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Chapter 20: Problem 35

An ideal gas initially at 300 K undergoes an isobaric expansion at \(2.50 \mathrm{kPa}\). If the volume increases from \(1.00 \mathrm{m}^{3}\) to \(3.00 \mathrm{m}^{3}\) and \(12.5 \mathrm{kJ}\) is transferred to the gas by heat, what are (a) the change in its internal energy and (b) its final temperature?

Short Answer

Expert verified
a) The change in internal energy of the gas is \( 7.50 \times 10^{3} \, J \). b) The final temperature of the gas is \( 900 \, K \).

Step by step solution

01

Calculate the Work Done

The work done by the gas during the isobaric process can be calculated using the formula \( W = pdv \), where \( p \) is the pressure and \( dv \) is the change in volume. Given that the pressure is \( p = 2.50 \, kPa = 2.50 \times 10^{3} \, Pa \) and the change in volume is \( dv = V_f - V_i = 3.00 \, m^3 - 1.00 \, m^3 = 2.00 \, m^3 \), the work done is \( W = 2.50 \times 10^{3} \, Pa \times 2.00 \, m^3 = 5.00 \times 10^{3} \, J \).
02

Calculate the Change in Internal Energy

The change in internal energy, \( 螖U \), can be calculated using the first law of thermodynamics: \( 螖U = Q - W \), where \( Q \) is the heat transferred to the system and \( W \) is the work done by the system. The question states that \( Q = 12.5 \, kJ = 12.5 \times 10^{3} \, J \). Therefore, \( 螖U = 12.5 \times 10^{3} \, J - 5.00 \times 10^{3} \, J = 7.50 \times 10^{3} \, J \).
03

Calculate the Final Temperature

In the ideal gas law, \( PV = nRT \), where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. Since the pressure is constant in this process, the relation becomes \( Vi/Ti = Vf/Tf \), where \( Vi \), \( Ti \), \( Vf \), and \( Tf \) are the initial and final volumes and temperatures respectively. Solving for the final temperature yields \( Tf = (Vf / Vi) \times Ti = (3.00 \, m^3 / 1.00 \, m^3) \times 300 \, K = 900 \, K \).

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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 relation between pressure (P), volume (V), temperature (T), and the amount of gas (n) in moles. It is commonly expressed as the equation
\( PV = nRT \)
where R represents the ideal gas constant. This law combines the empirical Boyle鈥檚 law, Charles's law, Avogadro鈥檚 law, and Gay-Lussac's law into a single comprehensive equation. Understanding this law is crucial when studying the behavior of gases under various conditions. For an isobaric expansion, as in the exercise, pressure remains constant, which simplifies the manipulation of this formula to relate initial and final states of temperature and volume.
Internal Energy Change
The internal energy change, denoted as \( \Delta U \), in a thermodynamic system is the total change in energy within a system. For an ideal gas undergoing an isobaric (constant pressure) process, the internal energy change is associated closely with changes in temperature. This is because, for an ideal gas, the internal energy is a function of temperature only. According to the exercise, the internal energy change is calculated taking into account the energy added to the system and the work done by the gas, which leads to the equation
\( \Delta U = Q - W \)
Here, Q represents the heat added to the system, and W the work done by the system.
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. The change in the internal energy of the system is equal to the heat added to the system minus the work done by the system on its surroundings. This law mathematically represented as
\( \Delta U = Q - W \)
corresponds to the principle of conservation of energy. In the context of the textbook problem, when the gas expands isobarically and heat is transferred to it, part of that energy goes into doing work on the surroundings (pushing the piston outwards), and the rest increases the internal energy of the gas, which is reflected in a change in temperature.
Work Done by Gas
The work done by a gas during a thermodynamic process can be thought of as the energy transferred when the gas expands or contracts against an external pressure. For an isobaric process, the work done by the gas can be calculated using the equation
\( W = p\Delta V \)
where p represents the constant pressure, and \( \Delta V \) is the change in volume. In the given problem, the gas does work as it expands against the constant external pressure, requiring us to integrate the infinitesimally small amounts of work done at each increment of volume, which simplifies to a straightforward multiplication due to the constant pressure.

