Problem 67
Question
A sample of a gas enclosed in a cylinder with a piston has a volume of \(500.0 \mathrm{mL}\) at \(30.0^{\circ} \mathrm{C} .\) What is the sample volume at \(100.0^{\circ} \mathrm{C} ?\)
Step-by-Step Solution
Verified Answer
Answer: The volume of the gas sample at 100.0°C will be approximately 615.22 mL.
1Step 1: Convert the temperatures from Celsius to Kelvin
To compare the volume of the gas between two different temperatures, we must first convert the initial and final temperatures from Celsius to Kelvin. To do this, apply the formula:
\(T_{(K)} = T_{(\textdegree C)} + 273.15\)
Given:
- \(T_1 = 30.0^{\circ} \mathrm{C}\)
- \(T_2 = 100.0^{\circ} \mathrm{C}\)
Convert to Kelvin:
- \(T_1 = 30.0 + 273.15 = 303.15\,\mathrm{K}\)
- \(T_2 = 100.0 + 273.15 = 373.15\,\mathrm{K}\)
2Step 2: Apply Charles' Law to find the new volume of the gas
Now, we can use Charles' Law to find the volume of the gas at the given temperature:
\(V_1 / T_1 = V_2 / T_2\)
We know:
- \(V_1 = 500.0 \,\mathrm{mL}\)
- \(T_1 = 303.15\,\mathrm{K}\)
- \(T_2 = 373.15\,\mathrm{K}\)
Find \(V_2\):
- \(V_2 = V_1 \times (T_2 / T_1) = (500.0\,\mathrm{mL}) \times (\frac{373.15\,\mathrm{K}}{303.15\,\mathrm{K}})\)
- \(V_2 \approx 615.22\,\mathrm{mL}\)
The volume of the gas sample at \(100.0^{\circ} \mathrm{C}\) will be approximately \(615.22\,\mathrm{mL}\).
Key Concepts
Temperature ConversionGas LawsVolume-Temperature Relationship
Temperature Conversion
Understanding temperature conversion is vital in various scientific calculations, especially when dealing with gas laws. Celsius and Kelvin are two scales used to measure temperature. While Celsius is commonly used for daily temperature readings, Kelvin is the SI unit for temperature and is used in scientific equations to avoid negative values that Celsius can present.
To convert Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. The formula for the conversion is:
\[ T_{(K)} = T_{(\textdegree C)} + 273.15 \]
Remember, since the size of one degree is the same on both scales, this conversion doesn't change the temperature difference, it only shifts the scale. This is crucial for calculations in gas laws, which require a consistent temperature scale, and for preventing errors that might occur if negative temperatures were used.
To convert Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. The formula for the conversion is:
\[ T_{(K)} = T_{(\textdegree C)} + 273.15 \]
Remember, since the size of one degree is the same on both scales, this conversion doesn't change the temperature difference, it only shifts the scale. This is crucial for calculations in gas laws, which require a consistent temperature scale, and for preventing errors that might occur if negative temperatures were used.
Gas Laws
Gas laws describe the behavior of gases and how they react to changes in temperature, pressure, and volume. These are foundational to understanding how gases will act under different conditions. The primary gas laws are Boyle's Law, Charles' Law, Gay-Lussac's Law, and Avogadro's Law.
- Boyle's Law: Defines the inverse relationship between pressure and volume at a constant temperature.
- Charles' Law: Describes how the volume of a gas is directly proportional to its temperature, when pressure is held constant.
- Gay-Lussac's Law: States that the pressure of a gas is directly proportional to its temperature, when the volume is held constant.
- Avogadro’s Law: Shows that volume is directly proportional to the number of gas molecules, or moles, at a constant temperature and pressure.
Volume-Temperature Relationship
The volume-temperature relationship for gases is best explained by Charles' Law, which states that the volume of a gas is directly proportional to its absolute temperature (measured in Kelvin), provided the pressure remains constant. Mathematically, it is expressed as:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where V represents volume, T represents temperature, and the subscripts 1 and 2 refer to the initial and final states of the gas, respectively. To calculate the new volume after a change in temperature, we must ensure both temperatures are in Kelvin. Then, we can rearrange the equation to solve for the unknown volume:
\[ V_2 = V_1 \times \frac{T_2}{T_1} \]
From the step-by-step solution provided, we see this law in action as the temperature of the gas increases leading to an increase in volume. It offers a predictable pattern of how gases expand when heated and contract when cooled, critical for multiple applications, from weather balloons to respiratory ventilation systems.
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where V represents volume, T represents temperature, and the subscripts 1 and 2 refer to the initial and final states of the gas, respectively. To calculate the new volume after a change in temperature, we must ensure both temperatures are in Kelvin. Then, we can rearrange the equation to solve for the unknown volume:
\[ V_2 = V_1 \times \frac{T_2}{T_1} \]
From the step-by-step solution provided, we see this law in action as the temperature of the gas increases leading to an increase in volume. It offers a predictable pattern of how gases expand when heated and contract when cooled, critical for multiple applications, from weather balloons to respiratory ventilation systems.
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