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Chapter 10: Problem 100

A 500 mL incandescent light bulb is filled with \(1.5 \times 10^{-5}\) mol of xenon to minimize the rate of evaporation of the tungsten filament. What is the pressure of xenon in the light bulb at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The pressure of xenon in the light bulb is approximately \(7.33 \times 10^{-4}\) atm.

Step by step solution

01

Understanding the Given Data

We have a 500 mL light bulb containing xenon at 25掳C, which is 298 K considering that Kelvin = Celsius + 273. The amount of xenon given is \(1.5 \times 10^{-5}\) mol. We need to calculate the pressure inside the bulb.
02

Using the Ideal Gas Law

The ideal gas law formula is \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume in liters, \(n\) is the number of moles, \(R\) is the ideal gas constant (0.0821 L路atm/(K路mol)), and \(T\) is the temperature in Kelvin.
03

Converting Units

The volume of the light bulb is 500 mL. We need to convert this to liters to match the units of the ideal gas constant, so \(V = 0.500\) L. The temperature in Kelvin has already been determined as 298 K.
04

Solving for Pressure

Rearrange the ideal gas law to solve for pressure: \(P = \frac{nRT}{V}\). Substituting in the known values, \(P = \frac{(1.5 \times 10^{-5} \, \text{mol}) \times (0.0821 \, \text{L}\cdot\text{atm/K}\cdot\text{mol}) \times (298 \, \text{K})}{0.500 \, \text{L}}\).
05

Calculating the Result

Calculate \(P\) by completing the multiplications and division: \(P = \frac{1.5 \times 10^{-5} \times 0.0821 \times 298}{0.500}\), which gives \(P = 7.32689 \times 10^{-4}\) atm when calculated precisely.

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Key Concepts

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

Pressure Calculation
Pressure is a measure of how much force the gas exerts on the walls of its container. In this exercise, we use the ideal gas law to determine the pressure of xenon inside a light bulb. The formula to use is:\[ P = \frac{nRT}{V} \]where:
  • \( P \) is the pressure, the value we are looking to find.
  • \( n \) is the number of moles of xenon gas.
  • \( R \) is the ideal gas constant, which we will discuss in a bit.
  • \( T \) is the temperature in Kelvin.
  • \( V \) is the volume of the gas in liters.
Using this equation, we can compute the pressure if we know the other variables. The calculations in this problem have provided us with a pressure of \( 7.32689 \times 10^{-4} \) atm. This means that the xenon gas inside the incandescent bulb applies this magnitude of pressure on the walls.
Moles of Gas
The amount of gas present is quantified by moles. One mole represents a quantity of \( 6.022 \times 10^{23} \) particles, such as atoms or molecules. In this case, we have \( 1.5 \times 10^{-5} \) moles of xenon. This value is crucial as it directly affects the outcome of the ideal gas law calculation. Understanding moles helps in allocating how much gas is present for a given volume or pressure. In our calculation, this number helps establish the proportion between the amount of gas and the space it occupies in combination with the gas constant and temperature.
Temperature Conversion
Temperature plays a significant role in pressure calculation and must be expressed in Kelvin for the ideal gas law. Celsius alone is not suitable because the law requires absolute temperature for correct results. The conversion from Celsius to Kelvin is straightforward:\[ \text{Kelvin} = \text{Celsius} + 273 \] In this specific exercise, the temperature provided is \( 25^{\circ}C \), which converts to \( 298\, K \) when you add 273. This accurate Kelvin value ensures that the pressure calculation adheres to the standard conditions for gas law calculations.
Gas Constant
The gas constant \( R \) is a crucial factor in the ideal gas law. It provides the link between moles, temperature, and volume to find pressure. The constant value typically used is:\[ R = 0.0821 \text{ L路atm/(K路mol)} \]This particular unit set is selected because it matches the conditions we are working with in the ideal gas law equation: volume in liters, pressure in atmospheres, and temperature in Kelvin. Without this constant, the relationship between each quantity would not hold. Ensure the correct units are consistently used across the equation for precise calculations.
Volume Conversion
Volume needs to be expressed in liters for consistency with the gas constant's units. In our exercise, the original volume of the light bulb is given in milliliters. Hence, conversion to liters is necessary:\[ \text{Volume in liters} = \text{Volume in milliliters} / 1000 \]So, the light bulb with a 500 mL volume converts to 0.500 liters when divided by 1000. This conversion is important because the gas constant requires volume in liters, and accurate conversion ensures the calculations using the ideal gas law yield the correct results.

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Most popular questions from this chapter

Imagine that the reaction \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) occurs in a container that has a piston that moves to maintain a constant pressure when the reaction occurs at constant temperature. Which of the following statements describes how the volume of the container changes due to the reaction: (a) the volume increases by \(50 \%,(\mathbf{b})\) the volume increases by \(33 \%\), (c) the volume remains constant, (d) the volume decreases by \(33 \%,(\mathbf{e})\) the volume decreases by \(50 \% .\)

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2},\) which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at normal body temperature, \(37^{\circ} \mathrm{C},\) and \(101.33 \mathrm{kPa}\) when \(10.0 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at \(100 \mathrm{kPa}\) and \(298 \mathrm{~K},\) to completely oxidize \(15.0 \mathrm{~g}\) of glucose.

Consider the sample of gas depicted here. What would the drawing look like if the volume and temperature remained constant while you removed enough of the gas to decrease the pressure by a factor of \(2 ?\) [Section 10.3\(]\) (a) It would contain the same number of molecules. (b) It would contain half as many molecules. (c) It would contain twice as many molecules. (d) There is insufficient data to say.

Calcium hydride, \(\mathrm{CaH}_{2}\), reacts with water to form hydrogen gas: $$ \mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g) $$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate \(145 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas if the pressure of \(\mathrm{H}_{2}\) is \(110 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C} ?\)

Radon (Rn) is the heaviest (and only radioactive) member of the noble gases. How much slower is the root-mean-square speed of \(\mathrm{Rn}\) than He at \(300 \mathrm{~K} ?\)
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