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Most popular questions from this chapter

During periods of high activity, the Sun has more sunspots than usual. Sunspots are cooler than the rest of the luminous layer of the Sun's atmosphere (the photosphere). Paradoxically, the total power output of the active Sun is not lower than average but is the same or slightly higher than average. Work out the details of the following crude model of this phenomenon. Consider a patch of the photosphere with an area of \(5.10 \times 10^{14} \mathrm{m}^{2} .\) Its emissivity is 0.965 (a) Find the power it radiates if its temperature is uniformly \(5800 \mathrm{K},\) corresponding to the quiet Sun. (b) To represent a sunspot, assume that \(10.0 \%\) of the area is at \(4800 \mathrm{K}\) and the other \(90.0 \%\) is at \(5890 \mathrm{K}\). That is, a section with the surface area of the Earth is \(1000 \mathrm{K}\) cooler than before and a section nine times as large is 90 K warmer. Find the average temperature of the patch. (c) Find the power output of the patch. Compare it with the answer to part (a). (The next sunspot maximum is expected around the year \(2012 .\) )

Two speeding lead bullets, each of mass \(5.00 \mathrm{g},\) and at temperature \(20.0^{\circ} \mathrm{C},\) collide head-on at speeds of \(500 \mathrm{m} / \mathrm{s}\) each. Assuming a perfectly inelastic collision and no loss of energy by heat to the atmosphere, describe the final state of the two-bullet system.

A \(3.00-\mathrm{g}\) copper penny at \(25.0^{\circ} \mathrm{C}\) drops \(50.0 \mathrm{m}\) to the ground. (a) Assuming that \(60.0 \%\) of the change in potential energy of the penny-Earth system goes into increasing the internal energy of the penny, determine its final temperature. (b) What If? Does the result depend on the mass of the penny? Explain.

(a) In air at \(0^{\circ} \mathrm{C},\) a \(1.60-\mathrm{kg}\) copper block at \(0^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over a sheet of ice at \(0^{\circ} \mathrm{C} .\) Friction brings the block to rest. Find the mass of the ice that melts. To describe the process of slowing down, identify the energy input \(Q,\) the work input \(W,\) the change in internal energy \(\Delta E_{\text {int }},\) and the change in mechanical energy \(\Delta K\) for the block and also for the ice. (b) A \(1.60-\mathrm{kg}\) block of ice at \(0^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over a sheet of copper at \(0^{\circ} \mathrm{C}\) Friction brings the block to rest. Find the mass of the ice that melts. Identify \(Q, W, \Delta E_{\text {int }},\) and \(\Delta K\) for the block and for the metal sheet during the process. (c) A thin 1.60-kg slab of copper at \(20^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over an identical stationary slab at the same temperature. Friction quickly stops the motion. If no energy is lost to the environment by heat, find the change in temperature of both objects. Identify \(Q, W, \Delta E_{\text {int }},\) and \(\Delta K\) for each object during the process.

An aluminum rod \(0.500 \mathrm{m}\) in length and with a crosssectional area of \(2.50 \mathrm{cm}^{2}\) is inserted into a thermally insulated vessel containing liquid helium at \(4.20 \mathrm{K}\). The rod is initially at \(300 \mathrm{K}\). (a) If half of the rod is inserted into the helium, how many liters of helium boil off by the time the inserted half cools to \(4.20 \mathrm{K} ?\) (Assume the upper half does not yet cool.) (b) If the upper end of the rod is maintained at \(300 \mathrm{K},\) what is the approximate boil-off rate of liquid helium after the lower half has reached \(4.20 \mathrm{K} ?\) (Aluminum has thermal conductivity of \(31.0 \mathrm{J} / \mathrm{s} \cdot \mathrm{cm} \cdot \mathrm{K}\) at \(4.2 \mathrm{K} ;\) ignore its temperature variation. Aluminum has a specific heat of \(0.210 \mathrm{cal} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) and density of \(2.70 \mathrm{g} / \mathrm{cm}^{3} .\) The density of liquid helium is \(0.125 \mathrm{g} / \mathrm{cm}^{3} .\) )
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