Question

Assume that the volume of a fixed amount of gas in a rigid container does not change. Calculate the pressure the gas would exert if the temperature were changed as shown in the following table. $$ \begin{array}{|c|c|c|c|} \hline \begin{array}{c} \text { Initial } \\ \text { Pressure } \end{array} & \begin{array}{c} \text { Initial } \\ \text { Temperature } \end{array} & \begin{array}{c} \text { Final } \\ \text { Temperature } \end{array} & \begin{array}{c} \text { Final } \\ \text { Pressure } \end{array} \\ \hline 255 \text { torr } & 225 \mathrm{~K} & 315 \mathrm{~K} & ? \\ \hline 895 \text { torr } & 25^{\circ} \mathrm{C} & 206 \mathrm{~K} & ? \\ \hline 2.74 \mathrm{~atm} & 150^{\circ} \mathrm{C} & 23^{\circ} \mathrm{C} & ? \\\ \hline \end{array} $$

Short answer

The final pressures are 354.9 Torr, 600.4 Torr and 1.90 atm for the first, second and third row respectively.

Step-by-step solution

Convert temperatures to Kelvin (If needed)

For the first row, no conversion is required since the temperature is already given in Kelvin. For the second row, the initial given temperature is in Celsius, to convert this to Kelvin, simply add 273 to the given Celsius temperature (therefore, 25 + 273 = 298K). For the third row, the initial is 150C (which is 423K when converted) and the final temperature is 23C (or 296K when converted).

Substitute known values into the formula

Now for each row, substitute the given initial pressure (P1), initial temperature (T1), and final temperature (T2) into the formula and solve for the final pressure (P2). The formula to use is \(P_2 = P_1(T_2/T_1)\).

Solve for the Final Pressure

First row: \(P_2 = 255(315/225) = 354.9 Torr\). Second row: \(P_2 = 895(206/298) = 600.4 Torr\). Third row: \(P_2 = 2.74(296/423) = 1.90 atm\)

Key concepts

Gas Pressure Calculation

When you're asked to calculate the gas pressure in a problem like this, it's all about understanding how pressure relates to temperature under constant volume conditions. This scenario is aligned with Gay-Lussac's Law, which is part of the ideal gas laws. The formula we use is: \(P_2 = P_1 \frac{T_2}{T_1}\). In this formula:

• \(P_1\) is the initial pressure.
• \(T_1\) is the initial temperature (in Kelvin).
• \(T_2\) is the final temperature (in Kelvin).
• \(P_2\) is the final pressure we need to find.

By substituting the known values into this equation, you can find how pressure will change with varying temperatures. This is valuable in situations where the temperature of a gas changes, but the volume remains constant, a common scenario in scientific calculations.
For instance, in the first row of the given data, using an initial pressure of 255 torr and temperatures of 225 K and 315 K, you can determine the final pressure using the steps: calculate \(P_2\) by plugging in the numbers: \(P_2 = 255 \times \frac{315}{225} = 354.9\) torr. This calculated pressure tells us how the pressure adjusts when the gas is heated.

Temperature Conversion

One crucial skill in dealing with gas problems is converting temperatures to the appropriate units. Often, temperatures are given in Celsius, which cannot be directly used in most gas law equations that require the absolute temperature, or Kelvin.
To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. The extra 0.15 can usually be ignored in basic problems for simplicity. Here are some important conversions from the problem:

• In the second row, 25°C is converted to Kelvin: 25 + 273 = 298 K.
• For the third row, convert 150°C to Kelvin: 150 + 273 = 423 K.

Remembering to convert temperatures to Kelvin is vital as it ensures you are using an absolute scale where zero represents absolute absence of thermal energy, which aligns with how the gas particles behave at a molecular level.

Rigid Container

In chemistry and physics problems, a rigid container means that the container does not change its shape, size, or volume. This implies that when gases are encapsulated within these containers, the volume is constant, and no matter how the pressure or temperature inside the container changes, the volume will remain the same.
This concept simplifies the gas laws since it allows us to use formulas derived under constant volume conditions, like the ideal gas law variations. By keeping the volume constant, pressures and temperatures become the primary variables to worry about, particularly when temperature changes are involved.
Using a rigid container eliminates any complications from volume changes, allowing direct application of formulas like Gay-Lussac's Law — reassured by the fact that any pressure changes result solely from temperature variations. This makes calculations straightforward and reduces potential errors when calculating pressure.