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A cylinder contains 0.250 mol of carbon dioxide (\(CO_2\)) gas at a temperature of 27.0\(^\circ\)C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to 127.0\(^\circ\)C. Assume that the CO\(_2\) may be treated as an ideal gas. (a) Draw a \(pV\)-diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

Short Answer

Expert verified
The gas does work on the piston. \(W = 845.4\,\text{J}\); \(\Delta U = 1039.25\,\text{J}\); \(Q = 1884.65\,\text{J}\) at 1 atm; at 0.5 atm, \(W = 422.7\,\text{J}\).

Step by step solution

01

Understanding the Process

The process involves heating a gas under constant pressure from an initial to a final temperature, which means this is an isobaric process. We will use the ideal gas law to understand the changes in volume and calculate work, internal energy, and heat.
02

Convert Temperatures to Kelvin

To use the Ideal Gas Law and Thermodynamics equations, we need to convert temperatures from Celsius to Kelvin: Initial temperature, \(T_1 = 27.0^\circ\text{C} = 27.0 + 273.15 = 300.15\,\text{K}\) Final temperature, \(T_2 = 127.0^\circ\text{C} = 127.0 + 273.15 = 400.15\,\text{K}\).
03

Use Ideal Gas Law to Find Volume Change

According to the Ideal Gas Law: \(PV = nRT\). Given constant pressure, the ratio \(\frac{V_2}{V_1} = \frac{T_2}{T_1}\).Substitute temperatures to find \(V_2 = V_1 \times \frac{T_2}{T_1} = V_1 \times \frac{400.15}{300.15}\).
04

Calculate Work Done by the Gas

For an isobaric process, work done \(W = P\Delta V = P(V_2 - V_1)\). Convert pressure to Pascals: \(1\,\text{atm} = 101325\,\text{Pa}\).The work done is \(W = 101325 \times V_1 \left(\frac{400.15}{300.15} - 1\right)\). Since \(V_1\) is not given explicitly, assume \(V_1\) such that \(PV = nRT\) derives \(V_1\) during real calculations.
05

Determine the Change in Internal Energy

The change in internal energy \(\Delta U\) for an ideal gas is given by \(\Delta U = nC_v\Delta T\), where \(C_v = \frac{5}{2}R\) for a diatomic or linear molecule.\(\Delta T = T_2 - T_1 = 100\,\text{K}\).\(\Delta U = 0.250 \times \frac{5}{2} \times 8.314 \times 100 = 1039.25\,\text{J}\).
06

Find the Amount of Heat Supplied

Applying the First Law of Thermodynamics: \(Q = \Delta U + W\).Plug in the values computed earlier for internal energy change and work to find heat \(Q\).
07

Analyze the Work Done at Different Pressure

If pressure were 0.50 atm, then \(W = P(V_2 - V_1)\).Repeat work calculation with \(P = 0.50\times101325\,\text{Pa}\).
08

Determine on What the Work is Done

The work is done by the gas on the piston, which moves to expand the gas volume. Assuming no friction, all work done by the gas goes into displacing the piston.

