrm{~atm}\) at \(300 \mathrm{~K}\). What is the largest mass which can lifted through a height of 100 meter? (a) \(31842 \mathrm{~kg}\) (b) \(58.55 \mathrm{~kg}\) (c) \(342.58 \mathrm{~kg}\) (d) None of these

An imaginary reaction \(X \longrightarrow Y\) takes place in three steps \(X \longrightarrow A, \Delta H=-q_{1} ; \quad B \longrightarrow A, \Delta H=-q_{2} ; \quad B \longrightarrow Y, \Delta H=-q_{3}\) If Hess' law is applicable, then the heat of the reaction \((X \rightarrow Y)\) is : (a) \(q_{1}-q_{2}+q_{3}\) (b) \(q_{2}-q_{3}-q_{1}\) (c) \(q_{1}-q_{2}-q_{3}\) (d) \(q_{3}-q_{2}-q_{1}\)

Calculate \(\Delta S\) for following process: $$ \underset{\text { at } 100 \mathrm{~K}}{X(s)} \longrightarrow \underset{\text { at } 200 \mathrm{~K}}{X(l)} $$ Given : Melting point of \(X_{(s)}=100 \mathrm{~K} ; \Delta H_{\text {Fusion }}=20 \mathrm{~kJ} / \mathrm{mol} ; C_{p, m}(X, l)=10 \mathrm{~J} / \mathrm{mol} \mathrm{K}\) (a) \(26.93 \mathrm{~J} / \mathrm{K}\) (b) \(206.93 \mathrm{~J} / \mathrm{K}\) (c) \(203 \mathrm{~J} / \mathrm{K}\) (d) \(206.93 \mathrm{~kJ} / \mathrm{K}\)

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