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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

The engine of a Ferrari F355 F1 sports car takes in air at 20.0\(^\circ\)C and 1.00 atm and compresses it adiabatically to 0.0900 times the original volume. The air may be treated as an ideal gas with \(_\Upsilon\) = 1.40. (a) Draw a \(pV\)-diagram for this process. (b) Find the final temperature and pressure.

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 \(\times\) 10\(^5 \)Pa and occupies a volume of 2.50 \(\times\) 10\(^{-3}\) m\(^3\). (a) Find the initial temperature of the gas in kelvins. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.

Three moles of argon gas (assumed to be an ideal gas) originally at 1.50 \(\times\) 10\(^4\) Pa and a volume of 0.0280 m\(^3\) are first heated and expanded at constant pressure to a volume of 0.0435 m\(^3\), then heated at constant volume until the pressure reaches 3.50 \(\times\) 10\(^4\) Pa, then cooled and compressed at constant pressure until the volume is again 0.0280 m\(^3\), and finally cooled at constant volume until the pressure drops to its original value of 1.50 \(\times\) 10\(^4\) Pa. (a) Draw the \(pV\)-diagram for this cycle. (b) Calculate the total work done by (or on) the gas during the cycle. (c) Calculate the net heat exchanged with the surroundings. Does the gas gain or lose heat overall?

On a warm summer day, a large mass of air (atmospheric pressure 1.01 \(\times\) 10\(^5\) Pa) is heated by the ground to 26.0\(^\circ\)C and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why?) Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 0.850 \(\times\) 10\(^5\) Pa. Assume that air is an ideal gas, with \(\Upsilon\) = 1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C\(^\circ\) per 100 m of altitude, is called the dry \(adiabatic\) \(lapse\) \(rate\).)

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?

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