Question

You have \(3.6 \mathrm{L}\) of \(\mathrm{H}_{2}\) gas at \(380 \mathrm{mm} \mathrm{Hg}\) and \(25^{\circ} \mathrm{C}\). What is the pressure of this gas if it is transferred to a 5.0 -I. flask at \(0.0^{\circ} \mathrm{C} ?\)

Short answer

The pressure is approximately 224 mm Hg.

Step-by-step solution

Understand the Problem

We need to find the new pressure of hydrogen gas when it's transferred from an initial state to a new state. The initial conditions are a volume of 3.6 L, pressure of 380 mm Hg, and temperature of 25°C. The final conditions given include a new volume of 5.0 L and temperature of 0°C. We will use the combined gas law to find the new pressure.

Convert Temperatures to Kelvin

To apply the gas laws correctly, we must use temperatures in Kelvin. The initial temperature is 25°C, which converts to 298 K (by adding 273 to the Celsius temperature). The final temperature is 0°C, which converts to 273 K.

Write the Combined Gas Law

The combined gas law is expressed as: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\), where \(P_1, V_1, T_1\) are the initial pressure, volume, and temperature, and \(P_2, V_2, T_2\) are the final pressure, volume, and temperature.

Substitute Known Values

We substitute the known values into the combined gas law: \(\frac{380 \text{ mm Hg} \times 3.6 \text{ L}}{298 \text{ K}} = \frac{P_2 \times 5.0 \text{ L}}{273 \text{ K}}\).

Solve for the Unknown Pressure \(P_2\)

Rearrange the equation to solve for \(P_2\): \(P_2 = \frac{380 \times 3.6 \times 273}{298 \times 5.0}\). Calculate the value: \(P_2 \approx 223.9 \text{ mm Hg}\).

Key concepts

Pressure-Volume-Temperature relationship

In the world of gases, understanding the pressure-volume-temperature relationship is essential to predicting how gases will behave under different conditions. This relationship is captured neatly by the Combined Gas Law. The foundational principle is that the pressure, volume, and temperature of a gas are interconnected. When any one of these variables changes, the others respond in a predictable manner, provided the amount of gas remains constant.

Imagine gas inside a sealed balloon. If you squeeze the balloon, you reduce its volume, causing the pressure inside to rise if the temperature remains unchanged. Similarly, if you heat the gas, its temperature increases, often causing the balloon to expand unless the pressure increases to compensate.

The Combined Gas Law links these three properties through the equation:\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]This formula shows that the ratio of pressure and volume over temperature is constant before and after a change, making it possible to solve for unknowns when conditions change.

Gas Laws

Gas laws form the backbone of understanding gaseous behavior through mathematical expressions. These laws are formulated from observations of gases under varying conditions of temperature, pressure, and volume.

There are several essential gas laws to know:

• Boyle's Law: At constant temperature, the pressure and volume of a gas are inversely proportional \( (P \propto \frac{1}{V}) \).
• Charles's Law: At constant pressure, the volume of a gas is directly proportional to its temperature \( (V \propto T) \).
• Gay-Lussac's Law: At constant volume, the pressure of a gas is directly proportional to its temperature \( (P \propto T) \).

The Combined Gas Law brings these individual expressions together. It helps solve problems where not just one, but multiple parameters like temperature, volume, and pressure change simultaneously.

Understanding how these laws interrelate helps predict and explain phenomena such as why heated air can lift hot air balloons or why a canister might explode if overheated.

Temperature conversion

Temperature conversion is a crucial step when working with gas laws because different temperature scales can dramatically alter calculations. To ensure consistency, we use Kelvin in gas law equations because it is an absolute scale with no negative numbers, making mathematical operations more straightforward and logical.

The conversion between Celsius and Kelvin is simple: \[K = °C + 273.15\]
For example, to convert 25°C to Kelvin, you add 273.15, resulting in 298.15K. Similarly, 0°C converts directly to 273.15K. Using Kelvin eliminates the confusion of different starting points and ensures consistency in calculations.

Always remember, when applying the gas laws, to convert your temperatures to Kelvin first. This practice helps avoid errors and makes the results more reliable.