Chapter 5: Problem 74
At constant pressure a certain gas at \(0^{\circ} \mathrm{C}\), was cooled until its volume was reduced to half. The temperature at this stage (1) \(-130.5^{\circ} \mathrm{C}\) (2) \(-140.0^{\circ} \mathrm{C}\) (3) \(-136.5^{\circ} \mathrm{C}\) (4) \(-120^{\circ} \mathrm{C}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gas Laws
This means that if the temperature of the gas increases, the volume will also increase, provided the pressure is constant. Conversely, if the temperature decreases, the volume will decrease. This relationship can be expressed mathematically as
\overset{\text{direct proportionality}}{\text{V_1 / T_1 = V_2 / T_2}}
Other primary gas laws include:
- Boyle's Law: At constant temperature, the volume of a gas is inversely proportional to its pressure\( \text{P_1 V_1 = P_2 V_2} \)
- Gay-Lussac's Law: At constant volume, the pressure of a gas is directly proportional to its absolute temperature\( \text{P_1 / T_1 = P_2 / T_2} \)
- Avogadro's Law: At constant temperature and pressure, the volume of a gas is directly proportional to the number of molecules\( \text{V_1 / n_1 = V_2 / n_2} \)
Understanding these laws helps predict and explain the behavior of gases in different scenarios.
Temperature Conversion
To convert Celsius to Kelvin, simply add 273.15:
\[T_{(K)} = T_{(°C)} + 273.15\]
For example, if a gas is initially at 0°C, you convert it to Kelvin as follows:
\[T_{(K)} = 0 + 273.15 = 273.15 \text{K}\]
To convert from Kelvin back to Celsius, subtract 273.15:
\[T_{(°C)} = T_{(K)} - 273.15\]
For instance, if the temperature in Kelvin is 136.575K, it converts to Celsius like this:
\[T_{(°C)} = 136.575 - 273.15 = -136.575 \text{°C}\]
Temperature conversion is an essential step in solving problems involving Charles's Law and other gas laws. Always double-check your conversions to ensure accuracy in your final calculations.
Volume-Temperature Relationship
Here’s a step-by-step example of how Charles's Law is applied:
- Step 1: Identify the initial and final conditions for volume and temperature. For example, a gas initially at 0°C and a certain volume (<\(V_1\)) is cooled to reduce its volume by half (<\(V_2 = V_1 / 2\)).
- Step 2: Convert the initial temperature from Celsius to Kelvin:
\(T_1 = 0^{\text{°}} \text{C} + 273.15 = 273.15 \text{K}\) - Step 3: Apply Charles's Law. Substitute the known values into the equation:
\(\frac{V_1}{273.15} = \frac{V_1/2}{T_2}\) - Step 4: Solve for the final temperature in Kelvin (\T_2):
\(T_2 = 273.15 / 2 = 136.575 \text{K}\) - Step 5: Convert the final temperature back to Celsius:
\(T_2 = 136.575 \text{K} - 273.15 = -136.575^{\text{°}} \text{C}\)
This process illustrates how changes in temperature affect the volume of a gas, as long as the pressure remains constant. Charles's Law helps us predict this behavior accurately.