Problem 100 - Chemistry The Central Science

Question

To minimize the rate of evaporation of the tungsten filament, \(1.4 \times 10^{-5}\) mol of argon is placed in a \(600-\mathrm{cm}^{3}\) light-bulb. What is the pressure of argon in the lightbulb at \(23^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The pressure of argon inside the lightbulb at \(23^\circ C\) is approximately \(5.38 \times 10^{-5}\) atm.

Step 1: Convert the temperature to Kelvin

To calculate the pressure of argon, we need the temperature in Kelvin. To convert Celsius to Kelvin, add 273.15 to the given temperature. \(T(K) = 23^\circ C + 273.15\) \(T(K) = 296.15 K\)

Step 2: Convert volume to liters

The volume of the lightbulb is given in cubic centimeters (cm³), but we need the volume in liters (L) to use the Ideal Gas Law. To convert cm³ to L, divide the volume by 1000. \(V(L) = 600 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3}\) \(V(L) = 0.600 \text{ L}\)

Step 3: Find the universal gas constant value

The universal gas constant (R) has various different units. We will use the unit that has L atm / mol K to be consistent with our given values. \(R = 0.0821 \frac{\text{L atm}}{\text{mol K}}\)

Step 4: Apply the Ideal Gas Law formula

Now, we have all the information we need to apply the Ideal Gas Law formula and solve for pressure (P). \(PV = nRT\) Rearrange the formula to solve for P: \(P = \frac{nRT}{V}\) Plug in the values: \(P = \frac{(1.4 \times 10^{-5} \text{ mol})(0.0821 \frac{\text{L atm}}{\text{mol K}})(296.15 \text{ K})}{0.600 \text{ L}}\)

Step 5: Calculate the pressure

Now, we simply perform the calculation in the formula. \(P = \frac{(1.4 \times 10^{-5})(0.0821)(296.15)}{0.600}\) \(P = 5.38 \times 10^{-5} \text{ atm}\) The pressure of argon inside the lightbulb at \(23^\circ C\) is approximately \(5.38 \times 10^{-5}\) atm.

Key Concepts

Gas Pressure Calculation

When calculating the pressure of a gas in a container like a lightbulb, a helpful tool is the Ideal Gas Law. This law provides a relationship between four critical variables: pressure (P), volume (V), number of moles (n), and temperature (T). These are all tied together using the universal gas constant, R. The formula is expressed as \[ PV = nRT \] This equation allows us to determine the unknown pressure if we have the other values.

• Pressure (P) is what we are solving for in this exercise.
• Volume (V) is the space the gas occupies.
• Number of Moles (n) represents the amount of gas.
• Temperature (T) needs to be in Kelvin for the equation to work.
• Universal Gas Constant (R), which is usually \(0.0821 \frac{\text{L atm}}{\text{mol K}}\).

When using the Ideal Gas Law, it's crucial to have all the units in the correct form: liters for volume, Kelvin for temperature, and atm for pressure. Once you plug in the values, rearrange the formula to solve for pressure: \[ P = \frac{nRT}{V} \] By carefully substituting the appropriate values, you can calculate the gas pressure, as demonstrated above.

Temperature Conversion to Kelvin

Temp conversion is a straightforward process. It's important in gas-related calculations using the Ideal Gas Law because the temperature needs to be in Kelvin. The Kelvin scale is an absolute temperature scale starting at absolute zero.
The conversion from Celsius to Kelvin is super easy: simply add 273.15 to the Celsius temperature.
This stems from the need to ensure all kinetic energy measurements are positive.

• Given Temperature: Start with your temperature in Celsius.
• Add 273.15: This accounts for the offset between the Celsius and Kelvin scales.

For example, in the exercise, the temperature is given as \(23^ \circ C\). To convert it:\[T(K) = 23^ \circ C + 273.15 = 296.15K\] This conversion ensures compatibility with the Ideal Gas Law, allowing the accurate calculation of other variables like pressure and volume.

Volume Conversion to Liters

Volume conversion is another essential step when applying the Ideal Gas Law. The reason? The equation uses liters for volume.
Cubic centimeters (cm³) are a common measurement, but they must be converted into liters for calculations.
The conversion between these units is simple and can be easily remembered:

• 1 liter equals 1000 cubic centimeters.
• To convert cm³ to L, divide the number by 1000.

For instance, in our problem, the volume of the lightbulb is 600 cm³. By converting:\[V(L) = 600 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3} = 0.600 \text{ L}\] This conversion is necessary to use these values in the Ideal Gas Law accurately. Always ensure your volume measurements are in liters to avoid any miscalculation or mismatch when calculating pressure or other properties.