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A quantity of air is taken from state \(a\) to state \(b\) along a path that is a straight line in the \(p V\) -diagram (Fig. \(\mathbf{P 1 9 . 3 5}\) ). (a) In this process, does the temperature of the gas increase, decrease, or stay the same? Explain. (b) If \(\quad V_{a}=0.0700 \mathrm{~m}^{3}\), \(V_{b}=0.1100 \mathrm{~m}^{3}, p_{a}=1.00 \times 10^{5} \mathrm{~Pa}\) and \(p_{b}=1.40 \times 10^{5} \mathrm{~Pa},\) what is the work \(W\) done by the gas in this process? Assume that the gas may be treated as ideal.

Short Answer

Expert verified
The temperature of the gas increases, and the work done by the gas is 3500 Joules.

Step by step solution

01

Understanding Ideal Gas Behavior

To answer part (a), it is necessary to know that for an ideal gas, if the process is such that the volume and the pressure both increase, the temperature of the gas must also increase. Since the path in the \(pV\) diagram is a straight line sloping upwards, both pressure (\(p\)) and volume (\(V\)) are increasing. Therefore, the temperature of the gas increases.
02

Determine the Work Performed by the Gas

In answering part (b), use the formula for work done in a \(pV\)-diagram that is \(W = \frac{1}{2}(p_a+p_b)(V_b-V_a)\), where \(p_a\) is the initial pressure, \(p_b\) is the final pressure, \(V_a\) is the initial volume and \(V_b\) is the final volume. Insert the given values into the formula, making sure they're all in the correct units.
03

Calculation

Perform the calculations to get the work done by the gas. \(W = \frac{1}{2}[(1.00 \times 10^{5} Pa + 1.40 \times 10^{5} Pa)(0.1100 m^{3} - 0.0700 m^{3})]\). Thus, \(W = 3500 Joules\)

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Key Concepts

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

pV diagram
A pV diagram represents the relationship between the pressure (\(p\)) and volume (\(V\)) of a gas during a thermodynamic process. These diagrams are incredibly useful for visualizing the state changes a gas goes through. Imagine a graph where pressure is on the y-axis and volume is on the x-axis, with various paths depicting different processes.
In the given problem, the path from state \(a\) to state \(b\) is a straight line sloping upwards. This indicates both pressure and volume increase. The slope or shape of the path informs us about the nature of the process. Here, an increasing trend in both parameters suggests a change in temperature, confirming a non-isothermal process. Understanding these path characteristics is critical in predicting temperature changes in gases.
thermodynamic work
Thermodynamic work involves energy transfer in a system when it undergoes a volume change under external pressure. In the context of pV diagrams, work done by or on the gas can be visualized as the area beneath the curve in the graph.
For our problem, the work done by the gas during the process from state \(a\) to state \(b\) can be calculated using the formula:\[W = \frac{1}{2}(p_a+p_b)(V_b-V_a)\]This formula is applied to straight-line or constant-path processes in a pV diagram. Here, the calculation involves using the initial and final pressures and volumes provided, allowing us to determine the work the gas performed as it expanded from state \(a\) to state \(b\). The result, \(3500\) Joules, implies the energy consumed in work due to expansion.
temperature change
When discussing temperature changes in a gas undergoing a transition, the Ideal Gas Law is the key link between pressure, volume, and temperature. For an ideal gas, the relationship is indicated by:\[PV = nRT\]where \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature.
In the exercise, as the process follows a straight-line path in the pV diagram with both parameters rising, the correlation between pressure and volume means there must also be a temperature increase. An increase in either pressure or volume under constant remaining quantities directly implies an increase in temperature. This aspect of the gas behavior underpins many foundational principles in thermodynamics.
gas laws
Gas laws are a collection of principles that govern the behavior of gases, describing how pressure, volume, and temperature interrelate. These laws include Boyle's Law, Charles's Law, and Gay-Lussac's Law, among others.
  • Boyle's Law states that pressure and volume are inversely related at a constant temperature.
  • Charles's Law indicates that volume is directly proportional to temperature at a constant pressure.
  • Gay-Lussac's Law reveals that pressure is directly proportional to temperature at constant volume.
In the context of an ideal gas, these relationships merge into the Ideal Gas Law (\(PV = nRT\)). This law enables us to predict how a gas will behave under varying conditions, which is essential for calculating work done, changes in pressure, or temperature in thermodynamic processes. By analyzing the path on a pV diagram, one can outline which specific gas laws are at play.

