Chapter 5: Problem 110
A volume of air is taken from the earth's surface, at \(19^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\), to the stratosphere, where the temperature is \(-21^{\circ} \mathrm{C}\) and the pressure is \(1.00 \times 10^{-3}\) atm. By what factor is the volume increased?
Short Answer
Expert verified
The volume increases by a factor of approximately 863.
Step by step solution
01
Convert Celsius to Kelvin
We need to convert the temperatures from Celsius to Kelvin. The formula to convert is: \( K = C + 273.15 \). \(19^{\circ} \mathrm{C} = 19 + 273.15 = 292.15 \mathrm{~K} \) and \(-21^{\circ} \mathrm{C} = -21 + 273.15 = 252.15 \mathrm{~K} \).
02
Apply the Combined Gas Law
The combined gas law states \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \). We know \(P_1 = 1.00 \mathrm{~atm}\), \(T_1 = 292.15 \mathrm{~K}\), \(P_2 = 1.00 \times 10^{-3} \mathrm{~atm}\), and \(T_2 = 252.15 \mathrm{~K}\).
03
Rearrange the Formula to Solve for Volume Change
Rearrange the gas law to solve for the change in volume: \( \frac{V_2}{V_1} = \frac{P_1 \cdot T_2}{P_2 \cdot T_1} \).
04
Substitute the Known Values
Substitute the known values into the equation: \[ \frac{V_2}{V_1} = \frac{1.00 \times 252.15}{1.00 \times 10^{-3} \times 292.15} \].
05
Calculate the Volume Increase Factor
Calculate the result of the substitution: \( \frac{V_2}{V_1} = \frac{252.15}{0.29215} \approx 863 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Temperature Conversion
In situations involving gas laws, converting temperature from Celsius to Kelvin is crucial. Kelvin is the standard unit in gas law calculations because it avoids negative numbers, ensuring equations work correctly. The conversion is simple, achieved by adding 273.15 to the Celsius temperature. For example, converting from Celsius to Kelvin for the temperatures in our problem:
- At the Earth's surface: from \( 19^{\circ} \mathrm{C} \) to \( 19 + 273.15 = 292.15 \, \mathrm{K} \).
- In the stratosphere: from \( -21^{\circ} \mathrm{C} \) to \( -21 + 273.15 = 252.15 \, \mathrm{K} \).
Volume Change
Volume change of gases under varying conditions is directly related to pressure and temperature changes. This is expressed using the Combined Gas Law. The formula \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \) allows us to calculate how much the volume of a gas changes when taken from one condition to another. In practice:
- Start by understanding the initial and final conditions: pressure \( P \), volume \( V \), and temperature \( T \).
- Use the formula rearranged as \( \frac{V_2}{V_1} = \frac{P_1 \cdot T_2}{P_2 \cdot T_1} \) to solve for the volume change factor.
Pressure
Pressure is a key factor in gas laws, affecting how gases behave under different conditions. Measured in atm, it represents force applied over a unit area. In the exercise, the transition from 1 atm on the surface to \(10^{-3}\) atm in the stratosphere shows a significant drop in pressure:
- Higher pressures compress gases, reducing volume.
- Lower pressures allow gases to expand, increasing volume.
Gas Laws
Gas laws provide the framework for understanding the behavior of gases under various conditions. The main laws include Boyle's Law, Charles's Law, and Gay-Lussac's Law. However, for complex changes, the Combined Gas Law is used as it incorporates:
- Boyle's principle that pressure and volume are inversely related when temperature is constant.
- Charles's principle that volume and temperature are directly related when pressure is constant.
- Gay-Lussac's principle that pressure and temperature are directly related when volume is constant.
Stratosphere Conditions
The stratosphere, the layer of Earth’s atmosphere above the troposphere, has unique conditions. It starts about 10 km above the ground and extends up to 50 km. Conditions here affect how gases behave:
- Lower atmospheric pressure means gases expand more.
- Temperatures in the stratosphere rise with altitude, after an initial drop, due to ozone absorption of ultraviolet radiation.