Question

A gas has a volume of \(2.84 \mathrm{~L}\) at \(1.00 \mathrm{~atm}\) and \(0{ }^{\circ} \mathrm{C}\). At what temperature does it have a volume of \(7.50 \mathrm{~L}\) at \(520 \mathrm{mmHg}\) ?

Short answer

The temperature is approximately \(220.6^{\circ}\text{C}.\)

Step-by-step solution

Convert Pressure Units

First, convert the pressure from mmHg to atm, as the initial pressure is given in atm. Use the conversion: \(1 \text{ atm} = 760 \text{ mmHg}\). Calculate the pressure in atm as follows:\[ P_2 = \frac{520 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.684 \text{ atm} \]

Apply Combined Gas Law

Use the combined gas law, which relates the initial and final states of a gas under varying pressure, volume, and temperature. The formula is: \[ \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} \]Given:- \(P_1 = 1.00 \text{ atm}\), \(V_1 = 2.84 \text{ L}\), \(T_1 = 273 \text{ K} \text{ (converted from } 0^{\circ} \text{C)}\) - \(P_2 = 0.684 \text{ atm}\), \(V_2 = 7.50 \text{ L}\)- \(T_2\) is unknown.

Rearrange and Solve for Temperature

First rearrange the equation to solve for \(T_2\):\[ T_2 = \frac{P_2 \cdot V_2 \cdot T_1}{P_1 \cdot V_1} \]Substitute the known values:\[ T_2 = \frac{0.684 \text{ atm} \times 7.50 \text{ L} \times 273 \text{ K}}{1.00 \text{ atm} \times 2.84 \text{ L}} \approx 493.6 \text{ K} \]

Convert Temperature to Celsius

Convert the temperature from Kelvin to Celsius by subtracting 273:\[ T_2 = 493.6 \text{ K} - 273 = 220.6^{\circ} \text{C} \]

Key concepts

Gas Laws

Gas laws help us to understand how gases behave under different conditions. The Combined Gas Law is particularly useful for relating volumes, pressures, and temperatures of a gas sample when changes occur.
The formula for the Combined Gas Law is:

• \( \frac{P_1 \, V_1}{T_1} = \frac{P_2 \, V_2}{T_2} \)

Here, \(P\) is pressure, \(V\) is volume, and \(T\) is temperature. When using this formula, it's crucial to keep the units consistent, especially when it comes to temperature (always use Kelvin for gas laws). By rearranging this equation, we can solve for any unknown variable if the others are known.
This law is a blend of Boyle's Law, Charles's Law, and Gay-Lussac's Law, showing their direct relationships in a single equation.

Temperature Conversion

Temperature plays a critical role in gas behavior, and gases are typically measured using the Kelvin scale in scientific calculations. To convert Celsius to Kelvin, simply add 273.15.
Kelvin is essential because it starts at absolute zero, the point where particles have minimal kinetic energy.
For instance, in the exercise, the initial temperature was given in Celsius (\(0^{\circ} \text{C}\)), which converted to Kelvin becomes \(273 \text{ K}\).
Conversely, when we need to express the final temperature back in Celsius from Kelvin, like after solving a gas law problem, we subtract 273. Without converting temperatures properly, the calculations could miss the mark, especially since the equations rely heavily on thermodynamic principles closely tied to Kelvin readings.

Pressure Conversion

Conversion of pressure units is often a necessary step in solving gas law problems, particularly when given different pressure units. It's common to encounter pressure measurements like atmospheres (atm) and millimeters of mercury (mmHg), or Torr.
For accurate calculations in gas law equations, pressures must be in uniform units. In our exercise, the final pressure was given in mmHg (\(520 \text{ mmHg}\)) and was converted to atm:\(\text{where } 1 \, \text{atm} = 760 \, \text{mmHg}\).
The conversion provided the following result: \(P_2 = \frac{520 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.684 \text{ atm}\).
Such conversions are straightforward but crucial, as they allow different sets of data to harmonize, ensuring the gas laws are applied correctly and the final answers are both precise and reliable.