Question

For each of the following sets of volume/temperature data, calculate the missing quantity. Assume that the pressure and the amount of gas remain constant. a. \(V=25 \mathrm{mL}\) at \(25^{\circ} \mathrm{C} ; V=?\) at \(0^{\circ} \mathrm{C}\) b. \(V=10.2 \mathrm{L}\) at \(100 .^{\circ} \mathrm{C} ; V=?\) at \(100 \mathrm{K}\) c. \(V=551 \mathrm{mL}\) at \(75^{\circ} \mathrm{C} ; V=1.00 \mathrm{mL}\) at? \(^{\circ} \mathrm{C}\)

Short answer

The missing quantities for each set of data are as follows: a. V2 ≈ 22.97 mL b. V2 ≈ 2.73 L c. T2 ≈ -272.4°C

Step-by-step solution

Convert temperatures to Kelvin

We need to convert Celsius to Kelvin using the formula: K = °C + 273 \(T_1 = 25^{\circ} \mathrm{C} + 273 = 298 \mathrm{K}\) \(T_2 = 0^{\circ} \mathrm{C} + 273 = 273 \mathrm{K}\)

Apply Charles's Law

Use the formula \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), and plug in the given values and the converted temperatures. \(\frac{25 \mathrm{mL}}{298 \mathrm{K}} = \frac{V_2}{273 \mathrm{K}}\)

Solve for V2

Cross multiply and isolate V2. \(V_2 = \frac{25 \mathrm{mL} \times 273 \mathrm{K}}{298 \mathrm{K}} = 22.97 \mathrm{mL}\) V2 = 22.97 mL (approx.) b. \(V_1 = 10.2 \mathrm{L}\) at \(T_1 = 100 .^{\circ} \mathrm{C}\); \(V_2 = ?\) at \(T_2 = 100 \mathrm{K}\) To find the missing volume V2, we will follow the steps similar to part a:

Convert temperatures to Kelvin

We need to convert Celsius to Kelvin using the formula: K = °C + 273 \(T_1 = 100^{\circ} \mathrm{C} + 273 = 373 \mathrm{K}\) \(T_2 = 100 \mathrm{K}\)

Apply Charles's Law

Use the formula \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), and plug in the given values and the converted temperatures. \(\frac{10.2 \mathrm{L}}{373 \mathrm{K}} = \frac{V_2}{100 \mathrm{K}}\)

Solve for V2

Cross multiply and isolate V2. \(V_2 = \frac{10.2 \mathrm{L} \times 100 \mathrm{K}}{373 \mathrm{K}} = 2.73 \mathrm{L}\) V2 = 2.73 L (approx.) c. \(V_1 = 551 \mathrm{mL}\) at \(T_1 = 75^{\circ} \mathrm{C}\); \(V_2 = 1.00 \mathrm{mL}\) at \(T_2=?^{\circ} \mathrm{C}\) To find the missing temperature T2, we will follow these steps:

Convert temperatures to Kelvin

We need to convert Celsius to Kelvin using the formula: K = °C + 273 \(T_1 = 75^{\circ} \mathrm{C} + 273 = 348 \mathrm{K}\) We will find T2 in Kelvin and then convert it back to Celsius.

Apply Charles's Law

Use the formula \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), and plug in the given values and the converted temperature T1. \(\frac{551 \mathrm{mL}}{348 \mathrm{K}} = \frac{1.00 \mathrm{mL}}{T_2}\)

Solve for T2

Cross multiply and isolate T2. \(T_2 = \frac{1.00 \mathrm{mL} \times 348 \mathrm{K}}{551 \mathrm{mL}} = 0.631 \mathrm{K}\)

Convert T2 back to Celsius

We will now convert T2 from Kelvin back to Celsius using the formula: °C = K - 273 \(T_2 = 0.631 \mathrm{K} - 273 = -272.369^{\circ} \mathrm{C}\) T2 ≈ -272.4°C

Key concepts

Gas Laws

Gas laws are a set of relationships that describe how the properties of gases—such as pressure, volume, and temperature—are interconnected. These relationships help us predict how a gas will behave under certain conditions. One of the fundamental gas laws is Charles's Law. Charles's Law explains how the volume of a gas changes with temperature, provided the pressure and amount of gas remain constant. According to the law:

• The volume of a gas varies directly with its temperature when measured in Kelvin.
• This means that as the temperature of a gas increases, its volume also increases, and vice versa.

Understanding Charles's Law is crucial for solving problems where you need to find the missing volume or temperature of a gas, as seen in our given data sets. Always remember that the changes described by gas laws are only strictly accurate when conditions such as constant pressure and amount of gas are met.

Temperature Conversion

Temperature conversion is a necessary step when working with gas laws since many laws require temperature to be in the Kelvin scale. The Kelvin scale is used in scientific calculations because it starts at absolute zero, which is the lowest theoretical temperature where all molecular motion ceases. To convert Celsius to Kelvin, you simply add 273 to the Celsius temperature:

• For example, to convert 25°C to Kelvin, you calculate: 25 + 273 = 298 K.
• This conversion is essential to apply Charles's Law correctly.

Having temperatures in Kelvin ensures that the proportionality factor in Charles's Law remains consistent. Remember, Kelvin does not have degrees, so we just write 'K' after the number. Ensure all temperatures in gas law calculations are converted to Kelvin to avoid incorrect results.

Volume and Temperature Relationship

The relationship between volume and temperature is at the heart of Charles's Law. When we say volume and temperature have a direct relationship under Charles's Law, we mean that they change in the same direction:

• If the temperature increases, the gas molecules move more energetically, needing more space, causing the volume to increase.
• Conversely, if the temperature drops, the gas then contracts, and the volume decreases.

This relationship can be mathematically represented by the formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V \) represents volume and \( T \) represents temperature in Kelvin. This formula allows us to solve for an unknown volume or temperature, given that the other variables are provided. Understanding this relationship is crucial for calculations involving gases under non-standard conditions.

Kelvin to Celsius Conversion

Understanding how to convert between Kelvin and Celsius is equally important when interpreting the results of gas law experiments. Since gas laws often give results in Kelvin, we may need to convert these back to the Celsius scale to answer real-world questions or provide results in a more familiar format:

• To convert Kelvin back to Celsius, subtract 273 from the Kelvin temperature.
• For instance, if you have a temperature of 348 K, converting to Celsius gives: 348 - 273 = 75°C.

This conversion step comes into play when you have solved for a temperature in Kelvin and need to express the result in everyday terms. Always make sure conversions are done accurately, as mistakes in conversion can lead to significant errors in chemical calculations or interpretations.