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(II) An ideal gas expands isothermally \((T=410 \mathrm{~K})\) from a volume of \(2.50 \mathrm{~L}\) and a pressure of 7.5 atm to a pressure of \(1.0 \mathrm{~atm} .\) What is the entropy change for this process?

Short Answer

Expert verified
The entropy change \(\Delta S\) for this process is approximately 0.954 J/K.

Step by step solution

01

Identify the Formula for Entropy Change

Since the process is isothermal for an ideal gas, the entropy change can be calculated using the formula:\[\Delta S = nR\ln\left(\frac{V_f}{V_i}\right)\]where \(n\) is the number of moles, \(R\) is the ideal gas constant \(8.314 \, \text{J/mol} \, \text{K}\), and \(V_f\) and \(V_i\) are the final and initial volumes respectively.
02

Calculate the Initial Number of Moles

Using the ideal gas law, we can find the number of moles \(n\) of the gas. The equation is:\[PV = nRT\]Substitute \(P = 7.5 \, \text{atm} = 7.5 \, \times 101.325 \, \text{kPa}\), \(V = 2.50 \, \text{L} = 0.00250 \, \text{m}^3\) and \(T = 410 \, \text{K}\), to find \(n\).\[7.5 \, \text{atm} \times 0.00250 \, \text{m}^3 = n \times 8.314 \, \frac{\text{J}}{\text{mol} \, \text{K}} \times 410 \, \text{K}\]\[ n = \frac{7.5 \times 101.325 \times 0.00250}{8.314 \times 410}\approx 0.0568 \, \text{mol}\]
03

Calculate the Final Volume

Since during isothermal process \( PV = P_fV_f\), replace \(P_f = 1.0 \, \text{atm}\) and \(P_i = 7.5 \, \text{atm}\), and given \(V_i = 2.50 \, \text{L}\), find \(V_f\):\[ P_i V_i = P_f V_f \Rightarrow V_f = \frac{P_i V_i}{P_f} = \frac{7.5 \times 2.50}{1.0} \approx 18.75 \, \text{L}\]
04

Calculate Entropy Change

Using the formula for the entropy change derived in Step 1, substitute in the values \(n = 0.0568 \, \text{mol}\), \(R = 8.314 \, \text{J/mol} \, \text{K}\), \(V_i = 2.50 \, \text{L}\), \(V_f = 18.75 \, \text{L}\):\[\Delta S = 0.0568 \cdot 8.314 \cdot \ln\left(\frac{18.75}{2.50}\right) \]\[ \Delta S \approx 0.0568 \times 8.314 \times \ln(7.5) \approx 0.0568 \times 8.314 \times 2.015 \approx 0.954 \, \text{J/K}\]

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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 foundational principle in thermodynamics that governs the behavior of ideal gases. It is often represented by the equation:
  • \( PV = nRT \)
where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant, approximately \( 8.314 \, \text{J/mol} \, \text{K} \)
  • \( T \) is the absolute temperature in Kelvin
For the ideal gas law to apply, the gas should have low density and high temperature conditions. This law proves useful in determining unknown variables when you have some of the other variables' values. In the context of your exercise, it helps us determine the number of moles of the gas using the initial conditions of pressure and volume, with the process held at a constant temperature.
Entropy Change
Entropy is a thermodynamic quantity that measures a system's thermal energy per temperature that isn't available for doing mechanical work. In simpler terms, it measures the disorder or randomness in a system. For an isothermal expansion of an ideal gas, the change in entropy \( \Delta S \) can be calculated using:
  • \( \Delta S = nR\ln\left(\frac{V_f}{V_i}\right) \)
Here:
  • \( \Delta S \) is the change in entropy
  • \( n \) is the number of moles of the gas
  • \( R \) is the ideal gas constant
  • \( V_f \) and \( V_i \) are the final and initial volumes respectively
Entropy change gives insight into how much energy disperses in a process, which is particularly important in understanding the feasibility and directionality of processes. In your exercise, entropy change was evaluated by determining the final volume during the isothermal process, helping quantify how the disorder in the system increased.
Isothermal Process
An isothermal process is a type of thermodynamic process where the temperature remains constant (
  • \( T = \text{constant} \)
) throughout the entire process. In such processes, while temperature remains unchanged, other properties like pressure and volume can change. Thus, the product of pressure and volume remains constant for an ideal gas:
  • \( PV = \text{constant} \)
This idea comes into play in the exercise when the gas expands from a higher pressure to a lower pressure, maintaining a constant temperature. When dealing with isothermal expansions, you see a rise in volume but a drop in pressure, highlighting how the gas occupies more space without altering the thermal energy of the system. Such processes are essential when studying cycles, like those in engines, where fluctuations in pressure and volume occur without changes in heat content due to temperature.
Entropy Formula
The formula to calculate changes in entropy during processes like isothermal expansions is fundamental in thermodynamics. Entropy change is determined using the logarithmic relationship between final and initial states of the volume,
  • giving: \( \Delta S = nR\ln\left(\frac{V_f}{V_i}\right) \).
This formula indicates that entropy is not solely a function of the current state of the system but is path-dependent—particularly the path of heat energy disbursement. If the final volume is larger, the entropy increases due to higher disorder. When applying this formula, ensuring accurate knowledge of the initial and final volumes, and temperature consistency is crucial, as seen in your exercise. This reveals how energy dispersal changes with different processes, central to understanding real-world thermodynamic phenomena.

