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An ideal gas undergoes a reversible isothermal expansion at 132C. The entropy of the gas increases by 46.2 J/K. How much heat is absorbed?

Short Answer

Expert verified
The heat absorbed by the gas is approximately 18711 J.

Step by step solution

01

Convert Temperature to Kelvin

First, convert the given temperature from Celsius to Kelvin. The formula to do this is K = C + 273. So, T=132C+273=405K
02

Understand the Given Values

We know that the entropy change ΔS=46.2J/K. The heat absorbed or expelled in a process is given by Q=TΔS. In this case, because we’re dealing with heat absorption and the entropy increase ΔS>0, we know that Q>0.
03

Calculate the Heat Absorbed

Substitute the values of T and ΔS into the equation. This gives us Q=TΔS=405K46.2J/K=18711J.

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Key Concepts

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

Reversible Isothermal Expansion
In the realm of thermodynamics, reversible isothermal expansion is a fascinating concept that involves the expansion of an ideal gas at a constant temperature. For this process to take place isothermally, the system must be in thermal equilibrium with its surroundings, maintaining a constant temperature throughout the expansion. This means that any heat added to the system is immediately used to do work, ensuring that temperature remains unchanged. This process is termed 'reversible' because it can be reversed by infinitesimal changes in external conditions, without any net energy change in the universe. This ideal condition is important because it provides a benchmark for the most efficient possible process, where energy losses are minimized.
Entropy Change
Entropy, in simple terms, is a measure of disorder or randomness within a system. During the reversible isothermal expansion, the entropy of the system changes because the gas particles spread out and occupy a larger volume at the same temperature.In thermodynamics, the change in entropy (ΔS) is an important parameter because it helps predict the direction of thermal energy flow. For the given exercise, we see an entropy change of 46.2J/K, indicating an increase in the system's disorder. Entropy change is often associated with the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time. Hence, as the gas expands, the increase in entropy suggests that the process is spontaneous.
Ideal Gas Law
The Ideal Gas Law is an essential equation in understanding the behavior of gases under various conditions. It is expressed as PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.This equation aptly describes the relationship between these variables for an ideal gas, which is a hypothetical gas that perfectly follows the kinetic molecular theory. In our context of isothermal expansion, the temperature T remains constant, which implies that any increase in volume V must be accompanied by a corresponding decrease in pressure P to maintain the equality.The Ideal Gas Law thus provides a vital link in calculating various properties of gases when considering entropy changes or heat absorption.
Heat Absorption
In a reversible isothermal expansion, heat absorption is a key aspect because it balances the increase in internal energy needed to perform work against the surroundings at a constant temperature.To calculate the amount of heat absorbed, one can use the relationship Q=TΔS. Here, T represents the temperature in Kelvin, and ΔS denotes the change in entropy. For the provided example, substituting T=405K and ΔS=46.2J/K gives us the heat absorbed, Q=18711J. This equation highlights the direct proportionality between heat and entropy in such scenarios, resonating with the thermodynamic paradigm at work.This calculation not only illustrates the linkage between heat energy and entropy changes but also emphasizes the efficiency of an isothermal process, where energy is carefully balanced between the system and its surroundings.

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Most popular questions from this chapter

Four moles of an ideal gas are caused to expand from a volume V1 to a volume V2=3.45V1.(a) If the expansion is isothermal at the temperature T=410 K, find the work done on the expanding gas. ( b ) Find the change in entropy, if any. ( c ) If the expansion is reversibly adiabatic instead of isothermal, what is the entropy change?

A mixture of 1.78 kg of water and 262 g of ice at 0C is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass, is 1:1 at 0C.(a) Calculate the entropy change of the system during this process. (b) The system is then returned to the first equilibrium state, but in an irreversible way (by using a Bunsen burner, for instance). Calculate the entropy change of the system during this process. (c) Show that your answer is consistent with the second law of thermodynamics.

An ideal gas undergoes an isothermal expansion at 77C increasing its volume from 1.3 to 3.4 L. The entropy change of the gas is 24 J/K. How many moles of gas are present?

An inventor claims to have created a heat pump that draws heat from a lake at 3.0C and delivers heat at a rate of 20 kW to a building at 35C, while using only 1.9 kW of electrical power. How would you judge the claim?

A car engine delivers 8.18 kJ of work per cycle. (a) Before a tune-up, the efficiency is 25.0%. Calculate, per cycle, the heat absorbed from the combustion of fuel and the heat exhausted to the atmosphere. (b) After a tune-up, the efficiency is 31.0%. What are the new values of the quantities calculated in (a)?

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