Question

What will be the new volume if \(125 \mathrm{mL}\) of He gas at \(100^{\circ} \mathrm{C}\) and 0.981 atm is cooled to \(25^{\circ} \mathrm{C}\) and the pressure is increased to 1.15 atm?

Short answer

The new volume of He gas when cooled down to \(25^\circ\mathrm{C}\) and the pressure is increased to 1.15 atm is approximately \(97.54\ \mathrm{mL}\).

Step-by-step solution

Convert temperatures to Kelvin

The Combined Gas Law requires temperatures to be in Kelvin. So, we must convert the given temperatures from Celsius to Kelvin, using the formula \(K = \mathrm{C} + 273.15\): \[ \begin{aligned} T_1 &= 100^\circ\mathrm{C} + 273.15 = 373.15\ \mathrm{K} \\ T_2 &= 25^\circ\mathrm{C} + 273.15 = 298.15\ \mathrm{K} \end{aligned} \]

Rearrange the Combined Gas Law to solve for V2

The Combined Gas Law is given by: \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \] To solve for the new volume (\(V_2\)), we need to isolate \(V_2\) in the equation: \[ V_2 = \frac{P_1V_1T_2}{P_2T_1} \]

Plug in the given values and calculate the new volume

Now, substitute the given values and the converted temperatures into the equation from Step 2: \[ V_2 = \frac{(0.981\ \mathrm{atm})(125\ \mathrm{mL})(298.15\ \mathrm{K})}{(1.15\ \mathrm{atm})(373.15\ \mathrm{K})} \] Calculate the new volume: \[ V_2 \approx 97.54\ \mathrm{mL} \] Therefore, the new volume of He gas when cooled down to \(25^\circ\mathrm{C}\) and the pressure is increased to 1.15 atm is approximately \(97.54\ \mathrm{mL}\).

Key concepts

Temperature Conversion to Kelvin

When dealing with gas law calculations, it's crucial to have all temperatures in Kelvin. The Kelvin scale is an absolute temperature scale based directly on thermodynamic principles. This means zero on the Kelvin scale, or absolute zero, corresponds to the point where all molecular motion stops.
To convert from Celsius to Kelvin, we add 273.15 to the Celsius temperature. This conversion is vital because most gas law equations, like the Combined Gas Law, use temperatures in Kelvin. Remember, Kelvin helps maintain consistency in calculations as it prevents negative temperatures, which can't exist in this context.
For example:

• If you have a temperature of 100°C, the Kelvin conversion would be: 100 + 273.15 = 373.15 K.
• For 25°C, it becomes: 25 + 273.15 = 298.15 K.

By converting temperatures to Kelvin right at the start, it ensures accuracy in calculating new conditions such as volume or pressure for gases.

Gas Law Calculations

Gas law calculations often involve relationships between pressure, volume, and temperature. One key formula that combines these three variables is the Combined Gas Law. This law is expressed as:\[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\] where:

• \( P_1 \) and \( P_2 \) are initial and final pressures.
• \( V_1 \) and \( V_2 \) are initial and final volumes.
• \( T_1 \) and \( T_2 \) are initial and final temperatures in Kelvin.

This equation allows us to calculate unknown variables when a gas experiences changes under non-constant conditions. To find a new volume \( V_2 \), rearrange to:\[V_2 = \frac{P_1V_1T_2}{P_2T_1}\]Substitute known values to determine the unknown. Each calculation step relies on accurate and consistent measurement units—key to understanding the behavior of gases under varying conditions.

Helium Gas Properties

Helium is a noble gas, known for its stability and non-reactivity due to having a full valence electron shell. This makes it ideal for experiments and calculations involving gas laws. Helium's low density and ability to remain inert even at higher energies make it unique among gases.
In the context of gas law problems, helium's properties ensure that calculations focus strictly on the variables of pressure, volume, and temperature without additional reactivity concerns. Moreover, because helium has a low molecular weight, it often escapes through small openings, illustrating its small atomic size.
Understanding these properties can explain why helium is used in various applications. This inert and lightweight gas features in balloons, cooling systems, and even breathing mixes for deep-sea divers. In gas law calculations, adopting helium can simplify the process as there are few to no chemical interactions interfering with theoretical results such as volume changes under different conditions.