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Most popular questions from this chapter

Refrigeration units can be rated in "tons." A 1 -ton air conditioning system can remove sufficient energy to freeze 1 British ton \((2000\) pounds \(=909 \mathrm{kg})\) of \(0^{\circ} \mathrm{C}\) water into \(0^{\circ} \mathrm{C}\) ice in one 24 -h day. If, on a \(35^{\circ} \mathrm{C}\) day, the interior of a house is maintained at \(22^{\circ} \mathrm{C}\) by the continuous operation of a 5 -ton air conditioning system, how much does this cooling cost the homeowner per hour? Assume the work done by the refrigeration unit is powered by electricity that costs \(\$ 0.10\) per kWh and that the unit's coefficient of performance is 15\(\%\) that of an ideal refrigerator. \(1 \mathrm{kWh}=3.60 \times 10^{6} \mathrm{J}\)

A car engine whose output power is 155 hp operates at about \(15 \%\) efficiency. Assume the engine's water temperature of \(95^{\circ} \mathrm{C}\) is its cold-temperature (exhaust) reservoir and \(495^{\circ} \mathrm{C}\) is its thermal "intake" temperature (the temperature of the exploding gas-air mixture). \((a)\) What is the ratio of its efficiency relative to its maximum possible (Carnot) efficiency? (b) Estimate how much power (in watts) goes into moving the car, and how much heat, in joules and in kcal, is exhausted to the air in \(1.0 \mathrm{~h}\).

(II) A heat engine exhausts its heat at \(340^{\circ} \mathrm{C}\) and has a Carnot efficiency of \(38 \%\). What exhaust temperature would enable it to achieve a Carnot efficiency of \(45 \% ?\)

(II) Assume that a \(65 \mathrm{~kg}\) hiker needs \(4.0 \times 10^{3} \mathrm{kcal}\) of energy to supply a day's worth of metabolism. Estimate the maximum height the person can climb in one day, using only this amount of energy. As a rough prediction, treat the person as an isolated heat engine, operating between the internal temperature of \(37^{\circ} \mathrm{C}\left(98.6^{\circ} \mathrm{F}\right)\) and the ambient air temperature of \(20^{\circ} \mathrm{C}\).

(II) Energy may be stored for use during peak demand by pumping water to a high reservoir when demand is low and then releasing it to drive turbines when needed. Suppose water is pumped to a lake 135 \(\mathrm{m}\) above the turbines at a rate of \(1.35 \times 10^{5} \mathrm{kg} / \mathrm{s}\) for 10.0 \(\mathrm{h}\) at night. (a) How much energy \((\mathrm{kWh})\) is needed to do this each night? \((b)\) If all this energy is released during a 14 -h day, at 75\(\%\) efficiency, what is the average power output?

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