/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 A \(10.0\) -L gas tank containin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(10.0\) -L gas tank containing 3.20 moles of ideal He gas at \(20.0^{\circ} \mathrm{C}\) is placed inside a completely evacuated, insulated bell jar of volume 35.0 L. A small hole in the tank allows the He to leak out into the jar until the gas reaches a final equilibrium state with no more leakage. (a) What is the change in entropy of this system due to the leaking of the gas? (b) Is the process reversible or irreversible? How do you know?

Short Answer

Expert verified
(a) The change in entropy is approximately 28.2 J/K. (b) The process is irreversible as it involves free expansion.

Step by step solution

01

Calculate Initial Conditions

First, we need to determine the initial conditions of the system. The initial volume of the gas (V_initial) is 10.0 L, and the number of moles (n) is 3.20 moles. The initial temperature (T_initial) is 20.0°C, which is converted to Kelvin as \( T_{initial} = 20.0 + 273.15 = 293.15\, \text{K} \). The ideal gas constant \( R \) is 8.314 J/(mol·K).
02

Calculate Final Conditions

The final volume of the gas will be the combined volume of the tank and the bell jar, so \( V_{final} = 10.0 + 35.0 = 45.0 \text{ L} \). Since it's an insulated system and no work is done, the temperature remains constant at 293.15 K.
03

Determine Change in Entropy

For an isothermal expansion of an ideal gas, the change in entropy \( \Delta S \) can be calculated using the formula: \[ \Delta S = nR \ln \left( \frac{V_{final}}{V_{initial}} \right) \]. Substituting in the values, we get: \( \Delta S = 3.20 \times 8.314 \times \ln \left( \frac{45.0}{10.0} \right) \). Calculating this gives \( \Delta S = 3.20 \times 8.314 \times \ln(4.5) \approx 28.2 \text{ J/K} \).
04

Assess Reversibility

The process is irreversible. We know this because the free expansion of a gas into a vacuum, like in this scenario where no work is done and external constraints are not applied, is a hallmark of an irreversible process.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 principle in chemistry and physics that describes the behavior of an ideal gas. This law is expressed by the formula:
  • \( PV = nRT \)
Where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume of the gas
  • \( n \) is the number of moles of gas
  • \( R \) is the ideal gas constant, approximately 8.314 J/(mol·K)
  • \( T \) is the absolute temperature in Kelvin
This law assumes that the gas particles are small and that they do not interact with each other. For many gases under standard conditions, this approximation works very well.
In the example given, you can see how initial conditions, such as temperature, volume, and moles, are vital. They help us determine changes in the gas behavior when conditions change, like in processes where volume changes but temperature remains the same.
Isothermal Process
An isothermal process is a change in a system where the temperature remains constant. This means that, while the system might expand or contract, the thermal energy is regulated so that the temperature doesn't change.
In mathematical terms, during an isothermal process, we can use the formula:
  • \( Q = W \)
Here:
  • \( Q \) is the heat added to the system
  • \( W \) is the work done by the system
For an ideal gas undergoing an isothermal expansion, the change in internal energy \( \Delta U \) is zero. Because \( T \) is constant, any work done is balanced by the heat added or removed.
In the provided exercise, you calculate the entropy change using the formula for an isothermal process. The fact that temperature remains constant, even as volume changes, is crucial in determining how entropy changes without a change in internal energy.
Irreversible Process
An irreversible process, as opposed to a reversible one, is a process where the system cannot return to its initial state without leaving changes in its surroundings. There is an increase in entropy and energy loss is inevitable.
Key indicators of irreversibility include:
  • Free expansion
  • Mixing of different gases
  • Uncontrolled chemical reactions
These processes are not at equilibrium at all points; they happen spontaneously and cannot be reversed step-by-step without external intervention.
In the given problem, the expansion of helium gas from the tank into the jar is deemed irreversible. This is because it is a spontaneous process into a vacuum where balance with the surroundings isn’t maintained. Hence, the process results in an increase in entropy.
Free Expansion
Free expansion refers to the process where a gas expands into an available space without performing external work and without the addition or removal of heat. This is also known as Joule expansion.
This process is characterized by:
  • Sudden and rapid expansion
  • Unresisted movement of gas particles
  • No energy exchange with the surroundings through work
During free expansion, ideal gases show no change in internal energy nor temperature, as the expansion involves no work nor heat exchange. This is different from other processes where heat or work could alter these conditions.
In the task presented, when helium gas leaks into the evacuated jar, it undergoes free expansion. The entropy change is calculated because the gas spreads into a larger volume, although temperature is held constant, signifying an isothermal free expansion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You make tea with 0.250 \(\mathrm{kg}\) of \(85.0^{\circ} \mathrm{C}\) water and let it cool to room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) before drinking it. (a) Calculate the entropy change of the water while it cools. (b) The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cols, assuming that all the heat lost by the water goes into the air. What is the total entropy change of the system tea \(+\) air?

An experimental power plant at the Natural Energy Laboratory of Hawaii generates electricity from the temperature gradient of the ocean. The surface and deep-water temperatures are \(27^{\circ} \mathrm{C}\) and \(6^{\circ} \mathrm{C},\) respectively. (a) What is the maximum theoretical efficiency of this power plant? (b) If the power plant is to produce 210 \(\mathrm{kW}\) of power, at what rate must he extracted from the warm water? At what rate must heat be absorbed by the cold water? Assume the maximum theoretical efficiency. (c) The cold water that enters the plant leaves it at a temperature of \(10^{\circ} \mathrm{C}\) . What must be the flow rate of cold water through the system? Give your answer in \(\mathrm{kg} / \mathrm{h}\) and in \(\mathrm{L} / \mathrm{h}\) .

You are designing a Carnot engine that has 2 \(\mathrm{mol}\) of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 atm. With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work. (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 -kg block of ice originally at \(0.0^{\circ} \mathrm{C},\) using only the heat rejected by the engine?

A Gasoline Engine. A gasoline engine takes in \(1.61 \times\) \(10^{4} \mathrm{J}\) of heat and delivers 3700 \(\mathrm{J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{J} / \mathrm{g} .(\mathrm{a})\) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 \(\mathrm{K}\) and 270 \(\mathrm{K}\) . (a) If in each cycle the refrigerator receives 415 \(\mathrm{J}\) of heat energy from the reservoir at 270 \(\mathrm{K}\) , how many joules of heat energy does it deliver to the reservoir at 320 \(\mathrm{K} ?\) (b) If the refrigerator completes 165 cycles each minute, what power input is required to operate it? (c) What is the coefficient of performance of the refrigerator?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.