Problem 12
Question
A balloon contains 146.0 mL of gas confined at a pressure of 1.30 atm and a temperature of 5.0ºC. If the pressure doubles and the temperature decreases to 2.0ºC, what will be the volume of gas in the balloon?
Step-by-Step Solution
Verified Answer
The volume of the gas in the balloon after the pressure doubles and the temperature decreases to 2.0ºC is approximately \(142.6\,\text{mL}\).
1Step 1: Understand the problem and write initial and final conditions of the gas
First, let's write down the initial and final conditions of the gas:
Initial state:
Volume = \(V_1 = 146.0\,\text{mL}\)
Pressure = \(P_1 = 1.30\,\text{atm}\)
Temperature = \(T_1 = 5.0\,^\circ\text{C}\)
Final state:
Volume = \(V_2=?\)
Pressure = \(P_2 = 2P_1 = 2(1.30)\,\text{atm}\)
Temperature = \(T_2 = 2.0\,^\circ\text{C}\)
We need to find the volume of the gas in its final state, \(V_2\).
2Step 2: Convert the temperatures to Kelvin
Since the Combined Gas Law requires the temperatures to be in Kelvin, we need to convert the Celsius temperatures to Kelvin. To do this, we can use the following formula: \(T(K) = T(^\circ\text{C}) + 273.15\)
Initial temperature in Kelvin:
\(T_1 = 5.0\,^\circ\text{C} + 273.15 = 278.15\,\text{K}\)
Final temperature in Kelvin:
\(T_2 = 2.0\,^\circ\text{C} + 273.15 = 275.15\,\text{K}\)
Now we have the temperatures in Kelvin:
\(T_1 = 278.15\,\text{K}\)
\(T_2 = 275.15\,\text{K}\)
3Step 3: Apply Combined Gas Law formula
The Combined Gas Law formula is given by: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)
Now we can substitute the values of the initial and final pressure, volume, and temperatures in this formula:
\(\frac{1.30\,\text{atm} \times 146.0\,\text{mL}}{278.15\,\text{K}} = \frac{(2\times1.30)\,\text{atm} \times V_2}{275.15\,\text{K}}\)
4Step 4: Solve for \(V_2\)
To solve for \(V_2\), we will first simplify the equation:
\(V_2 = \frac{(1.30\,\text{atm} \times 146.0\,\text{mL})\times 275.15\,\text{K}}{(2\times 1.30) \,\text{atm} \times 278.15\,\text{K}}\)
Then calculate the value of \(V_2\):
\(V_2 \approx 142.6\,\text{mL}\)
5Step 5: State the final answer
The volume of the gas in the balloon after the pressure doubles and the temperature decreases to 2.0ºC is approximately 142.6 mL.
Key Concepts
Understanding Temperature ConversionGas Volume Calculation Using the Combined Gas LawPressure and Temperature Relationship in Gases
Understanding Temperature Conversion
In the Combined Gas Law, temperature is an essential component because it affects the state of gases. Typically, temperatures must be expressed in Kelvin for gas law calculations. This conversion from Celsius to Kelvin ensures that calculations remain accurate, as Kelvin begins at absolute zero, the theoretical point where particles cease to move. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, if you have a temperature of 5.0°C, its conversion to Kelvin would work like this:
- Convert using \(T(K) = T(\, ^\circ\text{C}) + 273.15 \)
- 5.0°C becomes \(5.0 + 273.15 = 278.15\, \text{K}\)
Gas Volume Calculation Using the Combined Gas Law
Calculating gas volume under changing conditions requires a solid understanding of the Combined Gas Law. This law relates the pressure, volume, and temperature of a gas by showing how they interact. The formula is represented as \[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]where:
- \(P_1\), \(V_1\), and \(T_1\) are the initial pressure, volume, and temperature.
- \(P_2\), \(V_2\), and \(T_2\) are the final pressure, volume, and temperature.
Pressure and Temperature Relationship in Gases
The relationship between pressure and temperature within the context of gases is crucial to understanding the behavior of gases under various conditions. According to the Combined Gas Law, temperature changes can lead directly to changes in pressure, assuming the volume of the gas is constant.
Here's why this happens:
- As temperature increases, the kinetic energy of gas particles increases. This causes particles to collide more forcefully with the walls of their container, resulting in higher pressure.
- Conversely, a decrease in temperature reduces particle movement, leading to decreased pressure.
Other exercises in this chapter
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