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Chapter 10: Problem 100

A \(500 \mathrm{~mL}\) 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 gas in the light bulb at \(25^{\circ}\mathrm{C}\) can be found using the Ideal Gas Law, \(P = \frac{nRT}{V}\). Convert the given temperature to Kelvin, \(T(K) = 25 + 273.15 = 298.15\,\text{K}\). Convert volume to liters, \(V (L) = 500\,\text{mL} / 1,000 = 0.5\,\text{L}\). Substitute the known values into the equation: \(P = \frac{(1.5 \times 10^{-5}\,\text{mol})(0.0821 \, \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}})(298.15\,\text{K})}{0.5\,\text{L}}\). Calculate the pressure: \(P \approx 7.25 \times 10^{-4}\,\text{atm}\).

Step by step solution

01

Convert temperature to Kelvin

Since the given temperature is in Celsius, we have to convert it to Kelvin. To do so, we simply add 273.15 to the temperature in Celsius. \( T(K) = T(°C) + 273.15 \)
02

Convert volume to liters

The volume given is in milliliters (mL), and we need it to be in liters (L) because the Ideal Gas Law requires volume in liters. To convert the volume from mL to L, we simply divide by 1,000. \( V(L) = V(mL) / 1,000 \)
03

Apply the Ideal Gas Law

Now, we plug the known values into the Ideal Gas Law equation, PV = nRT, to find the pressure of the xenon gas. R is the ideal gas constant, which is 0.0821 L atm / (mol K). We know the moles of xenon gas (1.5 x 10⁻⁵ mol), the converted temperature in Kelvin, and the converted volume in liters. We need to rearrange the Ideal Gas Law formula to solve for pressure P. \( P = \frac{nRT}{V} \)
04

Calculate pressure

Now, plug in the known values and calculate the pressure of xenon gas. \( P = \frac{(1.5 \times 10^{-5} \,\text{mol})(0.0821 \, \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} )(25 + 273.15 \,\text{K})}{500\text{mL} / 1,000} \) Make sure the units are consistent, and then solve for pressure P in atmospheres.

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Most popular questions from this chapter

Hurricane Wilma of 2005 is the most intense hurricane on record in the Atlantic basin, with a low-pressure reading of 882 mbar (millibars). Convert this reading into (a) atmospheres, \((\mathbf{b})\) torr, and \((\mathbf{c})\) inches of \(\mathrm{Hg}\).

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\), molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-},\) forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$\mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s)$$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with $2.00 \mathrm{~L}\( of chlorine gas at a pressure of \)152.0 \mathrm{kPa}$ at \(21^{\circ} \mathrm{C},\) how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

A \(50.0 \mathrm{~g}\) sample of solid \(\mathrm{CO}_{2}\) (dry ice) is added at \(-100^{\circ} \mathrm{C}\) to an evacuated (all of the gas removed) container with a volume of \(5.0 \mathrm{~L}\). If the container is sealed and then allowed to warm to room temperature \(\left(25^{\circ} \mathrm{C}\right)\) so that the entire solid \(\mathrm{CO}_{2}\) is converted to a gas, what is the pressure inside the container?

On a single plot, qualitatively sketch the distribution of molecular speeds for (a) \(\mathrm{Kr}(g)\) at $-50^{\circ} \mathrm{C},(\mathbf{b}) \mathrm{Kr}(g)\( at \)0^{\circ} \mathrm{C}$ (c) \(\mathrm{Ar}(g)\) at \(0^{\circ} \mathrm{C}\). [Section \(\left.10.7\right]\)

A quantity of \(\mathrm{N}_{2}\) gas originally held at \(531.96 \mathrm{kPa}\) pressure in a 1.00 - \(\mathrm{L}\) container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5-\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(531.96 \mathrm{kPa}\) and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?
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