Question

A sample of hydrogen \(\left(\mathrm{H}_{2}\right)\) gas at \(127{ }^{\circ} \mathrm{C}\) has a pressure of \(2.00 \mathrm{~atm}\). At what temperature \(\left({ }^{\circ} \mathrm{C}\right)\) will the pressure of the \(\mathrm{H}_{2}\) decrease to \(0.25 \mathrm{~atm}\), if \(V\) and \(n\) are constant?

Short answer

-223^{\circ} C

Step-by-step solution

- Identify the given values

Given:Initial temperature, \ (T_{1}) = 127^{\circ} C = 127 + 273 = 400 K\Initial pressure, \(P_{1}\) = 2.00 atm\Final pressure, \(P_{2}\) = 0.25 atm\We need to find the final temperature, \(T_{2}\)

- Use the Pressure-Temperature Relationship

Because the volume (\(V\)) and the number of moles (\(n\)) are constant, we can use Gay-Lussac's law, which states that for a given mass and constant volume, the pressure of a gas is directly proportional to its absolute temperature:\[\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\]

- Rearrange the Equation

Rearrange the equation to solve for the final temperature, \(T_{2}\):\[ T_{2} = \frac{P_{2} \cdot T_{1}}{P_{1}} \]

- Substitute the Known Values and Solve

Substitute the known values into the equation and solve for \(T_{2}\):\[ T_{2} = \frac{0.25 \mathrm{~atm} \times 400 \mathrm{~K}}{2.00 \mathrm{~atm}} = \frac{100 \mathrm{~K}}{1} = 50 \mathrm{~K} \]Convert the final temperature from Kelvin to Celsius:\[ T_{2} = 50 \mathrm{~K} - 273 = -223^{\circ} C \]

Key concepts

Pressure-Temperature Relationship

When studying gases, it's important to understand how their properties change under different conditions. One key relationship is between pressure and temperature. According to Gay-Lussac's law, if the volume and the number of moles of gas are held constant, the pressure of a gas is directly proportional to its absolute temperature (measured in Kelvin). This means that as the temperature of a gas increases, its pressure also increases, and vice versa. Mathematically, Gay-Lussac's law is expressed as:

\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)

Here, \(P_1\) and \(P_2\) are the initial and final pressures, respectively, and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin. Understanding this relationship helps in solving problems where you need to calculate the new pressure or temperature of a gas after a change occurs.

Ideal Gas Law

The Ideal Gas Law is a major principle in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. The law is expressed with the formula:

\(\text{PV = nRT}\)

where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant (8.314 J/(mol·K) or 0.0821 atm·L/(mol·K)), and \(T\) is the temperature in Kelvin. This law is useful because it combines several different gas laws (Boyle's, Charles's, and Gay-Lussac's) into one comprehensive equation. When solving these kinds of problems, make sure that all units are consistent, particularly for temperature, which must be in Kelvin for accurate calculations.

Temperature Conversion

Converting temperatures is a crucial step when dealing with gas law problems because these laws typically require temperature to be in Kelvin. The Kelvin scale is an absolute scale where 0 K is the point at which particles have minimal thermal motion. To convert Celsius to Kelvin, you simply add 273 to the Celsius temperature. For example, 127°C becomes:

\(127 + 273 = 400\) K

Conversely, to convert from Kelvin back to Celsius, you subtract 273 from the Kelvin temperature. If you find the final temperature to be 50 K, converting it to Celsius would be:

\(50 - 273 = -223\)°C

These conversions ensure that you are using the correct units and help prevent calculation errors related to temperature.