/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 A cylinder contains \(0.250 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A cylinder contains \(0.250 \mathrm{~mol}\) of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\). The cylinder is provided with a friction less piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\). Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) -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
To summarize, once you have calculated the initial volume and final volume of the gas, you can draw the pV-diagram. You can then calculate the work done by the gas when its temperature raises from 27.0° C to 127.0° C under a constant pressure. This work is done on the surroundings. Then, calculate the change in internal energy and the amount of heat transferred to the gas. Redo the calculations of work for the lower pressure scenario.

Step by step solution

01

Calculate the Volume of the Gas Using Ideal Gas Law

First, apply the ideal gas law \( P V = n R T \) to find the volume of the gas initially. Here, \( P = 1.00 atm \), \( n = 0.250 mol \), \( R = 0.0821 L atm/(K mol) \) - the gas constant for these units, and \( T = 273.15 K + 27.0 °C = 300.15 K \). This gives us \( V = n R T / P \).
02

Draw the pV-Diagram

On the \( pV \)- diagram, the process of heating the gas at constant pressure is represented by a horizontal line whose initial coordinate is calculated in the previous step.
03

Calculate the Final Volume of the Gas

After heating the gas, the temperature rises to \( 273.15 K + 127.0°C = 400.15 K \). Using the ideal gas law again, we calculate the final volume, \( V' \), by the equation \( V' = n R T' / P \).
04

Calculate the Work Done by the Gas

The work done by the gas is given by the equation \( W = P(V'- V) \).
05

Identify Where the Work is Done

According to the laws of thermodynamics, the work is done on the surroundings. In this case, it is used to move the frictionless piston.
06

Calculate the Change in Internal Energy

The change in internal energy of the gas can be calculated using the equation ∆U=nCv ΔT where Cv is the molar heat capacity at constant volume and is equal to 3/2R for monoatomic gasses. Here, \( delta U = n C_v delta T \), where \( delta T = T' - T \).
07

Calculate Heat Transferred to the Gas

We can use the first law of thermodynamics to find the amount of heat transferred to the gas. It can be expressed as, ΔU=Q−W. Here, Q is the heat transferred to the gas. From the known values of ΔU and W, we can calculate Q.
08

Redo the Work Calculation at Lower Pressure

Given the second scenario where the pressure is 0.50 atm, repeat the steps for calculations of work using the formula \( W = P(V'- V) \).

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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 is a graphical representation of the changes in pressure (P) and volume (V) of a gas. For an isobaric process, like the one described in the exercise where carbon dioxide gas is heated at constant pressure, the diagram shows a horizontal line. This line represents the pressure remaining the same while the volume changes with temperature. The area under this line represents the work done by the gas as it expands.
Work Done by Gas
The work done by gas during a process can be visualized on a PV-diagram and calculated with the formula W = PΔV, where W is the work, P is the pressure, and ΔV is the change in volume. In the exercise, since the pressure is constant and the volume increases with heating, the work is positive, indicating that the gas does work on the surroundings by pushing the piston outward.
First Law of Thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, states that the energy of an isolated system is constant. Energy can be transformed from one form to another but cannot be created or destroyed. For the case of an ideal gas undergoing a thermodynamic process, the law is mathematically expressed as ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
Change in Internal Energy
The change in internal energy of a gas, denoted as ΔU, is related to the heat added to the system and the work done by it. For an ideal gas, this change can be calculated using the formula ΔU = nCVΔT, where n is the number of moles, CV is the molar heat capacity at constant volume, and ΔT is the change in temperature. It's important to note that internal energy is a function of temperature only for an ideal gas, and as temperature increases, so does internal energy.
Heat Transfer
In thermodynamics, heat transfer is the process by which heat energy is exchanged between a system and its surroundings. It can occur by conduction, convection, or radiation. In the context of the exercise, heat is transferred to the gas as it is heated, resulting in an increase in temperature and volume. According to the first law of thermodynamics, this heat transfer into the system can be quantified by the formula: Q = ΔU + W, directly linking heat to changes in internal energy and work done.
Molar Heat Capacity
Molar heat capacity is a property that indicates the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius at constant volume (CV) or pressure (CP). For an ideal gas, the molar heat capacity at constant volume, CV, can be used in combination with the change in temperature to calculate the change in internal energy. The specific heat capacities are different for monatomic, diatomic, and polyatomic gases due to their varying degrees of freedom in motion.

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Most popular questions from this chapter

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 cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at \(1.00 \times 10^{5} \mathrm{~Pa}\) and \(300 \mathrm{~K}\) and occupies a volume of \(1.50 \mathrm{~L}\). 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 \(300 \mathrm{~K}\). This continues until the pressure reaches \(2.50 \times 10^{4} \mathrm{~Pa}\). In the second process, the gas is compressed at constant pressure until it returns to its original volume of \(1.50 \mathrm{~L}\). Assume that the gas may be treated as ideal. (a) In a \(p V\) -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?

Helium gas undergoes an adiabatic process in which the Kelvin temperature doubles. By what factor does the pressure change?

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?

19.52 - A certain ideal gas has molar heat capacity at constant volume \(C_{V}\). A sample of this gas initially occupies a volume \(V_{0}\) at pressure \(p_{0}\) and absolute temperature \(T_{0}\). The gas expands isobarically to a volume \(2 V_{0}\) and then expands further adiabatically to a final volume \(4 V_{0}\). (a) Draw a \(p V\) -diagram for this sequence of processes. (b) Compute the total work done by the gas for this sequence of processes. (c) Find the final temperature of the gas. (d) Find the absolute value of the total heat flow \(|Q|\) into or out of the gas for this sequence of processes, and state the direction of heat flow.

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