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The gas in a constant-volume gas thermometer has a pressure of \(93.5 \mathrm{kPa}\) at \(105^{\circ} \mathrm{C} .\) (a) What is the pressure of the gas at \(50.0^{\circ} \mathrm{C} ?(\mathrm{b})\) At what temperature does the gas have a pressure of \(115 \mathrm{kPa} ?\)

Short Answer

Expert verified
(a) 79.82 kPa at 50°C. (b) 191.15°C at 115 kPa.

Step by step solution

01

Identify the given variables

Recognize the variables provided in the exercise. We have initial pressure \( P_1 = 93.5 \, \text{kPa} \) at temperature \( T_1 = 105^{\circ} \text{C} \) which converts to \( 378.15 \, \text{K} \); additionally, we need to find \( P_2 \) at \( T_2 = 50^{\circ} \text{C} \) converted to \( 323.15 \, \text{K} \).
02

Apply the Ideal Gas Law for Constant Volume

Use the equation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) to find the unknown pressure at the new temperature, where \( P_1 = 93.5 \, \text{kPa} \) and \( T_1 = 378.15 \, \text{K} \), \( T_2 = 323.15 \, \text{K} \).
03

Solve for new pressure at 50°C

Rearrange the equation to solve for \( P_2 \):\[ P_2 = P_1 \times \frac{T_2}{T_1} \]Substitute the known values: \( P_2 = 93.5 \, \text{kPa} \times \frac{323.15}{378.15} \) = 79.82 kPa.
04

Calculate temperature for given pressure

For part (b), let's find \( T \) when \( P = 115 \, \text{kPa} \). Use \( \frac{P_1}{T_1} = \frac{P}{T} \) and solve by substituting \( P_1 = 93.5 \, \text{kPa} \), \( T_1 = 378.15 \, \text{K} \), and \( P = 115 \, \text{kPa}\).
05

Solve for temperature at given pressure

Rearrange to find \( T \):\[ T = \frac{T_1 \times P}{P_1} \]Substitute the values:\( T = \frac{378.15 \, \text{K} \times 115 \, \text{kPa}}{93.5 \, \text{kPa}} \) = 464.3 K, converting to Celsius gives \( T = 191.15^{\circ} \text{C} \).

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Key Concepts

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

Constant Volume
In a constant volume scenario, such as a constant-volume gas thermometer, the volume of the gas remains unchanged even as other parameters like temperature and pressure vary. This scenario is key when understanding the behavior of gases under different temperature conditions without altering their volume. It simplifies the relationship between pressure and temperature, allowing us to use equations easily.

When dealing with gases at constant volume, the Ideal Gas Law becomes less complex, because the volume factor is fixed. This makes it easier to predict how pressure will change with temperature, assuming mass and gas type are also constant.
  • Key ideas to remember are:
    • Volume does not change.
    • Pressure and temperature have a direct relationship.
    • This condition is commonly used for precise temperature measurements.
Such conditions are ideal for experimentation since one major variable is held constant, allowing clearer observation of the effects of temperature on pressure.
Pressure-Temperature Relationship
The relationship between pressure and temperature in gases is vital to understand, especially under constant volume. This relationship means that if the temperature of a gas increases, the pressure will increase proportionally, provided the volume remains constant. This can be expressed with the formula:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
where \(P_1\) and \(T_1\) are the initial pressure and temperature, and \(P_2\) and \(T_2\) are the pressure and temperature after the change.
  • This is an example of Gay-Lussac's Law, which is a component of the Ideal Gas Law.
  • The relationship is direct, meaning both pressure and temperature move in the same direction.
  • This law highlights the need for precise conversion of temperature to Kelvin, as it is the absolute temperature scale needed for these calculations.
In practice, this relationship allows us to predict how altering the temperature of a gas can affect its pressure and vice versa. It's a perfect model when one cannot alter the volume, aiding engineers and scientists in their calculations across various fields.
Thermodynamics
Thermodynamics is the broader science that studies how heat and energy transfer within systems, like gases. The principles of thermodynamics are fundamental to understanding the Ideal Gas Law and its application. This law is a significant part of kinetic theory, which describes how the motion of particles is related to macroscopic properties like pressure and temperature.

In thermodynamics:
  • Energy conservation is crucial, indicating that energy changes can lead to temperature or phase changes in a gas.
  • The laws explain interactions and energy flows between systems, crucial for fields such as engineering, meteorology, and chemistry.
  • The First Law of Thermodynamics, for instance, concerns energy preservation, while the Second Law involves entropy and the direction of processes.
Through thermodynamics, one can better grasp why the Ideal Gas Law works under constant volume and pressure-temperature scenarios.

By comprehending these concepts, you'll understand how heat, work, and internal energy interplay to describe the behavior of gases under specified conditions, and its implications for practical tasks and scientific endeavors.

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Most popular questions from this chapter

If heat is transferred to 150 g of water at a constant rate for \(2.5 \mathrm{min},\) its temperature increases by \(13 \mathrm{C}^{\circ} .\) When heat is transferred at the same rate for the same amount of time to a \(150-g\) object of unknown material, its temperature increases by \(61 \mathrm{C}^{\circ}\). (a) From what material is the object made? (b) What is the heating rate?

Consider a double-paned window consisting of two panes of glass, each with a thickness of \(0.500 \mathrm{cm}\) and an area of \(0.725 \mathrm{m}^{2},\) separated by a layer of air with a thickness of \(1.75 \mathrm{cm}\) The temperature on one side of the window is \(0.00^{\circ} \mathrm{C}\); the temperature on the other side is \(20.0^{\circ} \mathrm{C}\). In addition, note that the thermal conductivity of glass is roughly 36 times greater than that of air. (a) Approximate the heat transfer through this window by ignoring the glass. That is, calculate the heat flow per second through \(1.75 \mathrm{cm}\) of air with a temperature difference of \(20.0 \mathrm{C}^{\circ} .\) (The exact result for the complete window is \(\left.19.1 \mathrm{J} / \mathrm{s} .\right)\) (b) Use the approximate heat flow found in part (a) to find an approximate temperature difference across each pane of glass. (The exact result is \(0.157 \mathrm{C}^{\circ}\) )

A \(226-\mathrm{kg}\) rock sits in full sunlight on the edge of a cliff \(5.25 \mathrm{m}\) high. The temperature of the rock is \(30.2^{\circ} \mathrm{C}\). If the rock falls from the cliff into a pool containing \(6.00 \mathrm{m}^{3}\) of water at \(15.5^{\circ} \mathrm{C}\), what is the final temperature of the rock- water system? Assume that the specific heat of the rock is \(1010 \mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\).

The Coldest Place in the Universe The Boomerang nebula holds the distinction of having the lowest recorded temperature in the universe, a frigid \(-272^{\circ} \mathrm{C}\). What is this temperature in kelvins?

Referring to the copper ring in the previous problem, imagine that initially the ring is hotter than room temperature, and that an aluminum rod that is colder than room temperature fits snugly inside the ring. When this system reaches thermal equilibrium at room temperature, is the rod (A, firmly wedged in the ring; or \(\mathbf{B},\) can it be removed easily)?

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