Question

A sample of air has a volume of 140.0 mL at 67°C. At what temperature would its volume be 50.0 mL at constant pressure?

Short answer

-151.67°C

Step-by-step solution

Understand the Problem

We need to find the temperature at which the volume of air changes from 140.0 mL to 50.0 mL at constant pressure. We will use the combined gas law for this purpose.

Write Down the Combined Gas Law

The combined gas law is \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]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.

Convert the Initial Temperature to Kelvin

The initial temperature is given in Celsius. To convert it to Kelvin, use the formula \(T(K) = T(°C) + 273.15\).Given \(T_1 = 67∘C\), \[T_1 = 67 + 273.15 = 340.15\ K\]

Plug in the Known Values and Solve for the Final Temperature

We have \(V_1 = 140.0\ mL\), \(T_1 = 340.15\ K\), and \(V_2 = 50.0\ mL\). Plug these into the combined gas law and solve for \(T_2\):\[\frac{140.0}{340.15} = \frac{50.0}{T_2}\]Cross-multiplying to solve for \(T_2\):\[T_2 = \frac{50.0 \times 340.15}{140.0} = 121.48 \ K \]

Convert the Final Temperature Back to Celsius (optional)

If needed, convert the final temperature back to Celsius using the formula \(T(°C) = T(K) - 273.15\). \[ T_2 = 121.48 - 273.15 = -151.67 \ °C \]

Key concepts

Gas Laws

The behavior of gases is described by a set of principles known as gas laws. These laws explain relationships between variables like volume, temperature, and pressure. One crucial gas law for understanding changes in gas states is the combined gas law.
The combined gas law is represented as: \(\frac{V_1}{T_1} = \frac{V_2}{T_2} \), where:

• V₁ = Initial volume
• T₁ = Initial temperature in Kelvin
• V₂ = Final volume
• T₂ = Final temperature in Kelvin

This law is particularly useful because it combines Charles's Law, Boyle's Law, and Gay-Lussac's Law into one formula. It helps us determine how one property of a gas will change if the others are held constant.

Temperature Conversion

Converting temperatures between different scales is a common task in chemistry. The most frequently used temperature scales are Celsius and Kelvin.
The formula to convert Celsius to Kelvin is simple: \(T(K) = T(°C) + 273.15 \). This formula is derived from the fact that 0 Kelvin (absolute zero) is equal to -273.15°C.
For instance, if you have a temperature of 67°C, you convert it to Kelvin like this:

• 67°C + 273.15 = 340.15 K

Converting back from Kelvin to Celsius also follows a straightforward formula: \(T(°C) = T(K) - 273.15 \). For example, if you have a temperature of 121.48 K, it converts back to Celsius as:

• 121.48 K - 273.15 = -151.67°C

Volume and Temperature Relationship

There is a direct relationship between the volume and temperature of a gas, which is defined by Charles's Law. Charles's Law states that the volume of a gas is directly proportional to its temperature (in Kelvin) if the pressure is held constant.
Mathematically, it can be written as: \(\frac{V_1}{T_1} = \frac{V_2}{T_2} \).
This means if you increase the temperature of a gas, its volume will increase, and vice versa, as long as the pressure remains constant.
For example, if you have a gas at 140.0 mL and 340.15 K, and its volume changes to 50.0 mL, you can use Charles's Law to find the new temperature:
\(\frac{140.0}{340.15} = \frac{50.0}{T_2} \)
Solving for \(T_2 \), we get:
\(T_2 = \frac{50.0 \times 340.15}{140.0} = 121.48 K \)
This shows how volume decreases with a decrease in temperature.

Kelvin to Celsius Conversion

Kelvin is the SI unit for temperature and is widely used in scientific calculations. It's essential to know how to convert temperatures between Kelvin and Celsius because experimental data often come in these units.
To convert Kelvin to Celsius, use the formula: \(T(°C) = T(K) - 273.15 \).
For instance, if you have a temperature of 340.15 K, you can convert it to Celsius like this:

• 340.15 K - 273.15 = 67°C

Similarly, if you have a Kelvin temperature calculated as 121.48 K, you can convert it to Celsius as follows:

• 121.48 K - 273.15 = -151.67 °C

This conversion is vital when you are interpreting temperature changes in reactions or when comparing experimental data with theoretical predictions.