Question

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

The final temperature is approximately \(-145.17 \degree C\).

Step-by-step solution

Understanding the Gas Law

The problem involves a gas undergoing a change in temperature and volume at constant pressure. This suggests the use of Charles's Law, which states that \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V \) is volume and \( T \) is temperature in Kelvin.

Convert Initial Temperature

Convert the initial temperature from Celsius to Kelvin. The formula for this is \( T(K) = T(\degree C) + 273.15 \). Thus, the initial temperature is \( 88 + 273.15 = 361.15 \text{ K} \).

Apply Charles's Law

Substitute the known values into Charles's Law: \( \frac{9.6}{361.15} = \frac{3.4}{T_2} \).

Solve for Final Temperature

Rearrange the equation to solve for \( T_2 \): \( T_2 = \frac{3.4 \times 361.15}{9.6} \). Calculate \( T_2 \approx 127.98 \text{ K} \).

Convert Final Temperature to Celsius

Convert the final temperature back to Celsius: \( T(\degree C) = T(K) - 273.15 \). Thus, \( 127.98 - 273.15 = -145.17 \degree C \).

Key concepts

constant-pressure

Charles's Law governs the behavior of gas under specific conditions. When we mention "constant-pressure," we're referring to situations where the pressure remains the same throughout the process. This constant-pressure is crucial because it allows Charles's Law to be applicable.

Charles's Law states that, at constant pressure, the volume of a gas is directly proportional to its Kelvin temperature.

This means:

• If you increase the temperature of a gas, it expands, increasing its volume.
• Conversely, if you decrease the temperature, the gas contracts, leading to a smaller volume.

Understanding the constant-pressure condition helps us predict how a gas will behave when only the temperature and volume change.

temperature conversion

Temperature conversion is a critical step when working with gas laws. Before utilizing any equations, you need to ensure the temperature is measured in Kelvin rather than Celsius.

• The Kelvin scale is an absolute scale where 0 K (Kelvin) is absolute zero, the hypothetical point where molecular motion stops.
• The Celsius scale is based on the freezing (0°C) and boiling points (100°C) of water.

To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.
For instance, if the initial temperature is 88°C:\[ T(K) = 88 + 273.15 = 361.15 \text{ K} \]This conversion is necessary because gas laws require temperature inputs in Kelvin to maintain proportionality and consistency in calculations.

gas volume change

Gas volume change under constant-pressure is directly related to changes in temperature. According to Charles's Law, when temperature decreases, the volume must also decrease proportionally, assuming constant pressure.
This relationship is described mathematically as:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]Where:

• \( V_1 \) and \( V_2 \) are the initial and final volumes.
• \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.

In this exercise, the initial volume \( V_1 \) is 9.6 L and the final volume \( V_2 \) is 3.4 L. As Charles's Law implies, since the final volume is smaller than the initial volume, the final temperature must also be lower than the initial temperature.
Calculations help us find the exact changes these variables undergo.

Kelvin and Celsius conversion

Understanding how to interconvert Kelvin and Celsius temperatures is vital. To navigate problems involving gas laws, one must be fluent in moving between these two temperature scales.
Here's how you convert:

• From Celsius to Kelvin: Add 273.15 to the Celsius value. For example, 88°C converts to:\[ T(K) = 88 + 273.15 = 361.15 \text{ K} \]
• From Kelvin back to Celsius: Subtract 273.15 from the Kelvin value. For instance, if an outcome gives 127.98 K, converting to Celsius yields:\[ T(\degree C) = 127.98 - 273.15 = -145.17 \degree C \]

Mastering these conversions ensures accurate problem-solving when handling gas law exercises, as Kelvin scale inputs are essential for the direct relationships described by these laws.