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2 moles of an ideal gas is expanded isothermally and reversibly from 1 litre of 10 litre at \(300 \mathrm{~K}\). The enthalpy change (in \(\mathrm{kJ}\) ) for the process is: (a) \(11.4 \mathrm{~kJ}\) (b) \(-11.4 \mathrm{~kJ}\) (c) \(0 \mathrm{~kJ}\) (d) \(4.8 \mathrm{~kJ}\).

Short Answer

Expert verified
(c) \(0 \mathrm{~kJ}\)

Step by step solution

01

Identify the Process

This is an isothermal and reversible expansion of an ideal gas. Since it's isothermal, the temperature remains constant at \(T = 300 \, \text{K}\).
02

Use the Enthalpy Formula for Isothermal Process

In an isothermal process for an ideal gas, the enthalpy change is zero because enthalpy depends on temperature, which does not change. Thus, \(\Delta H = 0 \, \text{kJ}\).
03

Verify the Calculation

Since the calculation or conclusion follows directly from the characteristic property of an isothermal process in ideal gases, no further calculations are needed. We verified that the change in temperature is zero, ensuring the enthalpy change is indeed zero.

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Key Concepts

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

Isothermal Process
An isothermal process is a fascinating concept in thermodynamics, where a system undergoes a transformation at a constant temperature. In practice, this means the gas within the system remains at one temperature throughout the entire process. For an ideal gas, this implies a balance between the heat added to the system and the work done by the system. As the process occurs slowly enough for thermal equilibrium to be maintained, no net temperature change can take place.
This characteristic is crucial because it determines how energy flows in and out of the system, without altering the thermal state. Understanding this idea helps in analyzing how ideal gases behave under specific conditions, such as slowly expanding or compressing in a cylinder without a temperature change.
**Key Points:**
  • The temperature remains constant during the process.
  • Any heat entering the system is used entirely to perform work.
  • For ideal gases, no net internal energy change occurs under isothermal conditions.
Enthalpy Change
Enthalpy is a vital concept in thermodynamics representing the total energy of a system, often denoted as \(H\). It includes both the internal energy and the product of pressure and volume. In the context of isothermal processes, the enthalpy change is particularly interesting.
Because enthalpy is influenced by temperature, which remains constant during an isothermal process, the enthalpy change, \(\Delta H\), is zero. This might initially seem counterintuitive because we often associate energy changes with changes in state or composition. However, for ideal gases, it's the constancy of temperature that ensures zero change in enthalpy.
**Key Insights:**
  • Enthalpy links internal energy, pressure, and volume.
  • In isothermal processes, constant temperature means \(\Delta H = 0\).
  • This is a defining feature of ideal gases undergoing isothermal change.
Reversible Expansion
Reversible expansion occurs when a system, such as a gas, is expanded by a process that can be reversed by an infinitesimal change. This means the process is carried out so that the system always remains in near-equilibrium. For an ideal gas, reversible processes are significant because they represent a maximum work-capacity approach.
In an isothermal reversible expansion specifically, the gas does work on the surroundings while maintaining an internal equilibrium. The heat absorbed from the surroundings equals the work done by the gas, keeping the temperature unchanged. The concept of reversibility ensures that no irreversible losses occur, making it an ideal scenario for efficiency.
**Steps to Understand Reversible Expansion:**
  • The process must be carried out very slowly to maintain equilibrium.
  • Work done is maximal, as the process is fully reversible.
  • Entropy remains unchanged, contributing to efficient energy transfer.
Thermodynamics
Thermodynamics is the field of physics that deals with the relationships between heat, work, and the forms of energy involved. It provides the theoretical foundation to analyze processes such as the one described in the original exercise—a common scenario in thermodynamics, involving ideal gases and isothermal transformations.
Understanding thermodynamics involves the laws governing energy conservation, entropy, and the distribution of thermal energy. In particular, the first law of thermodynamics helps us understand why the enthalpy remains constant in an isothermal process as the total energy is conserved. Meanwhile, the concept of reversibility ties closely into the second law, which pertains to entropy and efficiency.
**Primary Thermodynamics Concepts:**
  • First Law: Energy cannot be created or destroyed, only transferred or transformed.
  • Second Law: Total entropy of an isolated system can never decrease over time.
  • Thermodynamics allows the prediction and analysis of energy flow and transformation during processes.

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Most popular questions from this chapter

Calculate the enthalpy change for the combustion of cyclopropane at \(298 \mathrm{~K}\), if the enthalpy of formation \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{H}_{2} \mathrm{O}(1)\) and propene \((\mathrm{g})\) are \(-393.5,-385.8\) and \(20.42 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The enthalpy of isomerization of cyclopropane to propene is \(-33.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (a) \(1802 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(2091 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(2196 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) None

For which of the following changes \(\Delta \mathrm{H} \neq \Delta \mathrm{E} ?\) (a) \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) (b) \(\mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) (d) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}\)

The difference between heats of reaction at constant pressure and constant volume for the reaction: \(2 \mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l})+15 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 12 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(1)\) at \(25^{\circ} \mathrm{C}\) in \(\mathrm{kJ}\) is: (a) \(-7.43\) (b) \(+3.72\) (c) \(-3.72\) (d) \(+7.43\)

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of \(27^{\circ} \mathrm{C}\). If work done during the process is \(3 \mathrm{~kJ}\), then final temperature of the gas is \(\left(\mathrm{C}_{\mathrm{v}}=20 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(150 \mathrm{~K}\) (b) \(200 \mathrm{~K}\) (c) \(175 \mathrm{~K}\) (d) \(225 \mathrm{~K}\)

For which of the following processes will the entropy increase? (a) Reaction of magnesium with oxygen to form magnesium oxide (b) Reaction of nitrogen and hydrogen to form ammonia (c) Sublimation of dry ice (d) Condensation of steam

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