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Chapter 10: Problem 100

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 \(23^{\circ} \mathrm{C} ?\)

Short Answer

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

Step by step solution

01

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. \(T(K) = 23^\circ C + 273.15\) \(T(K) = 296.15 K\)
02

Convert volume to liters

The volume of the lightbulb is given in cubic centimeters (cm鲁), but we need the volume in liters (L) to use the Ideal Gas Law. To convert cm鲁 to L, divide the volume by 1000. \(V(L) = 600 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3}\) \(V(L) = 0.600 \text{ L}\)
03

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. \(R = 0.0821 \frac{\text{L atm}}{\text{mol K}}\)
04

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: \(P = \frac{(1.4 \times 10^{-5} \text{ mol})(0.0821 \frac{\text{L atm}}{\text{mol K}})(296.15 \text{ K})}{0.600 \text{ L}}\)
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 (cm鲁) 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 cm鲁 to L, divide the number by 1000.
For instance, in our problem, the volume of the lightbulb is 600 cm鲁. 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.

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Most popular questions from this chapter

Indicate which of the following statements regarding the kinetic-molecular theory of gases are correct. (a) The average kinetic energy of a collection of gas molecules at a given temperature is proportional to \(m^{1 / 2}\) . (b) The gas molecules are assumed to exert no forces on each other. (c) All the molecules of a gas at a given temperature have the same kinetic energy. (d) The volume of the gas molecules is negligible in comparison to the total volume in which the gas is contained. (e) All gas molecules move with the same speed if they are at the same temperature.

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(80^{\circ} \mathrm{C}\) if 1.00 mol occupies \(33.3 \mathrm{L},\) assuming that (a) \(\mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table \(10.3 .\) ) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4}\) ? Explain.

A 15.0 -L tank is filled with helium gas at a pressure of \(1.00 \times 10^{2}\) atm. How many balloons (each 2.00 L) can be inflated to a pressure of 1.00 atm, assuming that the temperature remains constant and that the tank cannot be emptied below 1.00 atm?

At constant pressure, the mean free path \((\lambda)\) of a gas molecule is directly proportional to temperature. At constant temperature, \(\lambda\) is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, \(\lambda\) is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it \(R_{\text { mfp }}\) , like the ideal-gas constant) and define units for \(R_{\operatorname{mfp}}\) .

The planet Jupiter has a surface temperature of 140 \(\mathrm{K}\) and a mass 318 times that of Earth. Mercury (the planet) has a surface temperature between 600 \(\mathrm{K}\) and 700 \(\mathrm{K}\) and a mass 0.05 times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.
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