During \(200 \mathrm{~J}\) work done on the system, \(140 \mathrm{~J}\) of heat is given out. Calculate the change in internal energy.

Short Answer

Expert verified
The change in internal energy is \(60\ J\).

Step by step solution

01

Understanding the First Law of Thermodynamics

The first law of thermodynamics states that the change in the internal energy of a system is equal to the work done on the system minus the heat lost by the system, often written as \(\Delta U = W - Q\).
02

Identifying the given quantities

Identify the work done on the system, \(W\), and the heat given out by the system, \(Q\). In this problem, work done \(W = 200\ J\) and heat given out \(Q = 140\ J\).
03

Calculating the change in internal energy

Substitute the given values into the first law of thermodynamics equation: \(\Delta U = W - Q\). So \(\Delta U = 200\ J - 140\ J = 60\ J\).

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle that serves as a bridge between physics and chemistry, and it's pivotal when studying energy changes in systems. It effectively captures the concept of energy conservation within the realm of thermodynamics, asserting that energy cannot be created nor destroyed, but rather, it can only be transferred or transformed from one form to another.

Mathematically, the law is expressed as \[\Delta U = W - Q\], where \(\Delta U\) denotes the change in internal energy of the system, \(W\) is the work done on the system, and \(Q\) is the heat exchanged with the surroundings. When a system undergoes a process, any energy supplied to it in the form of work or heat leads to a change in its internal energy.

This law is essential for solving problems related to energy within physical systems, enabling us to quantify internal energy changes as a direct result of work and heat transactions.
Work Done on the System
When we say 'work done on the system', we're referring to energy that has been transferred to the system through a force acting over a distance. In thermodynamics, this usually involves processes like compression, where external pressure forces the system into a smaller volume, thus increasing its energy.

In the language of physics, and particularly with respect to the direction of energy flow, work done on the system is considered positive. This means that the system has gained energy from its surroundings. Imagine someone pushing a piston into a cylinder of gas; the force exerted does work on the gas, which then adds up to the gas’s internal energy, assuming no heat is lost in the process.

Understanding this concept is crucial for evaluating the state functions of thermodynamic systems, especially when you want to calculate the energy involved in the various processes these systems may undergo.
Heat Exchange
Heat exchange is the transfer of thermal energy between a system and its environment due to a temperature difference. It's a natural process that seeks to bring the system and its surroundings into thermal equilibrium—the state where both have the same temperature. Heat can move in or out of the system, and in thermodynamics, we're careful to note the direction: heat exiting a system is considered negative, while heat entering is positive.

In our context, when we solve problems involving heat exchange, we quantify it in terms of joules (J) in the metric system or sometimes in calories for specific scientific applications. Just like with work, heat exchange is a pathway variable, meaning it describes the interaction between the system and its environment during a process.

A system that gives off heat to its surroundings would thus experience a decrease in internal energy, as is partially observed in our example problem, where the system releases 140 J of heat energy during the work process. To fully understand the internal energy variations, we must jointly consider both heat exchange and work done on the system.

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Most popular questions from this chapter

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

Calculate the standard free energy change for the reaction, $$ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} \text { at } 298 \mathrm{~K} $$ Given \(\Delta H^{\circ}=-92.4 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-198.3 \mathrm{JK}^{-1} .\) Also, comment on the result.

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 \(\Delta S^{0}=197.67 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Calculate the change of entropy, \(\Delta_{r} S^{\circ}\) at \(298 \mathrm{~K}\) for the reaction in which urea is formed from \(\mathrm{NH}_{3}\) and \(\mathrm{CO}_{2}\). \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) .\) The standard entropy of \(\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q), \mathrm{CO}_{2}(g), \mathrm{NH}_{3}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are \(174.0\), \(213.7,192.3\) and \(69.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively.

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.

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