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Chapter 12: Problem 14

A sample of \(\mathrm{CH}_{4}\) gas occupies a volume of \(575 \mathrm{~mL}\) at \(-25^{\circ} \mathrm{C}\). If pressure remains constant, what will be the new volume if temperature changes to: (a) \(298 \mathrm{~K}\) (b) \(32^{\circ} \mathrm{F}\) (c) \(45^{\circ} \mathrm{C}\)

Short Answer

Expert verified
a) 690 mL, b) 633 mL, c) 737 mL

Step by step solution

01

- Understand the problem

We need to find the new volume of \(\text{CH}_{4}\) gas when the temperature changes under constant pressure. This relates to Charles's Law, which states that the volume of a gas is directly proportional to its temperature in kelvin.
02

- Convert initial temperature to Kelvin

The initial temperature is \(-25^{\text{°C}}\). To convert this to Kelvin:\[ T_{1} = -25 + 273 = 248 \text{K} \]
03

- Apply Charles's Law for case (a)

Given: \(T_2 = 298 \text{K}, \ V_1 = 575 \text{mL}\)\Formula: \ \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)\ \(\frac{575 \text{mL}}{248 \text{K}} = \frac{V_2}{298 \text{K}}\)\ Solve for \(V_2\):\ \( V_2 = \frac{575 \text{mL} \cdot 298 \text{K}}{248 \text{K}} \ = 690 \text{mL} \)
04

- Convert temperature from Fahrenheit to Kelvin for case (b)

Initially, the temperature is given in Fahrenheit: \(32 \text{°F}\). Convert this to Celsius: \( T = (32 - 32) \cdot \frac{5}{9} = 0 \text{°C}\). Convert Celsius to Kelvin \( T_2 = 0 + 273 = 273 \text{K}\).
05

- Apply Charles's Law for case (b)

Using \(T_2 = 273 \text{K}\)\ \(\frac{575 \text{mL}}{248 \text{K}} = \frac{V_2}{273 \text{K}}\)\ Solve for \(V_2 \):\ \( V_2 = \frac{575 \text{mL} \cdot 273 \text{K}}{248 \text{K}} \ = 633 \text{mL} \)
06

- Convert temperature from Celsius to Kelvin for case (c)

The temperature is given as \(45^{\text{°C}} \). To convert to Kelvin: \( T_2 = 45 + 273 = 318 \text{K}\)
07

- Apply Charles's Law for case (c)

Using \(T_2 = 318 \text{K}\)\ \(\frac{575 \text{mL}}{248 \text{K}} = \frac{V_2}{318 \text{K}}\)\ Solve for \(V_2\):\ \( V_2 = \frac{575 \text{mL} \cdot 318 \text{K}}{248 \text{K}} \ = 737 \text{mL} \)

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Key Concepts

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

Gas Laws
Gas laws describe how gases behave under different conditions. One of the most important gas laws is Charles's Law. Charles's Law explains the relationship between the volume and temperature of a gas. This law states that, at constant pressure, the volume of a gas is directly proportional to its temperature when measured in Kelvin. This means that if the temperature of a gas increases, its volume will increase as well, and vice versa. When working with gas laws, it's crucial to keep the units consistent, especially temperature in Kelvin, as this avoids negative values and ensures accuracy.
Temperature Conversion
Temperature conversion is an essential skill when working with gas laws, as temperatures must often be converted to Kelvin. Many problems will provide temperatures in Celsius or Fahrenheit, but you'll need to convert these to Kelvin. To convert Celsius to Kelvin, simply add 273: \[ T(K) = T(°C) + 273 \] For example, \[-25°C + 273 = 248K \] Converting Fahrenheit to Kelvin requires a two-step process: first converting Fahrenheit to Celsius, then Celsius to Kelvin. The formula for converting Fahrenheit to Celsius is: \[ T(°C) = (T(°F) - 32) \cdot \frac{5}{9} \] For instance, to convert 32°F to Kelvin: \[ 32°F - 32 \cdot \frac{5}{9} = 0°C \] \[ 0°C + 273 = 273K \]
Volume-Temperature Relationship
The volume-temperature relationship under Charles's Law is best understood through the equation: \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\] Here, \(V_1\) and \(T_1\) are the initial volume and temperature, while \(V_2\) and \(T_2\) are the new volume and temperature. By rearranging this formula, you can solve for the new volume: \[V_2 = \frac{V_1 \cdot T_2}{T_1}\] For the methanol gas problem, we follow these steps for different temperature scenarios:
  • For 298K: \[V_2 = \frac{575 \cdot 298}{248} = 690 \text{mL}\]
  • For 32°F converted to 273K: \[V_2 = \frac{575 \cdot 273}{248} = 633 \text{mL}\]
  • For 45°C converted to 318K: \[V_2 = \frac{575 \cdot 318}{248} = 737 \text{mL}\]
This shows how the gas volume increases or decreases proportionally with temperature, validating Charles’s Law.

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Most popular questions from this chapter

A steel cylinder contains \(50.0 \mathrm{~L}\) of oxygen gas under a pressure of \(40.0 \mathrm{~atm}\) and at a temperature of \(25^{\circ} \mathrm{C}\). What was the pressure in the cylinder during a storeroom fire that caused the temperature to rise \(152^{\circ} \mathrm{C}\) ? (Be careful!)

Which of these occupies the greatest volume? (a) \(0.2 \mathrm{~mol}\) of chlorine gas at \(48^{\circ} \mathrm{C}\) and \(80 \mathrm{~cm} \mathrm{Hg}\) (b) \(4.2 \mathrm{~g}\) of ammonia at \(0.65 \mathrm{~atm}\) and \(-11^{\circ} \mathrm{C}\) (c) \(21 \mathrm{~g}\) of sulfur trioxide at \(55^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\)

How many grams of \(\mathrm{NH}_{3}\) are present in \(725 \mathrm{~mL}\) of the gas at STP?

A glass containing \(345 \mathrm{~mL}\) of a soft drink (a carbonated beverage) was left sitting out on a kitchen counter. If the \(\mathrm{CO}_{2}\) released at room temperature \(\left(25^{\circ} \mathrm{C}\right)\) and pressure \((1 \mathrm{~atm})\) occupies \(1.40 \mathrm{~L}\), at a minimum, what is the concentration (in ppm) of the \(\mathrm{CO}_{2}\) in the original soft drink? (Assume the density of the original soft drink is \(0.965 \mathrm{~g} / \mathrm{mL}\).)

In the future when commercial spacecraft have become safer, you decide to take a trip to a remote planet that has lots of laughing gas (nitrous oxide, \(\mathrm{N}_{2} \mathrm{O}\) ). On this planet the atmosphere is composed of oxygen gas (346 atm), nitrous oxide gas (93 atm), and nitrogen gas. If the total pressure on this planet is \(950 \mathrm{~atm}\), what is the pressure due to nitrogen gas?
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