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A fixed mass of an ideal gas is heated from 50 to \(80^{\circ} \mathrm{C}\) at a constant volume of \((a) 1 \mathrm{m}^{3}\) and \((b) 3 \mathrm{m}^{3} .\) For which case do you think the energy required will be greater? Why?

Short Answer

Expert verified
Answer: Yes, the energy required is the same for both cases.

Step by step solution

01

Identify the relevant formula

We want to find the change in internal energy for an ideal gas. The formula we will use is: \(\Delta U = n C_v \Delta T\) Where: \(\Delta U\) is the change in internal energy \(n\) is the number of moles of the gas \(C_v\) is the molar heat capacity at constant volume \(\Delta T\) is the change in temperature Since we are not given the values of \(n\) and \(C_v\), we can combine them into a single constant, \(k\), for the sake of comparing the two cases: \(\Delta U = k\Delta T\)
02

Calculate the change in temperature

The change in temperature is the same for both cases: \(\Delta T = 80^{\circ} C - 50^{\circ} C = 30^{\circ} C\)
03

Compare the energy required for both cases

Since the change in temperature and the constant \(k\) are the same for both cases, the change in internal energy (\(\Delta U\)) will be the same for both cases. Therefore, the energy required to heat the gas is the same for a constant volume of 1 m³ and 3 m³. The reason for this is that, in an ideal gas, the internal energy depends only on the temperature and is independent of the volume. When we change the volume, the pressure of the gas will change in such a way that the product of pressure and volume remains constant, but the internal energy will remain the same as long as the temperature remains the same. In conclusion, the energy required to heat a fixed mass of an ideal gas from 50°C to 80°C at a constant volume of 1 m³ and 3 m³ is the same.

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Key Concepts

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

Heat Capacity at Constant Volume
The heat capacity at constant volume (\(C_v\)) is a measure of how much heat energy is required to raise the temperature of a substance by one degree Celsius, while its volume remains unchanged. It provides an intrinsic understanding of how resistant a substance is to changing its thermal state, and for ideal gases, it's especially important because it helps explain their behavior under set conditions.

When we examine ideal gases under constant volume, no work is done by the gas since volume isn't increasing—therefore, all the heat energy goes into raising the temperature of the gas. This relationship can be expressed mathematically using the formula \(\Delta U = n C_v \Delta T\), where \(\Delta U\) is the change in internal energy, \(n\) refers to the number of moles of the gas, and \(\Delta T\) is the temperature change. This formula assumes no phase change occurs and the ideal gas does not perform external work.

Understanding the specific heat capacity helps us predict how much energy will be needed to achieve a certain temperature rise, a concept vital in processes like engine design, material science, and environmental control systems.
Ideal Gas Law
The ideal gas law is a fundamental equation that describes the relationship between pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the amount of substance (\(n\)) in moles for an ideal gas. The law is usually stated as \(PV = nRT\), where \(R\) is the gas constant. This equation allows us to understand how gases will respond to changes in pressure, volume, or temperature under ideal conditions.

For an ideal gas, the internal energy is dependent only on the temperature and not on the pressure or volume. This means that even if the volume changes, as long as the temperature is constant, the internal energy stays the same. That is why, in the given exercise, changing the volume of the gas does not affect the energy required to heat the gas as long as we keep the temperature change the same. The practical use of the ideal gas law extends to many applications in real life such as predicting the behavior of gases under different conditions, which is crucial in fields ranging from meteorology to mechanical engineering.
Temperature Change in Thermodynamics
In thermodynamics, temperature change plays a central role in understanding how systems interact with their surroundings. It is a direct indicator of the thermal energy exchange that occurs when a system undergoes a process, such as heating, cooling, or changing state. The temperature change (\(\Delta T\)) in a system is related to the change in internal energy (\(\Delta U\)) through the relationship involving heat capacity.

Using the relationship \(\Delta U = n C_v \Delta T\), it is clear that temperature change directly affects the internal energy of an ideal gas. Since internal energy is a function of temperature for ideal gases, even when the volume changes, as long as the temperature change remains consistent, the change in internal energy remains unaffected. This aspect results in the same amount of energy being required to heat our ideal gas from one temperature to another, regardless of the volume, as long as heat capacity remains constant. This is a key principle that underpins many thermodynamic processes including heating systems, refrigeration, and even biological processes.

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Most popular questions from this chapter

A room contains 75 kg of air at 100 kPa and \(15^{\circ} \mathrm{C}\) The room has a 250 -W refrigerator (the refrigerator consumes \(250 \mathrm{W} \text { of electricity when running }),\) a \(120-\mathrm{W} \mathrm{TV},\) a 1.8-kW electric resistance heater, and a 50-W fan. During a cold winter day, it is observed that the refrigerator, the TV, the fan, and the electric resistance heater are running continuously but the air temperature in the room remains constant. The rate of heat loss from the room that day is \((a) 5832 \mathrm{kJ} / \mathrm{h}\) (b) \(6192 \mathrm{kJ} / \mathrm{h}\) \((c) 7560 \mathrm{kJ} / \mathrm{h}\) \((d) 7632 \mathrm{kJ} / \mathrm{h}\) \((e) 7992 \mathrm{kJ} / \mathrm{h}\)

An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ideal gas at identical pressure but at different temperatures and masses. The wall separating the two compartments is removed and the two gases are allowed to mix. Assuming constant specific heats, find the simplest expression for the mixture temperature written in the form $$T_{3}=f\left(\frac{m_{1}}{m_{3}}, \frac{m_{2}}{m_{3}}, T_{1}, T_{2}\right)$$ where \(m_{3}\) and \(T_{3}\) are the mass and temperature of the final mixture, respectively.

An electronic device dissipating \(25 \mathrm{W}\) has a mass of \(20 \mathrm{g}\) and a specific heat of \(850 \mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\). The device is lightly used, and it is on for 5 min and then off for several hours, during which it cools to the ambient temperature of \(25^{\circ} \mathrm{C}\) Determine the highest possible temperature of the device at the end of the 5 -min operating period. What would your answer be if the device were attached to a 0.5 -kg aluminum heat sink? Assume the device and the heat sink to be nearly isothermal.

Air is contained in a piston-cylinder device at \(600 \mathrm{kPa}\) and \(927^{\circ} \mathrm{C},\) and occupies a volume of \(0.8 \mathrm{m}^{3} .\) The air undergoes and isothermal (constant temperature) process until the pressure in reduced to 300 kPa. The piston is now fixed in place and not allowed to move while a heat transfer process takes place until the air reaches \(27^{\circ} \mathrm{C}\) (a) Sketch the system showing the energies crossing the boundary and the \(P\) - \(V\) diagram for the combined processes. (b) For the combined processes determine the net amount of heat transfer, in \(\mathrm{kJ},\) and its direction. Assume air has constant specific heats evaluated at \(300 \mathrm{K}\).

A passive solar house that is losing heat to the outdoors at an average rate of \(50,000 \mathrm{kJ} / \mathrm{h}\) is maintained at \(22^{\circ} \mathrm{C}\) at all times during a winter night for \(10 \mathrm{h}\). The house is to be heated by 50 glass containers each containing \(20 \mathrm{L}\) of water that is heated to \(80^{\circ} \mathrm{C}\) during the day by absorbing solar energy. A thermostat-controlled \(15-\mathrm{kW}\) back-up electric resistance heater turns on whenever necessary to keep the house at \(22^{\circ} \mathrm{C}\). \((a)\) How long did the electric heating system run that night? (b) How long would the electric heater run that night if the house incorporated no solar heating? \(\quad\)

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