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Chapter 11: Problem 90

A container is filled with \(0.644 \mathrm{~g}\) of \(\mathrm{O}_{2}\) at \(5{ }^{\circ} \mathrm{C}\) and \(845 \mathrm{mmHg}\). What is the volume, in milliliters, of the container? \((11.8)\)

Short Answer

Expert verified
The volume of the container is approximately 413 mL.

Step by step solution

01

- Convert temperature to Kelvin

Convert the temperature from Celsius to Kelvin using the formula: \[ T(K) = T({}^\text{掳} \text{C}) + 273.15 \] So, the temperature is: \[ T(K) = 5 + 273.15 = 278.15 \text{ K} \]
02

- Convert pressure to atmospheres

Convert the pressure from mmHg to atm using the conversion factor: \[ 1 \text{ atm} = 760 \text{ mmHg} \] So, the pressure in atm is: \[ P = \frac{845 \text{ mmHg}}{760 \text{ mmHg/atm}} = 1.1118 \text{ atm} \]
03

- Use the Ideal Gas Law

Use the Ideal Gas Law formula to find the volume: \[ PV = nRT \] First, calculate the number of moles () of \text{ }_2\text{O}\text{ using the molar mass \text{(32 g/mol)}: \[ n = \frac{0.644 \text{ g}}{32 \text{ g/mol}} = 0.020125 \text{ mol} \]
04

- Calculate the volume

Rearrange the Ideal Gas Law to solve for the volume: \[ V=\frac{nRT}{P} \] where R (the gas constant) is given as 0.0821 L路atm/(K路mol). Substituting all values we get: \[ V = \frac{0.020125 \text{ mol} \times 0.0821 \text{ L路atm/(K路mol)} \times 278.15 \text{ K}}{1.1118 \text{ atm}} \] Therefore, the volume is: \[ V \thickapprox 0.413 \text{ L} \]
05

- Convert volume to milliliters

Convert the volume from liters to milliliters: \[ 1 \text{ L} = 1000 \text{ mL} \] So, the volume in milliliters is: \[ V = 0.413 \text{ L} \times 1000 \text{ mL/L} = 413 \text{ mL} \]

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Key Concepts

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

temperature conversion
Temperature conversion is key in gas law problems. We often need to convert temperatures from Celsius to Kelvin because the Kelvin scale is used in gas laws. The conversion formula is simple: \[ T(K) = T(掳C) + 273.15 \]For example, in this exercise, we needed to convert 5掳C to Kelvin. By adding 273.15, we get 278.15 K. It's essential always to use Kelvin when working with gas equations, as it ensures accuracy and standardization.
pressure conversion
Pressures in gas law calculations are commonly given in different units. To use the Ideal Gas Law, we convert pressure to atmospheres (atm). The conversion factor is: \[ 1 \text{ atm} = 760 \text{ mmHg} \]For instance, 845 mmHg can be converted by dividing by 760: \[ 845 \text{ mmHg} 梅 760 \text{ mmHg/atm} 鈮� 1.1118 \text{ atm} \]This step ensures the pressure fits the constants used in gas equations.
molar mass calculation
Molar mass helps convert mass to moles, an essential step in using the Ideal Gas Law. For O\textsubscript{2}, the molar mass is 32 g/mol. Using the given mass (0.644 g), we find the number of moles: \[ n = \frac {0.644 \text{ g}} {32 \text{ g/mol}} 鈮� 0.020125 \text{ mol} \]This conversion is crucial because the Ideal Gas Law (\[ PV = nRT \]) requires the amount of gas in moles.
gas constant
The gas constant R links pressure, volume, temperature, and moles in the Ideal Gas Law. Its value depends on the units used. Commonly, R = 0.0821 L路atm/(K路mol). This constant ensures that the units in the equation balance: \[ PV = nRT \]Understanding and using the correct value of R is vital for accurate calculations in gas law problems.
volume calculation
The final step is to find the volume using the Ideal Gas Law. Rearranging \[ PV = nRT \] to solve for V: \[ V = \frac {nRT} {P} \]By inserting the calculated values (n = 0.020125 mol, R = 0.0821 L路atm/(K路mol), T = 278.15 K, P = 1.1118 atm): \[ V = \frac {0.020125 \times 0.0821 \times 278.15} {1.1118} 鈮� 0.413 \text{ L} \]Finally, converting liters to milliliters: \[ 1 \text{ L} = 1000 \text{ mL} \] hence, \[ 0.413 \text{ L} = 413 \text{ mL} \] This results in the container's volume being approximately 413 mL.

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Most popular questions from this chapter

A sample of nitrogen \(\left(\mathrm{N}_{2}\right)\) has a volume of \(50.0 \mathrm{~L}\) at a pressure of \(760 . \mathrm{mmHg}\). What is the volume, in liters, of the gas at each of the following pressures, if there is no change in temperature and amount of gas? a. \(725 \mathrm{mmHg}\) b. \(2.0 \mathrm{~atm}\) c. \(0.500 \mathrm{~atm}\) d. 850 torr

A steel cylinder with a volume of \(15.0 \mathrm{~L}\) is filled with \(50.0 \mathrm{~g}\) of nitrogen gas at \(25^{\circ} \mathrm{C}\). What is the pressure, in atmospheres, of the \(\mathrm{N}_{2}\) gas in the cylinder? ( \(11.8\) )

When Zn reacts with HCl solution, the products are \(\mathrm{H}_{2}\) gas and \(\mathrm{ZnCl}_{2}\). A volume of \(425 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) gas is collected over water at a total pressure of \(758 \mathrm{mmHg}\) and \(16^{\circ} \mathrm{C}\). The vapor pressure of water at \(16{ }^{\circ} \mathrm{C}\) is \(14 \mathrm{mmHg}\). $$ \mathrm{Zn}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{ZnCl}_{2}(a q) $$ a. What was the partial pressure of the \(\mathrm{H}_{2}\) gas? b. How many moles of \(\mathrm{H}_{2}\) gas were produced in the reaction?

Butane undergoes combustion when it reacts with oxygen to produce carbon dioxide and water. What volume, in liters, of oxygen is needed to react with \(55.2 \mathrm{~g}\) of butane at \(0.850 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) ? $$ 2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow} 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) $$

Aluminum and oxygen react to form aluminum oxide. How many liters of oxygen at \(0{ }^{\circ} \mathrm{C}\) and \(760 \mathrm{mmHg}\) (STP) are required to completely react with \(5.4 \mathrm{~g}\) of aluminum? $$ 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) $$
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