Question

A gas has a volume of \(4.00 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\). What final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if \(n\) and \(P\) do not change? a. \(1.50 \mathrm{~L}\) b. \(1200 \mathrm{~mL}\) c. \(250 \mathrm{~L}\) d. \(50.0 \mathrm{~mL}\)

Short answer

a. -170.72°C; b. -191.21°C; c. 16798.73°C; d. -269.74°C.

Step-by-step solution

- Convert initial temperature to Kelvin

The initial temperature is 0°C. Use the formula to convert it to Kelvin: \[ T(K) = T(°C) + 273.15 \]. So, \[ T_i = 0 + 273.15 = 273.15 \text{ K} \].

- Set up the equation using Charles's Law

Recall Charles's Law: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]. In this problem, \[ V_1 = 4.00 \text{~L} \] and \[ T_1 = 273.15 \text{ K} \].

- Solve for the final temperature

Rearrange Charles's Law to solve for the final temperature, \[ T_2 = \frac{V_2 \times T_1}{V_1} \].

Step 4a - Calculate final temperature for 1.50 L

For \[ V_2 = 1.50 \text{~L} \], \[ T_2 = \frac{1.50 \text{ L} \times 273.15 \text{ K}}{4.00 \text{ L}} \approx 102.43 \text{ K} \]. Convert this back to Celsius by subtracting 273.15, so \[ T_2 = 102.43 - 273.15 \text{ °C} \approx -170.72 \text{ °C} \].

Step 5a - Calculate final temperature for 1200 mL

First, convert mL to L: \[ 1200 \text{ mL} = 1.200 \text{ L} \]. Then, \[ V_2 = 1.200 \text{~L} \], \[ T_2 = \frac{1.200 \text{ L} \times 273.15 \text{ K}}{4.00 \text{ L}} \approx 81.94 \text{ K} \]. Convert this back to Celsius: \[ T_2 = 81.94 - 273.15 \text{ °C} \approx -191.21 \text{ °C} \].

Step 6a - Calculate final temperature for 250 L

For \[ V_2 = 250 \text{~L} \], \[ T_2 = \frac{250 \text{ L} \times 273.15 \text{ K}}{4.00 \text{ L}} \approx 17071.88 \text{ K} \]. Convert this back to Celsius: \[ T_2 = 17071.88 - 273.15 \text{ °C} \approx 16798.73 \text{ °C} \].

Step 7a - Calculate final temperature for 50.0 mL

First, convert mL to L: \[ 50.0 \text{ mL} = 0.0500 \text{ L} \]. Then, \[ V_2 = 0.0500 \text{~L} \], \[ T_2 = \frac{0.0500 \text{ L} \times 273.15 \text{ K}}{4.00 \text{ L}} \approx 3.41 \text{ K} \]. Convert this back to Celsius: \[ T_2 = 3.41 - 273.15 \text{ °C} \approx -269.74 \text{ °C} \].

Key concepts

Gas Laws

Gas laws are fundamental principles that describe how gases behave under different conditions of temperature, pressure, volume, and amount. They help us understand and predict the behavior of gases in various situations. One of the well-known gas laws is Charles's Law. Charles's Law specifically deals with the relationship between the volume and temperature of a gas when the pressure and the amount of gas are constant. By understanding gas laws, we can solve practical problems involving gas-filled containers, such as those in the exercise above.
Imagine you have a balloon filled with air. When you heat it, the balloon expands. When you cool it, the balloon contracts. This observation is essentially Charles's Law in action.

Volume-Temperature Relationship

The volume-temperature relationship is a key aspect of Charles's Law. Charles's Law states that the volume of a gas is directly proportional to its temperature when the pressure and the amount of gas are constant. This means that if you increase the temperature, the volume of the gas will also increase, and vice versa.
Mathematically, this can be expressed as: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
Here, \( V_1 \) and \( V_2 \) are the initial and final volumes, respectively, while \( T_1 \) and \( T_2 \) are the initial and final temperatures. Let's see how this works in the context of the given problem:
- For an initial volume \( V_1 = 4.00 \text{ L} \) at an initial temperature \( T_1 = 273.15 \text{ K} \) (0°C), if we change the volume to various values (e.g., 1.50 L, 1200 mL, 250 L, 50 mL), we calculate the new temperature \( T_2 \) using the rearranged equation: \( T_2 = \frac{V_2 \times T_1}{V_1} \).

Kelvin to Celsius Conversion

Understanding how to convert temperatures between Kelvin and Celsius is crucial when working with gas laws. The Kelvin scale is an absolute temperature scale that starts at absolute zero. In gas law calculations, we always use temperatures in Kelvin to avoid negative values and ensure accurate results.
The conversion formula is simple: \( T(K) = T(°C) + 273.15 \)
For instance, in the given exercise, the initial temperature is 0°C, which converts to \( 0 + 273.15 = 273.15 \text{ K} \). After calculating the final temperature \( T_2 \) using Charles's Law, we convert it back to Celsius by subtracting 273.15, such as \( T_2 - 273.15 \). This conversion is straightforward but essential for accurate gas law problems.

Ideal Gas Law

The Ideal Gas Law is a more comprehensive equation that encompasses all the relationships among a gas's volume, temperature, pressure, and the amount of gas. The Ideal Gas Law is given by \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of the gas, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. This law is particularly useful because it combines several simpler gas laws.
Although Charles's Law deals only with the volume and temperature, the Ideal Gas Law can provide a more complete picture of a gas's behavior in various conditions. In the context of the exercise, assuming the pressure and amount of gas don't change means we are effectively using a simplified version of the Ideal Gas Law focusing on the volume-temperature relationship.