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A rigid 1.0 - \(\mathrm{L}\) container at \(75^{\circ} \mathrm{C}\) is fitted with a gas pressure gauge. A 1.0 -mol sample of ideal gas is introduced into the container. What would the pressure gauge in the container be reading in \(\mathrm{mmHg}\) ? Describe the interactions in the container that are causing the pressure. c Say the temperature in the container were increased to \(150^{\circ} \mathrm{C}\). Describe the effect this would have on the pressure, and, in terms of kinetic theory, explain why this change occurred.

Short Answer

Expert verified
The pressure gauge would read 2173.6 mmHg initially, increasing to 2637.2 mmHg at 150°C.

Step by step solution

01

Understand the problem

You have a rigid container with a volume of 1.0 L containing 1 mole of ideal gas at 75°C. We need to calculate the pressure in mmHg using the ideal gas law and determine how pressure changes if the temperature is increased to 150°C.
02

Convert units to Kelvin

First, convert temperatures from Celsius to Kelvin for use in calculations. Add 273.15 to each Celsius temperature. - Initial temperature: 75°C = 348.15 K - Increased temperature: 150°C = 423.15 K.
03

Use the ideal gas law to find initial pressure

Use the ideal gas law equation: \[ PV = nRT \]Where: - \(P\) is the pressure in atm,- \(V\) is the volume in liters,- \(n\) is the number of moles of gas,- \(R\) is the ideal gas constant (0.0821 L·atm/mol·K),- \(T\) is the temperature in Kelvin.Plug in the initial conditions: - \(n = 1.0\) mol, \(V = 1.0\) L, \(T = 348.15\) K. Calculate \(P\): \[ P = \frac{nRT}{V} = \frac{1.0 \times 0.0821 \times 348.15}{1.0} = 28.6 \text{ atm} \]
04

Convert pressure from atm to mmHg

Convert the pressure from atm to mmHg using the conversion factor (1 atm = 760 mmHg).\[ \text{Pressure in mmHg} = 28.6 \text{ atm} \times 760 \text{ mmHg/atm} = 2173.6 \text{ mmHg} \]
05

Analyze effect of temperature increase on pressure

The pressure in a rigid container depends on the number of molecules and their average kinetic energy. When the temperature increases to 150°C (423.15 K), the kinetic energy of gas molecules increases, leading to more frequent and stronger collisions with the container walls. This increases the pressure.
06

Calculate new pressure with increased temperature

Apply the ideal gas law for the new temperature:\[ P' = \frac{nRT'}{V} = \frac{1.0 \times 0.0821 \times 423.15}{1.0} = 34.7 \text{ atm} \]Convert this pressure to mmHg:\[ \text{Pressure in mmHg} = 34.7 \text{ atm} \times 760 \text{ mmHg/atm} = 2637.2 \text{ mmHg} \]
07

Summarize kinetic theory explanation

According to the kinetic molecular theory, increasing the temperature increases the average kinetic energy of the gas molecules. This results in more collisions per unit time with the container walls and with greater force, thereby increasing the pressure observed on the gauge.

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Key Concepts

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

Kinetic Molecular Theory
The Kinetic Molecular Theory is a fundamental concept that explains the behavior of gases. It is based on a few key principles that help us understand how gas molecules interact. First, gas molecules are in constant random motion. These molecules move in straight lines until they collide with either the container walls or each other.

These collisions are perfectly elastic, meaning that there is no loss of energy when they occur. The theory also assumes that gas molecules don't exert forces on each other except during collisions. Therefore, their volume is negligible compared to the container's volume.

When applying this theory to the ideal gas law, the focus is on the pressure, volume, and temperature of the gas. The behavior described by the Kinetic Molecular Theory provides an explanation for the observations made with the gas pressure in our described exercise.
Gas Pressure
Gas pressure is a result of the collisions between the gas molecules and the walls of the container. Imagine a swarm of tiny balls rapidly bouncing against the inside of a box. Each impact against the walls contributes to the overall pressure exerted by the gas.

The frequency and force of these impacts depend directly on both the speed of the gas molecules and their number. Pressure is directly proportional to the number of molecules in a given volume, as each molecule adds to the total impact against the walls.

In our ideal container, containing one mole of gas, the pressure was initially calculated using the ideal gas law. Using this law, you can find that more energetic collisions (due to increased kinetic energy) lead to higher pressure readings on the gauge. According to this, if you introduce more gas molecules into the same volume or increase the gas's temperature, you'll notice an increase in pressure.
Temperature and Pressure Relationship
The relationship between temperature and pressure in gases is one of direct proportionality. When the temperature of a gas increases, so does the kinetic energy of its molecules. With more energy, molecules move faster and collide with the container walls more frequently and with greater force.

This principle can be observed in the provided exercise. Initially, the gas was at 75°C, and the corresponding pressure was determined. When the temperature increased to 150°C, the kinetic energy of the gas molecules increased, and thus, so did the pressure.

This occurs because, given a constant volume and amount of gas, an increase in temperature will always lead to an increase in pressure. The ideal gas law equation \( P = \frac{nRT}{V} \)nicely captures this relationship by linking pressure directly to temperature.
Pressure Conversion
Pressure conversion is an essential skill in solving gas-related problems, as pressure can be expressed in various units. Commonly used units include atmospheres (atm), millimeters of mercury (mmHg), and pascals (Pa).

In the exercise, the initial and final pressures were calculated in atmospheres using the ideal gas law. However, the problem required the answer in mmHg. To convert from atm to mmHg, you multiply by 760 because 1 atm is equivalent to 760 mmHg. This conversion is crucial for ensuring that the answer is in the desired units and that it is comparable to everyday pressure measurements, which are often in mmHg.

Mastering pressure conversion allows you to apply theoretical calculations to practical situations, making this step an invaluable tool in understanding gas behaviors in various contexts.

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Most popular questions from this chapter

The density of air at \(20^{\circ} \mathrm{C}\) and 1.00 atm is \(1.205 \mathrm{~g} / \mathrm{L}\). If this air were compressed at the same temperature to equal the pressure at \(50.0 \mathrm{~m}\) below sea level, what would be its density? Assume the barometric pressure is constant at \(1.00 \mathrm{~atm} .\) The density of seawater is \(1.025 \mathrm{~g} / \mathrm{cm}^{3} .\)

If the rms speed of \(\mathrm{NH}_{3}\) molecules is found to be \(0.600 \mathrm{~km} / \mathrm{s}\), what is the temperature (in degrees Celsius)?

5.118 Raoul Pictet, the Swiss physicist who first liquefied oxygen, attempted to liquefy hydrogen. He heated potassium formate, \(\mathrm{KCHO}_{2}\), with \(\mathrm{KOH}\) in a closed 2.50-Lvessel. \(\mathrm{KCHO}_{2}(s)+\mathrm{KOH}(s) \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2}(g)\) If \(75.0 \mathrm{~g}\) of potassium formate reacts in a \(2.50-\mathrm{L}\) vessel, which was initially evacuated, what pressure of hydrogen will be attained when the temperature is finally cooled to \(25^{\circ} \mathrm{C} ?\) Use the preceding chemical equation and ignore the volume of solid product.

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