Question

Calculate the final Celsius temperature of sulfur dioxide gas if \(50.0 \mathrm{~mL}\) of the gas at \(20^{\circ} \mathrm{C}\) and 0.450 atm is heated until the pressure is 0.750 atm. Assume that the volume remains constant.

Short answer

The final temperature of the sulfur dioxide gas is approximately \( 215.43^{\circ} \mathrm{C} \).

Step-by-step solution

Identify Known Values

We know the initial temperature \( T_1 = 20^{\circ} \mathrm{C} \) and the initial pressure \( P_1 = 0.450 \) atm. The final pressure \( P_2 \) is 0.750 atm. The volume is constant, so \( V_1 = V_2 \). We need to find the final temperature \( T_2 \).

Convert Celsius to Kelvin

Temperature in gas law calculations should be in Kelvin. Convert \( T_1 = 20^{\circ} \mathrm{C} \) to Kelvin. \( T_1 = 20 + 273.15 = 293.15 \) K.

Use Gay-Lussac's Law

With the volume constant, Gay-Lussac's Law \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) applies. Solve for \( T_2 \): \( T_2 = \frac{P_2 \cdot T_1}{P_1} \).

Substitute Known Values

Substitute the known values into the equation: \( T_2 = \frac{0.750 \times 293.15}{0.450} \).

Perform the Calculation

Calculate \( T_2 \): \( T_2 = \frac{0.750 \times 293.15}{0.450} = 488.5833 \) K.

Convert Kelvin to Celsius

Convert \( T_2 \) from Kelvin to Celsius: \( T_2 = 488.5833 - 273.15 = 215.4333 \approx 215.43^{\circ} \mathrm{C} \).

Key concepts

Gas Laws

Gas laws describe how gases behave in response to changes in temperature, pressure, and volume. They are a set of mathematical relationships that help us understand and predict the behavior of gases under various conditions. These laws include Boyle's Law, Charles's Law, and Gay-Lussac's Law. Each one defines the relationship between two of the three variables when the third is held constant.
Let's explore Gay-Lussac's Law, which is particularly relevant to the given exercise. This law states that for a given mass and constant volume, the pressure of a gas is directly proportional to its temperature in Kelvin. Mathematically, it is expressed as:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
In simple terms, if you increase the temperature of a gas while keeping the volume constant, the pressure will also increase. Likewise, if you decrease the temperature, the pressure will drop. This is because heating a gas causes its particles to move more rapidly, which increases the frequency and force of their collisions with the container walls, thus increasing pressure.
By understanding Gay-Lussac's Law, you can solve problems involving changes in temperature or pressure when the other variable is known.

Temperature Conversion

When dealing with gas laws, it's crucial to express temperature in Kelvin rather than Celsius. This is because Kelvin is an absolute temperature scale where 0 K represents absolute zero, the point at which particles theoretically have minimal kinetic energy.
Converting from Celsius to Kelvin is straightforward. Simply add 273.15 to the Celsius temperature, as Kelvin starts at absolute zero, which corresponds to -273.15°C. For example, if the temperature is 20°C, the conversion is:
\[ T_{K} = 20^{\circ} C + 273.15 = 293.15 \text{ K} \]
This conversion is essential as gas laws, like Gay-Lussac's Law, depend on temperatures being in Kelvin for the calculations to remain consistent. This avoids any negative values which could complicate mathematical manipulations involving temperature-ratio calculations in gas laws.

Pressure Calculation

Pressure calculations are a key part of utilizing Gay-Lussac's Law effectively. When given an initial and final pressure, as well as an initial temperature, you can calculate the final temperature of a gas when the volume is constant.
Begin by organizing the known quantities, such as initial pressure \( P_1 \), final pressure \( P_2 \), and initial temperature in Kelvin \( T_1 \). With these, you can rearrange Gay-Lussac's equation to solve for the final temperature \( T_2 \):
\[ T_2 = \frac{P_2 \cdot T_1}{P_1} \]
Substitute the known values into the equation to find \( T_2 \). In our example, with \( P_1 = 0.450 \) atm, \( P_2 = 0.750 \) atm, and \( T_1 = 293.15 \) K, we calculate:
\[ T_2 = \frac{0.750 \times 293.15}{0.450} \approx 488.58 \text{ K} \]
Finally, convert \( T_2 \) back to Celsius if needed by subtracting 273.15 from the Kelvin temperature. This kind of calculation helps in predicting how changing one condition affects others in real-world situations involving gases.