Question

A sample of helium gas occupies a volume of \(29.4 \mathrm{mL}\) at \(18^{\circ} \mathrm{C} .\) Would the volume of this gas sample be doubled at \(36^{\circ} \mathrm{C} ?\) At what temperature would the volume of the gas sample be half as big? Assume the pressure remains constant.

Short answer

No, the volume of the gas sample is not doubled at \(36^\circ C\). Instead, it is approximately \(31.14 \) mL. The volume of the gas sample would be half as big at approximately \(-127.58 ^\circ C\).

Step-by-step solution

Convert temperatures to Kelvin scale

First, we need to convert temperatures given in Celsius to Kelvin scale. To do this, we will use the following conversion formula: \( K = 273.15 + C\) where K is the temperature in Kelvin and C is the temperature in Celsius. Temperature 1 (\(T_1\)) = \(18^\circ C\) = \(273.15 + 18\) K = \(291.15\) K Temperature 2 (\(T_2\)) = \(36^\circ C\) = \(273.15 + 36\) K = \(309.15\) K

Use Charles' law formula to determine the volume at 36°C

Now we will use Charles' law to determine the volume of the gas sample at \(T_2\). Charles' law states: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) We are given: Temperature 1 (\(T_1\)) = \(291.15\) K Volume 1 (\(V_1\)) = \(29.4\) mL Temperature 2 (\(T_2\)) = \(309.15\) K We need to find Volume 2 (\(V_2\)). Now, we will solve for \(V_2\): \( V_2 = \frac{V_1}{T_1} * T_2\) \( V_2 = \frac{29.4}{291.15} * 309.15 \) \( V_2 ≈ 31.14 \) mL No, the volume of the gas sample is not doubled at \(36^\circ C\).

Find the temperature at which the volume would be half as big

Given that the volume is halved, we have: Volume 3 (\(V_3\)) = \( \frac{29.4}{2} \) mL = 14.7 mL We will again use Charles' law to find temperature 3 (\(T_3\)): \( \frac{V_1}{T_1} = \frac{V_3}{T_3}\) Now, we will solve for \(T_3\): \( T_3 = \frac{V_3}{V_1} * T_1\) \( T_3 = \frac{14.7}{29.4} * 291.15 \) \( T_3 ≈ 145.57 \) K Finally, we will convert this temperature back to Celsius: \(C = T_3 - 273.15\) \(C ≈ -127.58 ^\circ C \) The volume of the gas sample would be half as big at approximately \(-127.58 ^\circ C\).

Key concepts

Gas Laws

Gas laws are essential principles that describe the behavior of gases under various conditions. One of the most fundamental laws is Charles' Law, which explains how gases change volume in relation to temperature, assuming pressure remains constant. Charles' Law is formulated as \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). Here, \( V_1 \) and \( V_2 \) are the initial and final volumes, while \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively. This law emphasizes that the volume of a gas expands as the temperature increases and contracts as it decreases. Consider it as a prediction tool for understanding how gases will behave if the temperature changes while pressure stays the same.

To apply Charles' Law, you only need the initial and final temperatures and the initial volume. From there, you can predict the final volume. This underlying principle is what allows us to solve exercises that seek to find unknown variables, such as the new volume of a gas sample after a temperature change. Remember, consistency is key: temperature must always be in Kelvin for these calculations.

Temperature Conversion

Understanding how to convert temperatures is crucial when dealing with gas laws, especially Charles' Law. Temperatures in gas law calculations need to be in Kelvin because Kelvin is the absolute temperature scale. The conversion formula between Celsius and Kelvin is simple: \( K = 273.15 + C \).

Let's break it down with an example: if the temperature is \( 18 ^\circ C \), converting to Kelvin gives us \( 291.15 \) K. Similarly, for \( 36 ^\circ C \), it converts to \( 309.15 \) K. This conversion ensures that calculations account for the zero point of temperature, as the Kelvin scale starts at absolute zero, where molecular motion ceases.

Without this conversion, any calculations with gas laws might end up incorrect, potentially leading to misunderstandings or error. To ensure accuracy, it is paramount to verify each temperature conversion step in your calculations.

Volume Calculation

Calculating volume changes is at the heart of problems that use Charles' Law, like determining whether the volume of a gas sample doubles at a new temperature. First, it's crucial to identify all given information, like initial volume and initial and final temperatures. Subsequently, temperature conversion to Kelvin follows.

Once you have the temperatures in Kelvin, apply Charles' Law to solve for the unknown, be it a new volume or temperature. Using the equation \( V_2 = \frac{V_1}{T_1} \times T_2 \), you can find the new volume \( V_2 \). For example, calculating the volume change of a helium gas sample from \( 291.15 \) K to \( 309.15 \) K:

• Select the known values: \( V_1 = 29.4 \) mL, \( T_1 = 291.15 \) K, and \( T_2 = 309.15 \) K.
• Replace them in the Charles' Law formula, resulting in a new volume \( V_2 \approx 31.14 \) mL.

If you need to find the temperature at which volume changes, such as when it halves, you rearrange and solve for temperature instead. This involves reversing the formula steps, exemplifying how versatile and powerful these calculations are in uncovering hidden relationships between volume and temperature.