Question

A fixed quantity of gas at \(25^{\circ} \mathrm{C}\) exhibits a pressure of \(99 \mathrm{kPa}\) and occupies a volume of \(4.00 \mathrm{~L}\) (a) Calculate the volume the gas will occupy if the pressure is increased to \(202.6 \mathrm{kPa}\) while the temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(100^{\circ} \mathrm{C}\) while the pressure is held constant.

Short answer

(a) The volume the gas would occupy when the pressure is increased to \(202.6 \text{kPa}\) while the temperature is held constant is approximately \(1.96 \text{L}\). (b) The volume the gas would occupy when the temperature is increased to \(100^{\circ} \text{C}\) while the pressure is held constant is approximately \(5.01 \text{L}\).

Step-by-step solution

Identify the variables and gather given information

We are given the initial conditions of a fixed quantity of gas: pressure \(P_1 = 99 \text{kPa}\), volume \(V_1 = 4.00 \text{L}\), and temperature \(T_1 = 25^{\circ} \text{C}\). We need to calculate the final volume \(V_2\), under two different conditions: (a) The pressure \(P_2 = 202.6 \text{kPa}\), temperature held constant. (b) The temperature increased to \(T_2 = 100^{\circ} \text{C}\), pressure held constant.

Convert temperature values from Celsius to Kelvin

To use the ideal gas law or Boyle's law equations, we need to convert the temperature values from Celsius to Kelvin. This can be done by adding \(273.15\) to each temperature value: For \(T_1\): \[T_1 = 25^{\circ} \text{C} + 273.15\text{K} = 298.15 \text{K}\] For \(T_2\): \[T_2 = 100^{\circ} \text{C} + 273.15\text{K} = 373.15 \text{K}\]

Use Boyle's law for part (a)

For part (a), where the pressure is changed and the temperature is held constant, we use Boyle's law to calculate the volume: \[P_1V_1 = P_2V_2\] We will solve for \(V_2\) by isolating the variable: \[V_2 = \frac{P_1V_1}{P_2}\] Then plug in the given values: \[V_2 = \frac{99 \text{kPa} \times 4.00 \text{L}}{202.6 \text{kPa}}\] And compute the final volume \(V_2\): \[V_2 \approx 1.96 \text{L}\] Thus, the volume the gas would occupy when the pressure is increased to \(202.6 \text{kPa}\), while the temperature is held constant, is approximately \(1.96 \text{L}\).

Use the ideal gas law for part (b)

For part (b), where the temperature is increased and the pressure is held constant, we have to find the final volume \(V_2\) using the ideal gas law equation: \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\] As the pressure remains constant (\(P_1 = P_2\)), we can simplify the equation: \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\] Next, we isolate \(V_2\): \[V_2 = \frac{V_1 \times T_2}{T_1}\] Plugging in the given values: \[V_2 = \frac{4.00 \text{L} \times 373.15 \text{K}}{298.15 \text{K}}\] And compute the final volume \(V_2\): \[V_2 \approx 5.01 \text{L}\] Thus, the volume the gas would occupy when the temperature is increased to \(100^{\circ} \text{C}\), while the pressure is held constant, is approximately \(5.01 \text{L}\).

Key concepts

Boyle's Law

Boyle's Law is a fundamental principle in the realm of gas laws. It establishes the relationship between pressure and volume of a gas when temperature is constant. Boyle's Law states that the pressure of a given mass of gas is inversely proportional to its volume, provided the temperature remains unchanged. Mathematically, it is expressed as:\[ P_1 V_1 = P_2 V_2 \]Here, \( P_1 \) and \( P_2 \) represent the initial and final pressures, while \( V_1 \) and \( V_2 \) denote the initial and final volumes respectively. If the pressure of a gas increases, its volume decreases, and vice versa, assuming the temperature remains constant. This principle was used in part (a) of the exercise to determine the new volume of a gas when its pressure was changed from 99 kPa to 202.6 kPa under constant temperature conditions. Using Boyle's Law ensures that changes in conditions do not violate the intrinsic properties of gases.Some key takeaways on Boyle's Law:

• It applies only to closed systems where the temperature remains constant.
• Demonstrates the inverse relationship between pressure and volume.
• Helpful in numerous practical applications such as breathing and hydraulic systems.

Ideal Gas Law

The Ideal Gas Law is a cornerstone in the study of gases, especially when multiple variables change. It combines several gas laws into one comprehensive equation:\[ PV = nRT \]In this law, \( P \) is the pressure, \( V \) is the volume, \( n \) signifies the amount of gas in moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This formula provides a way to calculate any state variable of the gas—pressure, volume, or temperature—given the others. It is versatile and can be applied when conditions shift from one state to another.For part (b) of the exercise, where the temperature changes while pressure remains constant, a similar proportional relationship was used from the Ideal Gas Law. Assuming pressure stability, the equation simplifies to: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] This formula allowed for the calculation of the final volume when temperature increased, showcasing how pivotal the Ideal Gas Law is for understanding gas behavior.Tips for using the Ideal Gas Law:

• Remember to always convert temperatures to Kelvin.
• Ensure units are consistent when substitutes into the equation.
• Useful in determining final states after multiple variable changes.

Kelvin temperature conversion

Converting temperatures from Celsius to Kelvin is an essential step in solving problems with gas laws. Gas laws require absolute temperature scales to ensure proportional relationships, and the Kelvin scale provides this by starting at absolute zero. The simple conversion formula is:\[ T(K) = T(^{\circ}C) + 273.15 \]This equation adds 273.15 to the Celsius temperature, aligning perfectly with the absolute temperature scale.In scenarios where gases are evaluated under varying conditions, such as in this exercise, using Kelvin maintains the integrity of calculations involving the Ideal Gas Law or any thermodynamic principles, since it avoids negative values that can skew results by making the relationships non-linear.Why Kelvin?

• Provides a true scale starting from absolute zero, making it essential for thermodynamics.
• Necessary for calculations involving gas laws to ensure proportional relationships hold.
• Avoids non-physical computations like negative volumes and pressures.

Make sure always to convert your temperatures when dealing with gas laws to prevent common mistakes and errors in your calculations.