Question

A sample of oxygen gas has a volume of 20.0 \(\mathrm{cm}^{3}\) at \(-10.0^{\circ} \mathrm{C}\) . What volume will this sample occupy if the temperature rises to \(110^{\circ} \mathrm{C} ?(\text {Chapter } 13)\)

Short answer

The final volume (V2) of the oxygen gas when the temperature rises to 110.0\(^{\circ}\mathrm{C}\) is approximately 29.12 cm³.

Step-by-step solution

Write down the given information and convert to Kelvin

To work with the gas laws, we need to convert the temperatures from Celsius to Kelvin. We can convert Celsius to Kelvin by adding 273.15. Initial volume (V1): 20.0 cm³ Initial temperature (T1): -10.0\(^{\circ}\mathrm{C}\) = (-10.0 + 273.15) K = 263.15 K Final temperature (T2): 110.0\(^{\circ}\mathrm{C}\) = (110.0 + 273.15) K = 383.15 K Now we have: V1 = 20.0 cm³ T1 = 263.15 K T2 = 383.15 K We need to find the final volume (V2) of the oxygen gas at T2.

Apply Charles' Law

Charles' Law states that for a given amount of gas (constant pressure), the volume is directly proportional to its temperature. Therefore, we can set up the following equation: \[\frac{V1}{T1} = \frac{V2}{T2}\] We are solving for V2, so we can rearrange the equation as: \[V2 = \frac{T2}{T1} \times V1\]

Plug in the values and solve for V2

Now we can substitute the given values of V1, T1, and T2 into the equation: \[V2 = \frac{383.15 K}{263.15 K} \times 20.0 cm^{3}\] \[V2 = 1.456 \times 20.0 cm^{3}\] \[V2 = 29.12 cm^{3}\]

Write the final answer

The final volume (V2) of the oxygen gas when the temperature rises to 110.0\(^{\circ}\mathrm{C}\) is approximately 29.12 cm³.

Key concepts

Gas Laws

When studying the behavior of gases, understanding the gas laws is fundamental. These laws describe the relationship between the pressure, temperature, volume, and amount of gas. Among the various gas laws, Charles' Law plays a crucial role, especially when examining how gas volume is affected by temperature changes under constant pressure.

One of the noteworthy points about gas laws is that they are based on ideal conditions, meaning that the gases are assumed to have perfectly elastic collisions and do not have any intermolecular forces. While real gases do not perfectly adhere to these laws, the laws provide a close approximation for most conditions.

Temperature and Volume Relationship

Charles' Law focuses specifically on the relationship between temperature and volume. According to this law, when the pressure of a gas is held constant, its volume is directly proportional to its absolute temperature (measured in Kelvin). In simple terms, as the temperature of a gas increases, its volume will also increase, provided the amount of gas and the pressure remain stable.

When visualized on a graph with temperature on the x-axis and volume on the y-axis, this relationship is represented by a straight line that passes through the origin. This linearity arises because temperature acts as a driving force for the gas particles to move and occupy more space, thus expanding the volume. Consequently, cooling a gas will lead to a decrease in its volume since the particles lose kinetic energy and move less.

Kelvin Temperature Scale

The Kelvin temperature scale is critical when working with gas laws as it provides an absolute scale for measuring temperature. This scale starts at absolute zero, which is the theoretical temperature where all particle motion ceases. The Kelvin scale is favored in scientific calculations because it avoids negative numbers, simplifies certain types of calculations, and directly correlates with the kinetic energy of particles.

One essential aspect to remember is that temperature in Kelvin is not expressed in degrees like Celsius (°C) or Fahrenheit (°F); it's simply 'Kelvin' (K). When converting from Celsius to Kelvin, which is necessary to apply Charles' Law, you add 273.15 to the Celsius temperature. This conversion allows us to work with proportional relationships between volume and temperature, ensuring accurate predictions of gas behavior under varying thermal conditions.