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A constant-volume gas thermometer (see Figures 12.3 and 12.4\()\) has a pressure of \(5.00 \times 10^{3} \mathrm{Pa}\) when the gas temperature is \(0.00^{\circ} \mathrm{C}\). What is the temperature (in \({ }^{\circ} \mathrm{C}\) ) when the pressure is \(2.00 \times 10^{3} \mathrm{Pa} ?\)

Short Answer

Expert verified
The temperature is \(-163.89^{\circ}\mathrm{C}\).

Step by step solution

01

Understand the Relationship

In a constant-volume gas thermometer, the relationship between pressure and temperature is given by the formula \( P_1/T_1 = P_2/T_2 \), where \( P \) is the pressure and \( T \) is the temperature in Kelvin.
02

Convert Celsius to Kelvin

The given initial temperature is \( 0.00^{\circ} \mathrm{C} \), which converts to Kelvin by adding 273.15. So, \( T_1 = 273.15 \, \mathrm{K} \).
03

Apply the Relation

We are given \( P_1 = 5.00 \times 10^{3} \mathrm{Pa} \) and \( P_2 = 2.00 \times 10^{3} \mathrm{Pa} \). Use the relationship \( P_1/T_1 = P_2/T_2 \) to find \( T_2 \).
04

Solve for \( T_2 \)

Rearrange the formula to solve for \( T_2 \): \[ T_2 = \frac{P_2}{P_1} \times T_1 = \frac{2.00 \times 10^{3} \mathrm{Pa}}{5.00 \times 10^{3} \mathrm{Pa}} \times 273.15 \]. Calculate this value.
05

Calculate Temperature in Kelvin

Calculate \( T_2 \) using the previous equation: \( T_2 = \frac{2.00}{5.00} \times 273.15 = 109.26 \, \mathrm{K} \).
06

Convert Kelvin Back to Celsius

Convert \( T_2 \) back to Celsius by subtracting 273.15. So, \( T_2^{\circ}\mathrm{C} = 109.26 - 273.15 = -163.89^{\circ}\mathrm{C} \).

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Key Concepts

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

Constant-Volume Gas Thermometer
A constant-volume gas thermometer is an essential tool for understanding temperature measurement in thermodynamics. It operates on the principle that, at a fixed volume, the pressure of a gas is directly proportional to its temperature in Kelvin.
This thermometer is built to maintain a constant volume while measuring the pressure of the gas inside as it changes with temperature. This characteristic makes it especially useful for exploring the pressure-temperature relationship.
By observing these changes in pressure at a constant volume, we can infer the temperature of the gas, or the surrounding environment, more accurately than with other types of thermometers. This precision is particularly important in scientific and experimental settings.
Pressure-Temperature Relationship
The pressure-temperature relationship is a core concept in thermodynamics, especially crucial in constant-volume scenarios. It is represented mathematically by the equation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), where \( P \) represents pressure and \( T \) the temperature in Kelvin.
This equation tells us that, at a constant volume, the ratio of the initial pressure over initial temperature equals the ratio of a new pressure over a new temperature. It's derived from the more general ideal gas law and reflects how gas particles behave under various thermal conditions.
This ratio allows us to solve for unknowns in various thermodynamic processes, making it a practical formula, not just a theoretical concept. By rearranging the equation, you can solve for unknown temperatures or pressures when one of them changes while the others are known.
Kelvin to Celsius Conversion
Understanding the conversion between the Kelvin and Celsius temperature scales is vital for accurately reading and applying thermodynamics principles.
Kelvin is the absolute temperature scale used predominantly in scientific calculations because it starts from absolute zero, where all molecular motion ceases. The conversion between Celsius and Kelvin is straightforward: add 273.15 to a Celsius temperature to convert it to Kelvin. So, \( T_{\text{K}} = T_{\text{C}} + 273.15 \).
Conversely, if you need to convert back to Celsius from Kelvin, subtract 273.15 from the Kelvin temperature, thus \( T_{\text{C}} = T_{\text{K}} - 273.15 \). This simple conversion is crucial for applying formulas like the pressure-temperature relationship effectively, which always uses Kelvin as the temperature unit.
Ideal Gas Law
The ideal gas law is a fundamental equation in thermodynamics that describes the behavior of an ideal gas. It combines various empirical gas laws into one comprehensive equation: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.
This law helps explain how gases will respond to changes in pressure, volume, or temperature, assuming no intermolecular forces. It serves as the basis for the equation used in a constant-volume thermometer, \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), which is effectively a form of the ideal gas law adapted for specific conditions.
Understanding and applying the ideal gas law is critical for students exploring the relationships between physical properties of gases in different thermodynamic scenarios.

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Most popular questions from this chapter

To help prevent frost damage, fruit growers sometimes protect their crop by spraying it with water when overnight temperatures are expected to go below freezing. When the water turns to ice during the night, heat is released into the plants, thereby giving a measure of protection against the cold. Suppose a grower sprays \(7.2 \mathrm{kg}\) of water at \(0^{\circ} \mathrm{C}\) onto a fruit tree. (a) How much heat is released by the water when it freezes? (b) How much would the temperature of a \(180-\mathrm{kg}\) tree rise if it absorbed the heat released in part (a)? Assume that the specific heat capacity of the tree is \(2.5 \times 10^{3} \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and that no phase change occurs within the tree itself.

An 85.0-N backpack is hung from the middle of an aluminum wire, as the drawing shows. The temperature of the wire then drops by \(20.0 \mathrm{C}^{\circ}\). Find the tension in the wire at the lower temperature. Assume that the distance between the supports does not change, and ignore any thermal stress.

Three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. Portion A has a mass \(m\) and a temperature of \(94.0^{\circ} \mathrm{C},\) portion \(\mathrm{B}\) also has a mass \(m\) but a temperature of \(78.0^{\circ} \mathrm{C},\) and portion \(\mathrm{C}\) has a mass \(m_{\mathrm{C}}\) and a temperature of \(34.0^{\circ} \mathrm{C} .\) What must be the mass of portion \(\mathrm{C}\) so that the final temperature \(T_{\mathrm{f}}\) of the three-portion mixture is \(T_{\mathrm{f}}=50.0^{\circ} \mathrm{C} ?\) Express your answer in terms of \(m ;\) for example, \(m_{\mathrm{C}}=2.20 \mathrm{m}\).

A 0.35-kg coffee mug is made from a material that has a specific heat capacity of \(920 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and contains \(0.25 \mathrm{kg}\) of water. The cup and water are at \(15^{\circ} \mathrm{C} .\) To make a cup of coffee, a small electric heater is immersed in the water and brings it to a boil in three minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

A copper-constantan thermocouple generates a voltage of \(4.75 \times 10^{-3}\) volts when the temperature of the hot junction is \(110.0^{\circ} \mathrm{C}\) and the reference junction is kept at a temperature of \(0.0^{\circ} \mathrm{C}\). If the voltage is proportional to the difference in temperature between the junctions, what is the temperature of the hot junction when the voltage is \(1.90 \times 10^{-3}\) volts?

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