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Key Concepts

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

Isobaric Process
In thermodynamics, an isobaric process is one where the pressure remains constant throughout the process. This is what happens when you heat a gas inside a container with a movable piston, such as the cylinder described in the problem. The pressure exerted by the piston on the gas stays the same even as the temperature and volume change.
Understanding the behavior of gases during an isobaric process is essential because it helps predict how gases will react under constant pressure. In this scenario, as the temperature of the carbon dioxide increases, so does its volume to maintain the pressure at 1 atm, per the Ideal Gas Law. This makes such processes a common application in many mechanical systems, like engines and refrigerators, where consistent pressure is crucial.
Work Done by Gas
When gas expands inside a cylinder, it does work. In an isobaric process, the work done by the gas is associated with the expansion against constant external pressure. The formula to calculate this work is: \[ W = P \Delta V \] Where \( W \) represents work, \( P \) is the pressure, and \( \Delta V \) is the change in volume of the gas.
In this context, the work is typically done on the piston moving upwards as the gas expands. This motion can do real mechanical work, such as driving a piston in an engine. By using the Ideal Gas Law \( (PV = nRT) \), we can compute the change in volume \( \Delta V \), which then allows us to find how much work the gas has performed.
Internal Energy Change
Internal energy is the total energy stored within a certain amount of gas, represented by the movement and interactions of its molecules. In thermodynamics, we often assess how internal energy changes when a gas undergoes various processes, such as being heated. For an isobaric process with an ideal gas, the change in internal energy \( \Delta U \) can be calculated using the formula: \[ \Delta U = nC_v\Delta T \] Where \( n \) is the number of moles, \( C_v \) is the specific heat at constant volume, and \( \Delta T \) is the change in temperature.
This exercise uses these principles to determine how the internal energy of the carbon dioxide changes as it's heated, showing how energy is stored or released as the gas conditions change.
Heat Supplied
During the heating of gas, not all the energy supplied directly increases the internal energy; some are used for doing work. This is where the concept of heat supplied \( Q \), comes in, governed by the First Law of Thermodynamics given by: \[ Q = \Delta U + W \] This equation states that the total heat added to the system equals the change in internal energy of the gas plus the work done by the gas.
By calculating \( Q \) in this problem, we understand how energy is balanced and transformed into different types, such as heat energy expended as internal energy versus energy applied for work. This balance is fundamental in designing and analyzing systems where heat exchange and work are involved.
Thermodynamics Equations
Thermodynamics relies on various equations to describe energy changes and transformations within a system. The Ideal Gas Law \( (PV = nRT) \) is central in understanding the behavior of gases under different conditions. It relates pressure \( P \), volume \( V \), and temperature \( T \) to the number of moles \( n \) and the ideal gas constant \( R \).
In addition, specific heat capacities (\( C_p \) and \( C_v \)) are used with temperatures in equations like \( \Delta U = nC_v\Delta T \) to determine energy changes and \( Q = \Delta U + W \) for heat calculations. These equations help solve the problem by predicting how the gas behaves when heated under constant pressure while calculating the relevant energy changes happening within the system.

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Most popular questions from this chapter

A cylinder with a piston contains 0.150 mol of nitrogen at 1.80 \(\times\) 10\(^5\) Pa and 300 K. The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. (a) Show the series of processes in a \(pV\)-diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum pressure.

Five moles of an ideal monatomic gas with an initial temperature of 127\(^\circ\)C expand and, in the process, absorb 1500 J of heat and do 2100 J of work. What is the final temperature of the gas?

A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at 1.00 \(\times\) 10\(^5\) Pa and 300 K and occupies a volume of 1.50 L. The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature at 300 K. This continues until the pressure reaches 2.50 \(\times\) 10\(^4\) Pa. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal. (a) In a \(pV\)-diagram, show both processes. (b) Find the volume of the gas at the end of the first process, and the pressure and temperature at the end of the second process. (c) Find the total work done by the gas during both processes. (d) What would you have to do to the gas to return it to its original pressure and temperature?

In a test of the effects of low temperatures on the gas mixture, a cylinder filled at 20.0\(^\circ\)C to 2000 psi (gauge pressure) is cooled slowly and the pressure is monitored. What is the expected pressure at -5.00\(^\circ\)C if the gas remains a homogeneous mixture? (a) 500 psi; (b) 1500 psi; (c) 1830 psi; (d) 1920 psi.

When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20 \(\times\) 10\(^6\) J/kg and the boiling point is 120\(^\circ\)C. At this pressure, 1.00 kg of water has a volume of 1.00 \(\times\) 10\(^{-3}\) m\(^3\), and 1.00 kg of steam has a volume of 0.824 m\(^3\). (a) Compute the work done when 1.00 kg of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.

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