Calculate the work done by an external agent in compressing \(1.12 \mathrm{~mol}\) of oxygen from a volume of \(22.4 \mathrm{~L}\) and \(1.32\) atm pressure to \(15.3 \mathrm{~L}\) at the same temperature.

Short Answer

Expert verified
The work done by an external agent in compressing the given quantity of oxygen is approximately 952.31 J.

Step by step solution

01

Identify Given Parameters

Identify the values of the initial volume \(V_{i}= 22.4 L\), final volume \(V_{f}= 15.3 L\), and pressure \(P = 1.32 atm\).
02

Conversion of Pressure Units

Always remember that in Physics, we use the SI unit system. Thus, convert pressure from atmospheres to pascals as: 1 atm = 101325 Pa. So, \( P = 1.32 atm = 1.32 * 101325 Pa = 134269.8 Pa\).
03

Volume Difference Calculation

Calculate the change in volume \(\Delta V = V_{f} - V_{i} = 15.3 L - 22.4 L = -7.1 L\). Convert volume from liters to cubic meters (1L = 0.001 m^3), therefore, \(\Delta V = -7.1 L = -0.0071 m^3\).
04

Work Done Calculation

Substitute all the values into the formula for work \( W = -P \Delta V\), yielding \( W = -134269.8 Pa * -0.0071 m^3 = 952.31 Joules\).

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Key Concepts

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

Thermodynamics
Thermodynamics is the science that deals with heat and temperature, and their relation to energy, work, radiation, and properties of matter. One of the key principles in thermodynamics is the concept of work done in processes which involve gas expansions or compressions.

When a gas undergoes a change in volume, work is done by or on the system. If the system does work on its surroundings, such as when a gas expands against an external pressure, that work is considered positive. Conversely, when work is done on the system, like when an external agent is used to compress a gas, the work is considered negative. This is crucial in understanding the nature of processes in thermodynamics, as it pertains to the conservation of energy and the First Law of Thermodynamics, which states that the total increase in the energy of a system is equal to the sum of the work done on it and the heat supplied to it.
Pressure-Volume Work
In the context of thermodynamics, pressure-volume work, or PV work, refers to the work done upon a system that results in a change of the system's volume.

This concept is frequently illustrated using gases in a piston where the work can be visually represented by the displacement of the piston as the gas expands or is compressed. The equation used to calculate this work is: \[ W = - P \Delta V \] where \(W\) represents the work done on the system (in joules), \(P\) is the constant external pressure (in pascals), and \(\Delta V\) is the change in volume (in cubic meters). The negative sign signifies that work positive when the gas does work on the surroundings (expanding) and negative when work is done on the gas (compressing).

In our exercise, we calculate the work done during the compression of a gas by an external agent, which is a practical example of pressure-volume work.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation that models the behavior of an ideal gas. It is typically stated as: \[ PV = nRT \], where \(P\) is the pressure of the gas, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the absolute temperature.

This law combines several simpler gas laws, like Boyle's Law, Charles's Law, and Avogadro's Law, into a single relationship. For instance, it implies that for a fixed amount of an ideal gas at constant temperature, the product of pressure and volume is constant, following Boyle's Law. It is important when considering the work done on or by a gas because the initial and final states of the gas can often be described using the ideal gas law, particularly when the temperature of the gas is held constant, as in the isothermal processes.

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Most popular questions from this chapter

A container holds a mixture of three nonreacting gases: \(n_{1}\) moles of the first gas with molar specific heat at constant volume \(C_{1}\), and so on. Find the molar specific heat at constant volume of the mixture, in terms of the molar specific heats and quantities of the three separate gases.

A quantity of ideal gas occupies an initial volume \(V_{0}\) at a pressure \(p_{0}\) and a temperature \(T_{0} .\) It expands to volume \(V_{1}(a)\) at constant pressure, \((b)\) at constant temperature, and \((c)\) adiabatically. Graph each case on a \(p V\) diagram. In which case is \(Q\) greatest? Least? In which case is \(W\) greatest? Least? In which case is \(\Delta E_{\text {int }}\) greatest? Least?

Consider that \(214 \mathrm{~J}\) of work are done on a system, and \(293 \mathrm{~J}\) of heat are extracted from the system. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of \((a) W,(b) Q\), and \((c) \Delta E_{\text {int }} ?\)

(a) Two \(50-\mathrm{g}\) ice cubes are dropped into \(200 \mathrm{~g}\) of water in a glass. If the water were initially at a temperature of \(25^{\circ} \mathrm{C}\), and if the ice came directly from a freezer at \(-15^{\circ} \mathrm{C}\), what is the final temperature of the drink? ( \(b\) ) If only one ice cube had been used in (a), what would be the final temperature of the drink? Neglect the heat capacity of the glass.

A certain substance has a molar mass of \(51.4 \mathrm{~g} / \mathrm{mol}\). When \(320 \mathrm{~J}\) of heat are added to a 37.1-g sample of this material, its temperature rises from \(26.1\) to \(42.0^{\circ} \mathrm{C}\). \((a)\) Find the specific heat of the substance. (b) How many moles of the substance are present? (c) Calculate the molar heat capacity of the substance.

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