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A 2.00 -mol sample of helium gas initially at \(300 \mathrm{K}\) and 0.400 atm is compressed isothermally to 1.20 atm. Noting that the helium behaves as an ideal gas, find (a) the final volume of the gas, (b) the work done on the gas, and (c) the energy transferred by heat.

Short Answer

Expert verified
The final volume of the gas is found using the ideal gas law from the given initial conditions and the final pressure. The work done on the gas is calculated using the formula for work in an isothermal process, while the energy transferred by heat is found from the first law of thermodynamics.

Step by step solution

01

Calculating the Initial Volume

By rearranging the ideal gas equation \(PV=nRT\), we can find the initial volume \(V_{i}\). Given initial pressure \(P_{i}=0.400\) atm (convert it to Pascal, so multiply by \(1.013 \times 10^{5}\)), number of moles n=2.00 mol, temperature \(T_{i}=300K\), and the gas constant \(R=8.314\, J/(mol \cdot K)\), we can substitute all these values into the formula to get \(V_{i} = \frac{nRT_{i}}{P_{i}} = \frac{2.00 \times 8.314 \times 300}{0.400 \times 1.013 \times 10^{5}}\).
02

Calculating the Final Volume

Using the fact that the process is isothermal, meaning \(T_{i} = T_{f} = T\), we use the ideal gas law again to solve for the final volume \(V_{f}\) with the final pressure \(P_{f}=1.20\) atm (convert to pascal). Thus, \(V_{f} = \frac{nRT}{P_{f}} = \frac{2.00 \times 8.314 \times 300}{1.20 \times 1.013 \times 10^{5}}\).
03

Calculating the Work done on the Gas

The work done on the gas during an isothermal process can be found by using the formula \(W = nRT \ln \frac{V_{i}}{V_{f}}\). Note, we use the natural logarithm \(\ln\) here. Substitute the known values in to calculate work \(W\).
04

Calculating the heat energy

For an ideal gas in an isothermal process, the change in internal energy \(\Delta U\) is zero. So, using the first law of thermodynamics \(\Delta U = Q - W\), we solve for Q (energy transferred by heat) to find \(Q = \Delta U + W = 0 + W = W\). In this case, the heat energy transferred is equal to the work done on the gas.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation that describes the behavior of an ideal gas. This equation is represented as \( PV = nRT \), where \( P \) is the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature in kelvins. In simple terms, the law connects the pressure, volume, and temperature of a fixed quantity of gas, assuming there are no interactions between the gas particles and they occupy no space. This idealization makes it easier to predict the behavior of gases under different conditions.

When solving problems like the one at hand, we use the ideal gas law to calculate unknown properties of a gas when the other properties are provided. As seen in the problem, understanding how to manipulate the formula to solve for the desired variable is key to finding solutions in exercises involving changes in pressure, volume or temperature of an ideal gas.
Isothermal Compression
Isothermal compression refers to the process of reducing the volume of a gas while maintaining its temperature constant. As the ‘iso’ prefix suggests, 'the same', isothermal processes occur at a consistent temperature. This is in contrast to adiabatic processes, where no heat is exchanged with the environment.

In isothermal compression of an ideal gas, since the temperature remains constant, according to the ideal gas law \( PV = nRT \), the product of pressure \( P \) and volume \( V \) remains constant throughout. If the volume decreases, the pressure must increase proportionally, and vice versa. This behavior is encapsulated in Boyle's Law, which is a special case of the ideal gas law for isothermal conditions. This concept aids in understanding the thermodynamic processes and solving problems such as calculating final volumes after a compression, as exemplified in the exercise.
Work Done by Gas
When we talk about work done by or on a gas in thermodynamics, we refer to the energy transferred by the gas as it expands or by the surroundings as the gas is compressed. The formula to calculate the work done during an isothermal process is \( W = -nRT \ln\left(\frac{V_{f}}{V_{i}}\right) \), where \( n \) is the number of moles, \( R \) is the universal gas constant, \( T \) is the temperature, \( V_{i} \) is the initial volume, and \( V_{f} \) is the final volume.

The minus sign indicates that work is done on the gas when \( V_{f} < V_{i} \), as in compression, and done by the gas during expansion \( V_{f} > V_{i} \). For instance, in the given problem, the work was calculated based on the natural logarithm of the ratio of the initial and final volumes. This calculation determines how much energy is needed to compress the gas at a constant temperature.
First Law of Thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. Instead, it can only be transformed from one form to another. Mathematically, it is expressed as \( \Delta U = Q - W \), where \( \Delta U \) is the change in the internal energy of the system, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.

In an isothermal process for an ideal gas, the internal energy is only a function of temperature. Therefore, if the temperature remains constant, there is no change in internal energy (\( \Delta U = 0 \)). Consequently, any work done on the gas (such as during compression) must be balanced by an equal amount of heat transferred out of the gas, or vice versa. This principle enabled the solution of part (c) in the given exercise, illustrating the direct relationship between heat transfer and work in thermodynamic processes.

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Most popular questions from this chapter

Systematic use of solar energy can yield a large saving in the cost of winter space heating for a typical house in the north central United States. If the house has good insulation, you may model it as losing energy by heat steadily at the rate \(6000 \mathrm{W}\) on a day in April when the average exterior temperature is \(4^{\circ} \mathrm{C},\) and when the conventional heating system is not used at all. The passive solar energy collector can consist simply of very large windows in a room facing south. Sunlight shining in during the daytime is absorbed by the floor, interior walls, and objects in the room, raising their temperature to \(38^{\circ} \mathrm{C} .\) As the sun goes down, insulating draperies or shutters are closed over the windows. During the period between 5: 00 P.M. and 7: 00 A.M. the temperature of the house will drop, and a sufficiently large "thermal mass" is required to keep it from dropping too far. The thermal mass can be a large quantity of stone (with specific heat \(850 \mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) ) in the floor and the interior walls exposed to sunlight. What mass of stone is required if the temperature is not to drop below \(18^{\circ} \mathrm{C}\) overnight?

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