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Chapter 3: Problem 106

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at \(27^{\circ} \mathrm{C}\) and a gage pressure of \(300 \mathrm{kPa}\). The gas is heated, and the gage pressure at the final state is \(367 \mathrm{kPa}\). Determine the final temperature, in \({ }^{\circ} \mathrm{C}\). The local atmospheric pressure is \(1 \mathrm{~atm}\).

Short Answer

Expert verified
The final temperature is \(77.3^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the given data

Initial temperature, \(T_1\), is \(27^{\,circ} \mathrm{C}\). Initial gage pressure, \(P_{g1}\), is \(300 \,\mathrm{kPa}\). Final gage pressure, \(P_{g2}\), is \(367 \,\mathrm{kPa}\). Atmospheric pressure, \(P_{atm}\), is \(1 \,\mathrm{atm}=101.325 \,\mathrm{kPa}\).
02

Convert the temperatures to Kelvin

Initial temperature, \(T_1\), needs to be converted to Kelvin. \[ T_1 = 27 + 273.15 = 300.15 \,\mathrm{K} \]
03

Calculate absolute pressures

Calculate the absolute pressures using the atmospheric pressure: \[ P_{1} = P_{g1} + P_{atm} = 300 \,\mathrm{kPa} + 101.325 \,\mathrm{kPa} = 401.325 \,\mathrm{kPa} \] \[ P_{2} = P_{g2} + P_{atm} = 367 \,\mathrm{kPa} + 101.325 \,\mathrm{kPa} = 468.325 \,\mathrm{kPa} \]
04

Apply the ideal gas law

Since the volume and the number of moles of gas are constant, we can use the relation: \[ \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \] Rearrange to find \(T_2\): \[ T_{2} = T_{1} \frac{P_{2}}{P_{1}} \] \[ T_{2} = 300.15 \frac{468.325}{401.325} = 350.45 \,\mathrm{K} \]
05

Convert the final temperature back to Celsius

To find the final temperature in \(^{\circ} \mathrm{C}\), subtract 273.15 from the Kelvin temperature: \[ T_{2} = 350.45 - 273.15 = 77.3 \,\mathrm{C} \]

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It helps us understand how temperature, energy, and work are related. In this exercise, we're applying its principles to find out how the temperature of an ideal gas changes when it's heated in a rigid tank. Knowing the initial and final pressures, along with following the laws of thermodynamics, allows us to determine the final temperature. By using various thermodynamic equations, like the Ideal Gas Law, we can predict how gases will behave under different conditions.
Pressure Conversion
Pressure conversion is essential to understand and correctly calculate the absolute pressure of a gas, particularly when it is initially given in a different form such as gage pressure. In our problem, the gage pressure needs to be converted to absolute pressure. Absolute pressure is the actual pressure of the gas, which includes atmospheric pressure. For example, the initial gage pressure of 300 kPa needs to be added to the atmospheric pressure of 101.325 kPa to get the total initial pressure:
\[ P_{1} = 300 \text{ kPa} + 101.325 \text{ kPa} = 401.325 \text{ kPa} \] This step is crucial because the ideal gas law requires absolute pressures.
Temperature Conversion
Temperature conversion is needed because the calculations in the ideal gas law are done using the Kelvin scale. Kelvin is an absolute temperature scale starting at absolute zero. To convert Celsius to Kelvin, you simply add 273.15. For this exercise:
Initial temperature, \[27 + 273.15 = 300.15 \text{ K} \] After solving the ideal gas equation, we found the temperature in Kelvin as 350.45 K. To convert it back to Celsius:
Final temperature, \[350.45 - 273.15 = 77.3 \text{ 鈩儅 \] Correct temperature conversion ensures that all the calculations are accurate.
Rigid Tank
A rigid tank means that the volume of the gas does not change, no matter what happens to the pressure or temperature of the gas inside. This is crucial when using the ideal gas law, as it simplifies the equation by keeping the volume constant. In this problem, since the tank does not expand or contract, it allows us to focus on the relationship between pressure and temperature only. This is why we can directly use the relation: \[ \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \] This keeps things simpler and allows for accurate calculations without adding additional variables.
Ideal Gas Behavior
Ideal gas behavior assumes that the gas particles are in constant motion, and interact only through elastic collisions. In this exercise, we treat the gas as 'ideal,' which means it follows the Ideal Gas Law: \[ PV = nRT \] Although real gases can deviate from this behavior, especially under high pressure or low temperature, ideal gas behavior is a useful approximation for our calculations. For our problem:
Since the number of moles and volume stay constant, we use the ratio form: \[ \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \] This helps us find the unknown final temperature provided we know initial and final pressures and the initial temperature.

