Problem 10 - General Chemistry: Principles and Modern Applications

Question

An 886 mL sample of \(\mathrm{Ne}(\mathrm{g})\) is at \(752 \mathrm{mmHg}\) and \(26^{\circ} \mathrm{C}\). What will be the new volume if, with the pressure and amount of gas held constant, the temperature is (a) increased to \(98^{\circ} \mathrm{C} ;\) (b) lowered to \(-20^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The new volume when the temperature is increased to \(98^{\circ} \mathrm{C}\) is about 1094.24 mL, and the new volume when the temperature is lowered to -20^{\circ} \mathrm{C} is about 751.54 mL.

Step 1 Convert temperatures to Kelvin

Convert all Celsius temperatures to Kelvin since the gas laws are based on the absolute (Kelvin) temperature scale. You can do this by adding 273.15 to the Celsius temperature.\nSo, for the initial temperature: \(26^{\circ}\mathrm{C} + 273.15 = 299.15 \mathrm{K}\) \n(a) For the final temperature when increased to \(98^{\circ}\mathrm{C}\): \(98^{\circ}\mathrm{C} + 273.15 = 371.15 \mathrm{K}\) (b) For the final temperature when lowered to \(-20^{\circ}\mathrm{C}\): \(-20^{\circ}\mathrm{C} + 273.15 = 253.15 \mathrm{K}\)

Step 2 Use Charles's Law to find the new volume

Use Charles's Law: \(V_1/T_1 = V_2/T_2\), where \(V_1\) and \(T_1\) denote the initial volume and temperature, and \(V_2\) and \(T_2\) denote the final volume and temperature. (a) Solve for the new volume \(V_2\) when the temperature is increased to \(98^{\circ}\mathrm{C} (or 371.15\mathrm{K}\): \((886\mathrm{mL}/299.15\mathrm{K}) = V_2/371.15\mathrm{K}\) \n\nThis yields \(V_2 = (886\mathrm{mL}/299.15\mathrm{K}) * 371.15\mathrm{K} \)\n (b) Solve for the new volume \(V_2\) when the temperature is lowered to \(-20^{\circ}\mathrm{C} (or 253.15\mathrm{K}\): \((886mL/299.15\mathrm{K}) = V_2/253.15\mathrm{K}\)\n\nThis yields \(V_2 = (886\mathrm{mL}/299.15\mathrm{K}) * 253.15\mathrm{K}\)

Step 3 Calculate the new volumes

Perform the calculations in Step 2 to get the new volumes. (a) For the increased temperature: \(V_2 = (886\mathrm{mL}/299.15\mathrm{K}) * 371.15\mathrm{K} = 1094.24 \mathrm{mL}\) (rounded to two decimal places) (b) For the decreased temperature: \(V_2 = (886\mathrm{mL}/299.15\mathrm{K}) * 253.15\mathrm{K} = 751.54 \mathrm{mL}\) (rounded to two decimal places)

Key Concepts

Gas Laws

Gas laws describe how gases behave under varying conditions of temperature, pressure, and volume. One of the fundamental gas laws is Charles's Law. This law states that, under constant pressure, the volume of a gas is directly proportional to its temperature when measured in Kelvin. This relationship can be expressed as \( V_1/T_1 = V_2/T_2 \), where \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin, respectively. This means that as the temperature of a gas increases, its volume increases, provided the pressure remains constant. Conversely, if the temperature decreases, the volume will decrease as well. A practical application of Charles's Law is understanding why hot air balloons rise when the air inside is heated; the increase in temperature causes the air's volume to expand, making the balloon buoyant.

Temperature Conversion

When dealing with gas laws, it is essential to use the Kelvin scale for temperatures because it is an absolute scale. The Kelvin temperature scale starts at absolute zero, where molecular motion ceases, and it makes the proportional relationships in gas laws simpler and more accurate. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. For example, in the exercise, the initial temperature is given as \( 26^{\circ}\mathrm{C} \), which converts to \( 26^{\circ}\mathrm{C} + 273.15 = 299.15 \mathrm{K} \). Meanwhile, at \( 98^{\circ}\mathrm{C} \), it becomes \( 371.15 \mathrm{K} \). If the temperature is lowered to \( -20^{\circ}\mathrm{C} \), it becomes \( 253.15 \mathrm{K} \). Using Kelvin ensures consistent and reliable calculations across various gas law applications.

Volume Calculation

Volume calculation using Charles's Law involves determining the final volume of a gas when the temperature changes, while pressure remains constant. You rearrange the formula \( V_1/T_1 = V_2/T_2 \) to solve for the unknown volume \( V_2 \) by multiplying both sides by \( T_2 \), yielding \( V_2 = V_1 \times (T_2 / T_1) \). In the exercise example, if you're increasing the temperature from \( 299.15 \mathrm{K} \) to \( 371.15 \mathrm{K} \), you would calculate the new volume as \( V_2 = (886 \mathrm{mL} / 299.15 \mathrm{K}) \times 371.15 \mathrm{K} = 1094.24 \mathrm{mL} \). Conversely, if the temperature is lowered to \( 253.15 \mathrm{K} \), you calculate \( V_2 = (886 \mathrm{mL} / 299.15 \mathrm{K}) \times 253.15 \mathrm{K} = 751.54 \mathrm{mL} \). This method allows you to predict how the volume of a gas will change under different temperature conditions.