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A gas absorbs \(45 \mathrm{~kJ}\) of heat and does \(29 \mathrm{~kJ}\) of work. Calculate \(\Delta E\).

Short Answer

Expert verified
The change in internal energy, \(\Delta E = 16 \, \mathrm{kJ}\).

Step by step solution

01

Write down the given values

We are given the following values: - \(q = 45 \,\mathrm{kJ}\) (heat absorbed by the gas) - \(w = -29 \,\mathrm{kJ}\) (work done by the gas; we use a negative sign since work is done by the system)
02

Apply the first law of thermodynamics

According to the first law of thermodynamics, the change in internal energy of a system can be calculated by using the formula: \[\Delta E = q + w\] We can plug in the values of \(q\) and \(w\): \[\Delta E = 45 \,\mathrm{kJ} - 29 \,\mathrm{kJ}\]
03

Solve for \(\Delta E\)

Now, we just need to perform the subtraction: \[\Delta E = 16 \,\mathrm{kJ}\] The change in internal energy, \(\Delta E\), is 16 kJ.

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Key Concepts

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

Internal Energy
Internal energy refers to the total energy contained within a thermodynamic system. It accounts for all forms of energy present internally, like kinetic and potential energies of molecules.
This concept is crucial in understanding energy changes, as it helps us see how energy is stored in materials. According to the First Law of Thermodynamics, the change in a system's internal energy (\( \Delta E \)) is determined by the heat (\( q \)) added to the system and the work (\( w \)) done by the system.
The formula used is:
\[ \Delta E = q + w \]
For instance, if heat is absorbed by a gas and it also performs work, the system loses some energy through work, but its internal energy still increases if the absorbed heat surpasses the energy loss due to work.
Heat Absorption
Heat absorption involves the transfer of thermal energy into a system. As heat is absorbed, the molecules within a substance increase their kinetic energy, causing a rise in its internal energy.
A key thing to remember:
  • If a system absorbs heat, the process is endothermic.
  • The energy addition can increase the temperature or change the system's phase.
In the exercise, the gas absorbs 45 kJ of heat. This is represented by \( q = 45 \, \mathrm{kJ} \) in the First Law equation. Keep in mind that absorbed heat is a positive value since it adds energy to the system.
Work Done by a Gas
When discussing work in thermodynamic systems, it often relates to the concept of a gas expanding or contracting.
Work done by a gas is associated with it expanding against external pressure.
  • This is considered work done by the system on the surroundings.
  • As a result, energy leaves the system, often represented by a negative sign for work.
In the formula for change in internal energy, the work done by gas is expressed as a negative value, \( w = -29 \, \mathrm{kJ} \).
This negative sign illustrates the energy leaving the system due to work output. Hence, if a gas does work, we subtract this energy from the heat absorbed to find the change in internal energy.

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Most popular questions from this chapter

Given the following data $$ \begin{aligned} 2 \mathrm{O}_{3}(g) & \longrightarrow 3 \mathrm{O}_{2}(g) & \Delta H &=-427 \mathrm{~kJ} \\ \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{O}(g) & \Delta H &=495 \mathrm{~kJ} \\ \mathrm{NO}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-199 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{NO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}_{2}(g) $$

The enthalpy change for the reaction $$ \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is \(-891 \mathrm{~kJ}\) for the reaction as written. a. What quantity of heat is released for each mole of water formed? b. What quantity of heat is released for each mole of oxygen reacted?

Calculate \(w\) and \(\Delta E\) when 1 mole of a liquid is vaporized at its boiling point \(\left(80 .{ }^{\circ} \mathrm{C}\right)\) and \(1.00\) atm pressure. \(\Delta H_{\text {vap }}\) for the liquid is \(30.7 \mathrm{~kJ} / \mathrm{mol}\) at \(80 .{ }^{\circ} \mathrm{C}\).

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Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is \(40 . \mathrm{cm}^{3}\). If the combustion of this mixture releases 950. J of energy, to what volume will the gases expand against a constant pressure of 650 . torr if all the energy of combustion is converted into work to push back the piston?

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