(a) What sign for \(\Delta S\) do you expect when the volume of 0.200 mol of an ideal gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from an initial volume of \(10.0 \mathrm{~L} ?\) (b) If the final volume is \(18.5 \mathrm{~L},\) calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

Short Answer

Expert verified
(a) The sign of \(\Delta S\) is positive when the volume of an ideal gas increases isothermally. (b) The entropy change for the given process is approximately 1.6628 J/K. (c) No, the temperature does not need to be specified to calculate the entropy change for an isothermal process involving an ideal gas.

Step by step solution

01

In this step, we will think conceptually about the entropy change when the volume of an ideal gas increases at a constant temperature. Entropy is a measure of the randomness or disorder of a system. When the volume of ideal gas increases, the particles have more space to occupy, and thus the randomness of the system increases. Therefore, when the volume increases isothermally for an ideal gas, the entropy change (\(\Delta S\)) is positive. #Step 2: Plug the given values into the formula for entropy change in part (b)#

To calculate the entropy change for part (b), we will use the formula \(\Delta S = nR\ln\frac{V_f}{V_i}\), where: \(n\) = number of moles = 0.200 mol \(R\) = gas constant = 8.314 J/(mol·K) \(V_i\) = initial volume = 10.0 L \(V_f\) = final volume = 18.5 L Keep in mind that the volumes should be converted to m³, so we need to multiply the liters by 0.001 to get m³. #Step 3: Calculate the entropy change in part (b)#
02

Using the values from step 2, we can calculate the entropy change: \[\Delta S = (0.200 \,\text{mol}) \times (8.314 \,\text{J/(mol·K)}) \times \ln\frac{(18.5 \times 0.001 \,\text{m³})}{(10.0 \times 0.001 \,\text{m³})}\] \[\Delta S \approx 0.200 \times 8.314 \times \ln\frac{18.5}{10.0}\] \[\Delta S \approx 1.6628\,\text{J/K}\] The entropy change is positive, and its value is approximately 1.6628 J/K. #Step 4: Answer part (c) regarding the temperature requirement#

In part (c), we are asked if the temperature needs to be specified to calculate the entropy change. Looking back at the formula for the entropy change, we see that: \[\Delta S = nR\ln\frac{V_f}{V_i}\] The formula does not include the temperature directly, so we do not need to specify the temperature when calculating the entropy change for an isothermal process involving an ideal gas. However, having a constant temperature is essential for the process, as it ensures that the entropy change formula applies only to the isothermal case.

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Most popular questions from this chapter

The standard entropies at \(298 \mathrm{~K}\) for certain group 14 elements are: \(\mathrm{C}(s,\) diamond $)=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Si}(s)=18.81 \mathrm{~J} /$ $\mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \quad\( a n d \)\quad \mathrm{Sn}(s)=51.818 \mathrm{~J} /$ mol-K. All but \(S\) n have the same (diamond) structure. How do you account for the trend in the \(S^{\circ}\) values?

Which of the following processes are spontaneous and which are nonspontaneous: (a) mixing of water and ethanol, \((\mathbf{b})\) dissolution of sugar in a cup of hot coffee, (c) formation of oxygen atoms from \(\mathrm{O}_{2}\) molecules at \(\mathrm{STP}\), (d) rusting of iron, (e) formation of glucose from \(\mathrm{CO}_{2}\) and $\mathrm{H}_{2} \mathrm{O}\( at \)\mathrm{STP} ?$

Indicate whether each statement is true or false. (a) The entropy of the universe increases for any spontaneous process. (b) The entropy change of the system is equal and opposite that of the surroundings for any irreversible process. (c) The entropy of the system must increase in any spontaneous process. (d) The entropy change for an isothermal process depends on both the absolute temperature and the amount of heat reversibly transferred.

The reaction $2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)$ is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{\circ}\). Did your classmate make a mistake in the calculation? Explain.

Consider the melting of ice (solid water) to liquid water at a pressure of \(101.3 \mathrm{kPa}\). (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

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