Problem 46
Question
A perfectly expandable balloon filled with helium gas at \(27^{\circ} \mathrm{C}\) and a pressure of \(720 \mathrm{~mm}\) of \(\mathrm{Hg}\) has a volume of \(100 \mathrm{l}\). The balloon rises to an altitude where the pressure is \(420 \mathrm{~mm}\) of \(\mathrm{Hg}\) and the temperature \(-53^{\circ} \mathrm{C}\). What is the change in the volume of the balloon? (a) 161 (b) \(25.71\) (c) 481 (d) \(15.71\)
Step-by-Step Solution
Verified Answer
The change in the volume of the balloon is approximately 17.83 liters, but this option is not available in the given choices.
1Step 1: Convert Temperatures to Kelvin
Since gas laws require temperature in Kelvin, convert the initial and final temperatures from Celsius to Kelvin.Initial Temperature, T1: \(27^{\rc}C = 27 + 273.15 = 300.15 K\)Final Temperature, T2: \(-53^{\rc}C = -53 + 273.15 = 220.15 K\)
2Step 2: Convert Pressures to Pascals
To ensure consistency in units, convert the initial and final pressures from mm of Hg to Pascals (Pa) using the conversion factor \(1 \text{mm} \text{Hg} = 133.322 Pa\).Initial Pressure, P1: \(720 \text{mm Hg} = 720 \times 133.322 = 96031.68 Pa\)Final Pressure, P2: \(420 \text{mm Hg} = 420 \times 133.322 = 55995.24 Pa\)
3Step 3: Apply Charles's Law and Boyle's Law Combined
Since the problem involves changes in pressure, volume, and temperature, use the combined gas law formula: \( \frac{P1 \times V1}{T1} = \frac{P2 \times V2}{T2} \).Rearrange the formula to solve for the final volume, V2: \( V2 = \frac{P1 \times V1 \times T2}{P2 \times T1} = \frac{96031.68 Pa \times 100 L \times 220.15 K}{55995.24 Pa \times 300.15 K} \)
4Step 4: Calculate the New Volume
Substitute the known values into the rearranged formula to calculate V2.\( V2 = \frac{96031.68 \times 100 \times 220.15}{55995.24 \times 300.15} \approx 117.83 L \)
5Step 5: Determine the Change in Volume
The change in volume is the final volume minus the initial volume.\( \Delta V = V2 - V1 = 117.83 L - 100 L = 17.83 L \).The change in volume is 17.83 liters. None of the choices provided in the question match this result, so there might be a mistake in the options.
Key Concepts
Charles's lawBoyle's lawIdeal gas lawTemperature conversionPressure conversion
Charles's law
Charles's Law pertains to the behavior of gases as they are subjected to temperature variations with a constant pressure. This law states that the volume of a given mass of an ideal gas is directly proportional to its temperature (measured in Kelvin) when the pressure is held constant.
For example, suppose we have a sealed balloon that is heated. As per Charles's law, the volume of the gas inside the balloon will increase because the temperature increases, if the pressure does not change. The mathematical relationship can be expressed as \(\frac{V1}{T1} = \frac{V2}{T2}\).
This law is crucial in understanding how gases expand or contract with temperature changes and explains why a gas-filled balloon expands when warmed and contracts when cooled.
For example, suppose we have a sealed balloon that is heated. As per Charles's law, the volume of the gas inside the balloon will increase because the temperature increases, if the pressure does not change. The mathematical relationship can be expressed as \(\frac{V1}{T1} = \frac{V2}{T2}\).
This law is crucial in understanding how gases expand or contract with temperature changes and explains why a gas-filled balloon expands when warmed and contracts when cooled.
Boyle's law
Boyle's Law deals with the correlation between the pressure and volume of a gas at constant temperature. Specifically, it asserts that the volume of a gas is inversely proportional to its pressure when temperature remains unchanged.
If you compress a gas by reducing its volume, the pressure will proportionally rise, assuming temperature is constant. Similarly, if you allow a gas to expand, its pressure drops. Boyle’s Law is represented by the equation \(P1 \times V1 = P2 \times V2\),
This inverse relationship is essential in many practical situations, such as when diving underwater where pressure increases with depth causing the volume of air in the diver's lungs to decrease.
If you compress a gas by reducing its volume, the pressure will proportionally rise, assuming temperature is constant. Similarly, if you allow a gas to expand, its pressure drops. Boyle’s Law is represented by the equation \(P1 \times V1 = P2 \times V2\),
This inverse relationship is essential in many practical situations, such as when diving underwater where pressure increases with depth causing the volume of air in the diver's lungs to decrease.
Ideal gas law
The Ideal Gas Law is a fundamental equation that bridges together Boyle's Law, Charles's Law, and Gay-Lussac's Law. It delivers a comprehensive description of the state of an ideal gas by relating its pressure (P), volume (V), temperature (T), and number of moles (n) through the equation \(PV = nRT\).
The 'R' here represents the ideal gas constant. This law allows us to solve for any one of the gas state variables if we know or can control the others.
In the context of combined gas law problems, the Ideal Gas Law is crucial because it provides a more holistic view that considers the amount of gas in addition to the pressure, volume, and temperature.
The 'R' here represents the ideal gas constant. This law allows us to solve for any one of the gas state variables if we know or can control the others.
In the context of combined gas law problems, the Ideal Gas Law is crucial because it provides a more holistic view that considers the amount of gas in addition to the pressure, volume, and temperature.
Temperature conversion
Understanding temperature conversion is imperative because gas laws require temperature to be in Kelvin. The Kelvin scale is an absolute temperature scale which starts at absolute zero—the coldest possible temperature.
To convert Celsius to Kelvin, you can use the formula \(K = ^{\rc}C + 273.15\).
This is vital for accuracy when applying gas laws, as using degrees Celsius would produce incorrect results. For example, 0 degrees Celsius corresponds to 273.15 Kelvin. Always use Kelvin in gas law equations to accurately describe gas behavior.
To convert Celsius to Kelvin, you can use the formula \(K = ^{\rc}C + 273.15\).
This is vital for accuracy when applying gas laws, as using degrees Celsius would produce incorrect results. For example, 0 degrees Celsius corresponds to 273.15 Kelvin. Always use Kelvin in gas law equations to accurately describe gas behavior.
Pressure conversion
Pressure conversion is a fundamental step in solving combined gas law problems because it's essential to use consistent units. Common units for measuring pressure include atmospheres (atm), Pascals (Pa), millimeters of mercury (mm Hg), and torr.
A typical conversion factor that is used frequently is \(1 \text{atm} = 101325 \text{Pa}\).
When converting from mm Hg to Pascals, as in the exercise provided, you use the conversion \(1 \text{mm} \text{Hg} = 133.322 \text{Pa}\).
Consistency in units ensures the accuracy of calculations when using the combined gas law or when switching between different gas law equations.
A typical conversion factor that is used frequently is \(1 \text{atm} = 101325 \text{Pa}\).
When converting from mm Hg to Pascals, as in the exercise provided, you use the conversion \(1 \text{mm} \text{Hg} = 133.322 \text{Pa}\).
Consistency in units ensures the accuracy of calculations when using the combined gas law or when switching between different gas law equations.
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