A system does \(40 \mathrm{~J}\) work on surrounding as well as gives out \(20 \mathrm{~J}\) energy. Calculate the change in internal energy.

Short Answer

Expert verified
The change in internal energy of the system is \(-60\ J\).

Step by step solution

01

Identify the Given Information

The exercise provides two key pieces of information: The system does work on the surroundings amounting to 40 J, and there is 20 J of energy given out by the system.
02

Apply the First Law of Thermodynamics

According to the first law of thermodynamics, the change in internal energy \(\Delta U\) of a system is equal to the heat added to the system \(Q\) minus the work done by the system \(W\). The formula is given by \(\Delta U = Q - W\).
03

Plug in the Known Values

In this situation, we can treat the energy given out by the system as heat that is lost, meaning \(Q = -20\ J\), and the work done by the system on the surroundings as \(W = 40\ J\).
04

Calculate the Change in Internal Energy

Substitute the known values into the equation to find \(\Delta U\).\[\Delta U = (-20\ J) - (40\ J) = -60\ J\]. The negative sign indicates that the internal energy of the system decreases.

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Key Concepts

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

Change in Internal Energy
The concept of change in internal energy is fundamental when studying thermodynamics. Imagine a system, like a gas in a piston, as appearing stationary, yet its molecules are constantly in motion. The sum of kinetic and potential energies of these molecules constitutes the system's internal energy, often symbolized by 'U'.

The internal energy can change, and these changes occur due to interactions with the environment. When energy is transferred to a system, either by heating it or by doing work on it, 'U' increases. Conversely, if the system loses energy by doing work on its surroundings or releasing heat, as stated in our exercise, 'U' decreases. The exercise clearly shows a system losing energy, emphasizing that the internal energy is not static but dynamic, changing in response to physical processes.
Thermodynamics Problems
Approaching thermodynamics problems usually involves a few critical steps: identifying known and unknown variables, applying relevant laws, and solving the equations.

Thermodynamics problems range from simple to complex, like calculating a change in internal energy, or predicting the behaviour of real-world engines. The key to solving these problems is a strong grasp of the main laws of thermodynamics, such as conservation of energy, often tested through exercises similar to the one provided.

In our example, clear identification of energy transfers—both work and heat—allows the application of the first law to solve for the change in internal energy. This structured approach is invaluable for tackling all kinds of problems in this domain.
Work and Heat in Thermodynamics
In thermodynamics, the concepts of work and heat are two primary ways energy is transferred to and from a system. Heat can be thought of as energy transferred due to temperature difference, and work as the energy transfer when a force moves an object over a distance.

In the context of our exercise, when the system does 40 J of work on the surroundings, it means energy is being expended to cause some physical change outside the system, be it lifting a weight or compressing a spring.

Heat, on the other hand, is depicted as the 20 J given out by the system, likely by cooling down. The negative sign for heat reflects that this energy is lost by the system. The interplay between work and heat is central in thermodynamic processes, and understanding their relationship is key for solving problems and analyzing energy changes within any given system.

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Most popular questions from this chapter

For the water gas reaction : $$ \mathrm{C}_{(\mathrm{s})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_{2(\mathrm{~g})} $$ the standard Gibbs energy of reaction (at \(1000 \mathrm{~K}\) ) is \(-8.1 \mathrm{~kJ} \mathrm{~mol}^{-\mathrm{i}}\). Calculate its equilibrium constant.

The enthalpy of vaporisation of liquid diethyl ether \(-\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\), is \(26.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at its boiling point \(\left(35.0^{\circ} \mathrm{C}\right)\). Calculate \(\Delta S\) for conversion of : (a) liquid to vapour, and \(\quad\) (b) vapour to liquid at \(35^{\circ} \mathrm{C}\).

$$ \begin{aligned} &\text { Calculate the value of } \Delta G \text { at } 700 \mathrm{~K} \text { for the reaction } n X \longrightarrow m B .^{-1}\\\ &\text { Given that value of } \Delta H=-113 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { and } \Delta S=-145 \mathrm{JK} \mathrm{mol}^{-1} \text { . } \end{aligned} $$

Calculate the equilibrium constant \(K_{\mathrm{p}}\) for the reaction given below if \(\Delta G^{\circ}=-10.632 \mathrm{~kJ}\) at \(300 \mathrm{~K}\) $$ \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} $$

Find out whether it is possible to reduce \(\mathrm{MgO}\) using carbon at \(298 \mathrm{~K}\). If not, at what temperature it becomes spontaneous. For reaction, \(\mathrm{MgO}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Mg}(s)+\mathrm{CO}(g), \Delta H^{\circ}=+491.18 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\Delta S^{0}=197.67 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

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