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Under constant-pressure conditions a sample of hydrogen gas initially at \(88^{\circ} \mathrm{C}\) and \(9.6 \mathrm{~L}\) is cooled until its final volume is \(3.4 \mathrm{~L}\). What is its final temperature?

Short Answer

Expert verified
The final temperature of the gas sample is \(127.81 K\).

Step by step solution

01

Convert the initial temperature from Celsius to Kelvin

Converting the initial temperature from Celsius to Kelvin can be done with this formula: \( K = ^{\circ}C + 273.15 \). Substitute \( 88^{\circ}C \) into the formula: \( K = 88 + 273.15 = 361.15 K \). The initial temperature in Kelvin is \(361.15 K\).
02

Apply Charles's Law to calculate the final temperature

Charles's Law can be written as: \(V_1 / T_1 = V_2 / T_2\), where \(V_1\) and \(T_1\) correspond to the initial volume and temperature, \(V_2\) is the final volume, and \(T_2\) is the final temperature we want to determine. Now it's time to substitute our values into the equation: \(9.6 L / 361.15 K = 3.4 L / T_2\).
03

Solve for the final temperature \(T_2\)

To solve for \(T_2\), multiply both sides of the equation by \(T_2\), and then divide by the left side to get: \(T_2 = (3.4 L * 361.15 K / 9.6 L)\). Calculate the value: \(T_2 = 127.81 K\).

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Key Concepts

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

Gas Laws
The gas laws are a set of principles that describe how gases behave under different conditions of temperature, volume, and pressure. Among these laws, Charles's Law is particularly significant when examining the relationship between volume and temperature. Charles's Law states that at constant pressure, the volume of a given mass of gas is directly proportional to its Kelvin temperature. This means that when the temperature of a gas increases, its volume will increase as well, provided the pressure remains unchanged
  • Charles's Law: \( V_1/T_1 = V_2/T_2 \)
  • The principle applies under constant pressure
  • Direct proportionality between volume and temperature in Kelvin
These relationships are fundamental for understanding the behavior of gases in everyday applications, such as breathing, ballooning, and even in car engines. By applying these basic principles, we can predict how gases will react to changes in their surroundings, which is invaluable in both practical and theoretical scenarios.
Temperature Conversion
When dealing with gas laws, converting temperatures from Celsius to Kelvin is an essential step. The Kelvin scale is an absolute temperature scale used in science because it begins at absolute zero, where no molecular activity occurs. This makes it highly suitable for calculations in thermodynamics.
To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature:
  • Formula: \( K = ^{\circ}C + 273.15 \)
  • Example: Converting \(88^{\circ}C\) results in \(361.15 K\)
Using Kelvin ensures uniformity in scientific calculations, particularly because it avoids the complications of negative temperatures, which can occur with Celsius or Fahrenheit scales in gas law equations. This conversion is not just a mathematical exercise; it allows the scientific community to have a shared understanding of temperature readings across different investigations and studies.
Volume and Temperature Relationship
The relationship between volume and temperature is a key component in understanding how gases respond to temperature changes. According to Charles's Law, if the temperature of a gas increases, its volume will increase if the pressure is held constant. Conversely, if the temperature decreases, the volume will also decrease.
  • This relationship is captured by the formula: \( V_1/T_1 = V_2/T_2 \)
  • Ensure temperatures are in Kelvin for accurate calculations
In the given problem, we start with conditions of a gas at a higher volume and temperature, and upon cooling, the volume is reduced. By using Charles's Law, we can determine the final temperature by recognizing this direct proportional reduction. It emphasizes that the volume change is a direct effect of the temperature change, highlighting the predictable nature of gas behavior. Understanding this provides clarity not just in controlled scenarios like laboratories but also in real-world situations such as weather balloons or air conditioning systems, where temperature changes lead to precise volume adjustments.

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Most popular questions from this chapter

Calculate the mass in grams of hydrogen chloride produced when \(5.6 \mathrm{~L}\) of molecular hydrogen measured at STP react with an excess of molecular chlorine gas.

Acidic oxides such as carbon dioxide react with basic oxides like calcium oxide (CaO) and barium oxide \((\mathrm{BaO})\) to form salts (metal carbonates). (a) Write equations representing these two reactions. (b) A student placed a mixture of \(\mathrm{BaO}\) and \(\mathrm{CaO}\) of combined mass \(4.88 \mathrm{~g}\) in a 1.46 - \(\mathrm{L}\) flask containing carbon dioxide gas at \(35^{\circ} \mathrm{C}\) and \(746 \mathrm{mmHg}\). After the reactions were complete, she found that the \(\mathrm{CO}_{2}\) pressure had dropped to \(252 \mathrm{mmHg}\). Calculate the percent composition by mass of the mixture. Assume volumes of the solids are negligible.

One way to gain a physical understanding of \(b\) in the van der Waals equation is to calculate the "excluded volume." Assume that the distance of closest approach between two similar atoms is the sum of their radii \((2 r) .\) (a) Calculate the volume around each atom into which the center of another atom cannot penetrate. (b) From your result in (a), calculate the excluded volume for 1 mole of the atoms, which is the constant \(b\). How does this volume compare with the sum of the volumes of 1 mole of the atoms?

A barometer having a cross-sectional area of \(1.00 \mathrm{~cm}^{2}\) at sea level measures a pressure of \(76.0 \mathrm{~cm}\) of mercury. The pressure exerted by this column of mercury is equal to the pressure exerted by all the air on \(1 \mathrm{~cm}^{2}\) of Earth's surface. Given that the density of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\) and the average radius of Earth is \(6371 \mathrm{~km},\) calculate the total mass of Earth's atmosphere in kilograms. (Hint: The surface area of a sphere is \(4 \pi r^{2},\) where \(r\) is the radius of the sphere.)

The volume of a sample of pure HCl gas was \(189 \mathrm{~mL}\) at \(25^{\circ} \mathrm{C}\) and \(108 \mathrm{mmHg} .\) It was completely dissolved in about \(60 \mathrm{~mL}\) of water and titrated with an \(\mathrm{NaOH}\) solution; \(15.7 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution were required to neutralize the \(\mathrm{HCl}\). Calculate the molarity of the \(\mathrm{NaOH}\) solution.

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