Question

A gas has a volume of 1.75 L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30 L at 210.0 kPa?

Short answer

The gas would occupy 1.30 L at approximately 260.92 K (or -12.23°C).

Step-by-step solution

- Convert Temperatures to Kelvin

The initial temperature is given in Celsius. Convert it to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \]Thus, \[ T_1 = -23 + 273.15 = 250.15 \, K \].

- Use the Combined Gas Law

The combined gas law relates pressure, volume, and temperature: \[ \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \].We need to solve for the final temperature \( T_2 \).

- Input Known Values

Substitute the known values into the equation. Given \( P_1 = 150.0 \, kPa \), \( V_1 = 1.75 \, L \), \( T_1 = 250.15 \, K \), \( P_2 = 210.0 \, kPa \), and \( V_2 = 1.30 \, L \), we have:\[ \frac{150.0 \, kPa \cdot 1.75 \, L}{250.15 \, K} = \frac{210.0 \, kPa \cdot 1.30 \, L}{T_2} \].

- Isolate and Solve for the Final Temperature

Rearrange the equation to solve for \( T_2 \):\[ T_2 = \frac{210.0 \, kPa \cdot 1.30 \, L \cdot 250.15 \, K}{150.0 \, kPa \cdot 1.75 \, L} \].Perform the calculations:\[ T_2 \approx 260.92 \, K \].

- Convert Temperature Back to Celsius (Optional)

If needed, convert the final temperature back to Celsius:\[ T(°C) = T(K) - 273.15 \].Thus,\[ T_2(°C) \approx 260.92 - 273.15 = -12.23 \, °C \].

Key concepts

Gas Laws

Gas laws are fundamental principles in chemistry that describe the behavior of gases. The three primary gas laws are Boyle's Law, Charles's Law, and Gay-Lussac's Law.
These laws individually relate the pressure, volume, and temperature of a gas.

• Boyle's Law states that the pressure of a gas is inversely proportional to its volume when the temperature is constant: \( P_1 V_1 = P_2 V_2 \).
• Charles's Law states that the volume of a gas is directly proportional to its temperature (in Kelvin) at constant pressure: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).
• Gay-Lussac's Law states that the pressure of a gas is directly proportional to its temperature (in Kelvin) at constant volume: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).

The combined gas law combines these three laws into one equation that allows calculations involving changes in pressure, volume, and temperature.
It is expressed as: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).By understanding and using these laws, we can predict and explain how gases will respond to changes in their environment.

Temperature Conversion

Temperature conversion is crucial to correctly use the gas laws, as they require temperatures in Kelvin.
The Kelvin scale is an absolute temperature scale, starting at absolute zero, meaning there are no negative Kelvin temperatures.
To convert from Celsius to Kelvin, use the formula:
\( T(K) = T(°C) + 273.15 \).
For example, in the exercise given, we converted -23°C to Kelvin:
\( T_1 = -23 + 273.15 = 250.15 \, K \).Converting back to Celsius, if needed, can be done using the reverse formula:
\( T(°C) = T(K) - 273.15 \). When working on gas law problems, always make sure your temperatures are in Kelvin before starting the calculations.
It helps avoid errors and ensures the formulas work correctly.

Pressure and Volume

Pressure and volume are key parameters when studying gases.

• Pressure is the force exerted by the gas particles per unit area on the walls of its container. It is generally measured in units like kPa (kilopascals), atm (atmospheres), or mmHg (millimeters of mercury).
• Volume is the amount of space that a gas occupies, usually measured in liters (L) or cubic meters (m³).

According to Boyle's Law, for a fixed amount of gas kept at a constant temperature, the volume of the gas is inversely proportional to its pressure:
\( P_1 V_1 = P_2 V_2 \). This relation tells us that increasing the pressure will reduce the volume and vice versa, assuming the temperature is constant.
The combined gas law we used in the example exercise includes both pressure (P) and volume (V) and allows us to solve for changes in one when the other quantities are known.