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Most popular questions from this chapter

Nitrogen gas in an expandable container is cooled from \(50.0^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\) with the pressure held constant at \(3.00 \times 10^{5} \mathrm{~Pa}\). The total heat liberated by the gas is \(2.50 \times 10^{4} \mathrm{~J}\). Assume that the gas may be treated as ideal. Find (a) the number of moles of gas; (b) the change in internal energy of the gas; (c) the work done by the gas. (d) How much heat would be liberated by the gas for the same temperature change if the volume were constant?

A steel cargo drum has a height of \(880 \mathrm{~mm}\) and a diameter of \(610 \mathrm{~mm}\). With its top removed it has a mass of \(17.3 \mathrm{~kg}\). The drum is turned upside down at the surface of the North Atlantic and is pulled downward into the ocean by a robotic submarine. On this day the surface temperature is \(23.0^{\circ} \mathrm{C}\) and the surface air pressure is \(p_{0}=101 \mathrm{kPa}\). The water temperature decreases linearly with depth to \(3.0^{\circ} \mathrm{C}\) at \(1000 \mathrm{~m}\) below the surface. As the drum moves downward in the ocean, the air inside the drum is compressed, reducing the upward buoyant force. (a) At what depth \(y_{\text {neutral }}\) is the barrel neutrally buoyant? (Hint: The pressure in the drum is equal to the sea pressure, which at depth \(y\) is \(p_{0}+\rho g y\) where \(\rho=1025 \mathrm{~kg} / \mathrm{m}^{3}\) is the density of seawater. The temperature at depth can be determined using the information above. Together with the ideal- gas law, you can derive a formula for the volume of air at depth \(y,\) and therefore a formula for the upward buoyant force as a function of depth.) (b) What is the volume of the air in the drum at depth \(y_{\text {neutral }} ?\)

A thermodynamic system undergoes processes in which \(|Q|=100 \mathrm{~J}\) and \(|W|=300 \mathrm{~J}\). Find \(Q\) and \(W,\) including whether the quantity is positive or negative, if the change in internal energy is (a) \(+400 \mathrm{~J},(\mathrm{~b})+200 \mathrm{~J},(\mathrm{c})-200 \mathrm{~J},\) and \((\mathrm{d})-400 \mathrm{~J} .\)

A quantity of \(2.00 \mathrm{~mol}\) of a monatomic ideal gas undergoes a compression during which the volume decreases from \(0.0800 \mathrm{~m}^{3}\) to \(0.0500 \mathrm{~m}^{3}\) while the pressure stays constant at a value of \(1.80 \times 10^{4} \mathrm{~Pa}\). (a) What is the work \(W ?\) Is work done by the gas or on the gas? (b) What is the heat flow \(Q\) ? Does heat enter or leave the gas? (c) What is the internal energy change for the gas? Does the internal energy of the gas increase or decrease?

Three moles of an ideal monatomic gas expands at a constant pressure of 2.50 atm; the volume of the gas changes from \(3.20 \times 10^{-2} \mathrm{~m}^{3}\) to \(4.50 \times 10^{-2} \mathrm{~m}^{3} .\) Calculate (a) the initial and final temperatures of the gas; (b) the amount of work the gas does in expanding; (c) the amount of heat added to the gas; (d) the change in internal energy of the gas.

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