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Most popular questions from this chapter

A rigid, well-insulated container with a volume of \(2 \mathrm{ft}^{3}\) holds \(0.12 \mathrm{lb}\) of ammonia initially at a pressure of \(20 \mathrm{lbf} / \mathrm{in}^{2}\) The ammonia is stirred by a paddle wheel, resulting in an energy transfer to the ammonia with a magnitude of 1 Btu. For the ammonia, determine the initial and final temperatures, each in \({ }^{\circ} \mathrm{R}\), and the final pressure, in lbf/in. \({ }^{2}\) Neglect kinetic and potential energy effects.

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) contained in a piston-cylinder arrangement, initially at 6 bar and \(400 \mathrm{~K}\), undergoes an expansion to a final temperature of \(298 \mathrm{~K}\), during which the pressure-volume relationship is \(p V^{1.2}=\) constant. Assuming the ideal gas model for the \(\mathrm{CO}_{2}\), determine the final pressure, in bar, and the work and heat transfer, each in \(\mathrm{kJ} / \mathrm{kg}\).

A system consisting of \(1 \mathrm{~kg}\) of \(\mathrm{H}_{2} \mathrm{O}\) undergoes a power cycle composed of the following processes: Process 1-2: Constant-volume heating from \(p_{1}=5\) bar, \(T_{1}=\) \(160^{\circ} \mathrm{C}\) to \(p_{2}=10\) bar. Process 2-3: Constant-pressure cooling to saturated vapor. Process 3-4: Constant-volume cooling to \(T_{4}=160^{\circ} \mathrm{C}\). Process 4-1: Isothermal expansion with \(Q_{41}=815.8 \mathrm{~kJ}\). Sketch the cycle on \(T-v\) and \(p-v\) diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency.

Water in a piston-cylinder assembly, initially at a temperature of \(99.63^{\circ} \mathrm{C}\) and a quality of \(65 \%\), is heated at constant pressure to a temperature of \(200^{\circ} \mathrm{C}\). If the work during the process is \(+300 \mathrm{~kJ}\), determine (a) the mass of water, in \(\mathrm{kg}\), and (b) the heat transfer, in kJ. Changes in kinetic and potential energy are negligible.

The following table lists temperatures and specific volumes of ammonia vapor at two pressures: $$ \begin{array}{lccc} {}{}{p=50 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}} & {}{c}{p=60 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}} \\ { 4 } T\left({ }^{\circ} \mathrm{F}\right) & v\left(\mathrm{ft}{ }^{3} / \mathrm{lb}\right) & T\left({ }^{\circ} \mathrm{F}\right) & v\left(\mathrm{ft}^{3} / \mathrm{lb}\right) \\ \hline 100 & 6.836 & 100 & 5.659 \\ 120 & 7.110 & 120 & 5.891 \\ 140 & 7.380 & 140 & 6.120 \end{array} $$ Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate (a) the specific volume at \(T=120^{\circ} \mathrm{F}, p=54 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\), in \(\mathrm{ft}^{3} / \mathrm{lb}\). (b) the temperature at \(p=60 \mathrm{lbf} / \mathrm{in}^{2}, v=5.982 \mathrm{ft}^{3} / \mathrm{lb}\), in \({ }^{\circ} \mathrm{F}\). (c) the specific volume at \(T=110^{\circ} \mathrm{F}, p=58 \mathrm{lbf} / \mathrm{in}^{2}\), in \(\mathrm{ft}^{3} / \mathrm{lb}\).
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