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What is the entropy change when 1 mole oxygen gas expands isothermally and reversibly from an initial volume of \(10 \mathrm{~L}\) to \(100 \mathrm{~L}\) at \(300 \mathrm{~K} ?\) (a) \(19.14 \mathrm{JK}^{-1}\) (b) \(109.12] \mathrm{K}^{-1}\) (c) \(29.12 \mathrm{JK}^{-1}\) (d) \(10 \mathrm{~K}^{-1}\)

Short Answer

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The entropy change (\

Step by step solution

01

Understanding entropy change for an ideal gas

The entropy change for an ideal gas during an isothermal process can be calculated using the formula: \(\Delta S = nR\ln\left(\frac{V_f}{V_i}\right)\), where \(\Delta S\) is the entropy change, \(n\) is the number of moles of the gas, \(R\) is the gas constant (8.314 J/(mol K)), \(V_i\) is the initial volume, and \(V_f\) is the final volume. In this exercise, we are given that \(n = 1\) mole of oxygen gas, \(V_i = 10\) L, and \(V_f = 100\) L.
02

Convert volumes from liters to cubic meters

Before applying the entropy change formula, we need to work with consistent volume units. The gas constant R is given in joules per mole per Kelvin, which requires volume in cubic meters. Hence, we convert liters to cubic meters: \(1\) L = \(0.001\) m^3. Thus, \(V_i = 10 \times 0.001 = 0.010\) m^3, and \(V_f = 100 \times 0.001 = 0.100\) m^3.
03

Calculate the entropy change

Now we can calculate the entropy change by substituting the values into the formula: \(\Delta S = nR\ln\left(\frac{V_f}{V_i}\right)\). That gives us \(\Delta S = 1 \times 8.314 \times \ln\left(\frac{0.100}{0.010}\right)\), which simplifies to \(\Delta S = 8.314 \times \ln(10)\).
04

Calculate the natural logarithm

The natural logarithm of 10 is a constant and can be approximated as \(\ln(10) \approx 2.303\). With this, the entropy change can be further calculated as \(\Delta S = 8.314 \times 2.303\).
05

Solve for the entropy change

Multiplying these values together gives the final entropy change: \(\Delta S = 8.314 \times 2.303 = 19.14\) J/K.

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Key Concepts

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

Isothermal process
In thermodynamics, an isothermal process is one that occurs at a constant temperature. This means that the system we are examining exchanges heat with its surroundings to maintain a uniform temperature throughout the process. When dealing with an ideal gas, like oxygen in this problem, an isothermal process is described by Boyle's Law, which states that the product of pressure and volume is a constant when temperature is held steady.

An isothermal expansion, such as the oxygen gas expanding from 10L to 100L, involves the gas doing work against the external pressure, and thus absorbing an identical amount of heat from the surroundings to preserve its temperature. The significance of an isothermal process in entropy calculations lies in the fact that, unlike adiabatic or polytropic processes, calculating the entropy change only involves the volumes and the gas constant, making the computation more straightforward.
Ideal gas entropy
Entropy, in simple terms, is a measure of the disorder or randomness of a system. It plays a pivotal role in determining the feasibility and direction of a thermodynamic process. For an ideal gas undergoing an isothermal process, the entropy change is directly related to the change in volume.

The key equation for entropy change in an ideal gas during an isothermal and reversible expansion or compression is \[\[\begin{align*}\Delta S = nR \ln\left(\frac{V_f}{V_i}\right)\end{align*}\]\], where:\begin{itemize}\item\(\Delta S\) is the entropy change\item\(n\) is the number of moles\item\(R\) is the universal gas constant\item\(V_i\) and \(V_f\) are the initial and final volumes, respectively.\end{FOOTER}.\end{itemize} To better understand entropy, imagine spreading a drop of ink into a glass of water. Initially, the ink is concentrated, but over time it disperses throughout the water, creating a more disordered state. In the context of gases, when a gas expands, its molecules become more dispersed, increasing the entropy of the system.
Gas constant
In thermodynamics, the gas constant, denoted as \(R\), is a fundamental parameter that appears in various equations relating the physical properties of gases. It features prominently in the ideal gas law, which relates the pressure, volume, temperature, and number of moles of an ideal gas. The value of the universal gas constant is approximately \(8.314\) J/(mol K).

Understanding the gas constant is crucial when performing calculations involving gases. It connects the macroscopic and microscopic worlds - relating quantities like temperature and volume at a scale we can observe directly, to the kinetic activity of molecules, which we cannot see. In entropy calculations, the gas constant allows us to quantify the entropy change associated with the expansion or compression of the gas, as it serves as a scaling factor between temperature and molecular randomness.

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Most popular questions from this chapter

According to the first law of thermodynamics, \(\Delta U=q+w .\) In special cases the statement can be expressed in different ways. Which of the following is not a correct expression? (a) At constant temperature: \(q=-\boldsymbol{w}\) (b) When no work is done: \(\Delta U=q\) (c) In gaseous system: \(\Delta U=q+P \Delta V\) (d) When work is done by the system: \(\Delta U=q+w\)

If enthalpy of an overall reaction \(X \rightarrow Y\) along one route is \(\Delta_{r} H\) and \(\Delta_{r} H_{1}, \Delta_{1} H_{2}, \Delta_{r} H_{3} \ldots\) representing enthalpies of reactions leading to same product \(y\) then \(\Delta_{r} H\) is (a) \(\Delta_{r} H=\Delta_{r} H_{1}+\Delta_{r} H_{2}+\Delta_{r} H_{3} \ldots\) (b) \(\Delta_{r} H=\Delta_{r} H_{1} \times \Delta_{r} H_{2} \times \Delta_{r} H_{3} \ldots\) (c) \(\Delta_{r} H=\Delta_{r} H_{1}+\Delta_{r} H_{2}-\Delta_{r} H_{3} \ldots\) (d) \(\Delta_{r} H=\frac{\Delta_{r} H_{1} \times \Delta_{r} H_{2} \times \Delta_{r} H_{3}}{2} \cdots\)

The molar heat capacity of water at constant pressure, \(C_{P}\) is \(75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\). When \(10 \mathrm{~kJ}\) of heat is supplied to \(1 \mathrm{~kg}\) water which is free to expand, the increase in temperature of water is (a) \(2.4 \mathrm{~K}\) (b) \(4.8 \mathrm{~K}\) (c) \(3.2 \mathrm{~K}\) (d) \(10 \mathrm{~K}\)

What will be \(\Delta H\) for the reaction, \(\mathrm{CH}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{C}+2 \mathrm{H}+2 \mathrm{Cl} ?\) \(\left(B \cdot E\right.\). of \(\mathrm{C}-\mathrm{H}\) and \(\mathrm{C}-\mathrm{Cl}\) bonds are \(416 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(325 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively) (a) \(832 \mathrm{~kJ}\) (b) \(1482 \mathrm{~kJ}\) (c) \(650 \mathrm{~kJ}\) (d) \(1855 \mathrm{~kJ}\)

For the reaction: \(\mathrm{H}_{2(g)}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{HCl} ;\) \(\Delta H=-44 \mathrm{kcal}\) What is the enthalpy of decomposition of HCl? (a) \(+44 \mathrm{kcal} / \mathrm{mol}\) (b) - \(44 \mathrm{kcal} / \mathrm{mol}\) (c) \(-22 \mathrm{kcal} / \mathrm{mol}\) (d) \(+22 \mathrm{kcal} / \mathrm{mol}\)

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