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Chapter 5: Problem 79

A piece of solid carbon dioxide, with a mass of \(7.8 \mathrm{~g}\), is placed in a 4.0-L otherwise empty container at \(27^{\circ} \mathrm{C}\). What is the pressure in the container after all the carbon dioxide vaporizes? If \(7.8 \mathrm{~g}\) solid carbon dioxide were placed in the same container but it already contained air at 740 torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide vaporizes?

Short Answer

Expert verified
The pressure in the container after all the CO2 vaporizes in the empty container is: \(P = \frac{(7.8\,\text{g} / 44.01\,\text{g/mol}) \times (0.0821\,\text{L}\cdot\text{atm/mol}\cdot\text{K}) \times (300.15\,\text{K})}{4.0\,\text{L}}\) When there is air present in the container, the partial pressure of CO2 is: Partial pressure of CO2 = \(\frac{(7.8\,\text{g} / 44.01\,\text{g/mol}) \times (62.36\,\text{L}\cdot\text{torr/mol}\cdot\text{K}) \times (300.15\,\text{K})}{4.0\,\text{L}}\) And the total pressure in the container is: Total pressure = 740 torr + Partial pressure of CO2 calculated above.

Step by step solution

01

Find moles of carbon dioxide

First, we need to find the number of moles of carbon dioxide (CO2) in the given mass of solid. To do this, we divide the given mass by the molar mass of CO2: Molar mass of CO2 = 12.01 g (C) + 2 * 16.00 g (O) = 44.01 g/mol Number of moles of CO2 = (Mass of CO2) / (Molar mass of CO2) n = (7.8 g) / (44.01 g/mol)
02

Convert temperature to Kelvin

We need to convert the given temperature from Celsius to Kelvin before using it in the ideal gas law equation: T(K) = T(°C) + 273.15 T(K) = 27°C + 273.15 K = 300.15 K
03

Calculate pressure in an empty container

Now we can use the ideal gas law to find the pressure in the 4.0-L empty container after all the carbon dioxide vaporizes: PV = nRT P = nRT / V P = (7.8 g / 44.01 g/mol) * (0.0821 L*atm/mol*K) * (300.15 K) / (4.0 L)
04

Calculate partial pressure of carbon dioxide with air present

When the container has air at 740 torr, we need to find the partial pressure of carbon dioxide after it vaporizes. We can use the same ideal gas law to find the pressure of carbon dioxide as we did in step 3: Partial pressure of CO2 = (7.8 g / 44.01 g/mol) * (62.36 L*torr/mol*K) * (300.15 K) / (4.0 L)
05

Calculate total pressure in the container with air present

To find the total pressure in the container when the air is present, we add the pressure of the air to the partial pressure of carbon dioxide: Total pressure = Pressure of air + Partial pressure of CO2 Total pressure = 740 torr + Partial pressure of CO2 Calculated in Step 4

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Most popular questions from this chapter

An organic compound containing only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\) yields the following data. i. Complete combustion of \(35.0 \mathrm{mg}\) of the compound produced \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) ii. A \(65.2-\mathrm{mg}\) sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 119 ), giving \(35.6 \mathrm{~mL} \mathrm{~N}_{2}\) at 740 . torr and \(25^{\circ} \mathrm{C}\). iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{~mL} / \mathrm{min}\). The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{~mL} / \mathrm{min}\). What is the molecular formula of the compound?

Without looking at a table of values, which of the following gases would you expect to have the largest value of the van der Waals constant \(b: \mathrm{H}_{2}, \mathrm{~N}_{2}, \mathrm{CH}_{4}, \mathrm{C}_{2} \mathrm{H}_{6}\), or \(\mathrm{C}_{3} \mathrm{H}_{8}\) ?

A 5.0-L flask contains \(0.60 \mathrm{~g} \mathrm{O}_{2}\) at a temperature of \(22^{\circ} \mathrm{C}\). What is the pressure (in atm) inside the flask?

The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot \(\mathrm{CuO}(s)\) : $$ \text { Compound } \underset{\text { Cvoss }}{\stackrel{\text { Hot }}{\longrightarrow}} \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ The product gas is then passed through a concentrated solution of \(\mathrm{KOH}\) to remove the \(\mathrm{CO}_{2}\). After passage through the \(\mathrm{KOH}\) solution, the gas contains \(\mathrm{N}_{2}\) and is saturated with water vapor. In a given experiment a \(0.253-g\) sample of a compound produced \(31.8 \mathrm{~mL} \mathrm{~N}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\) and 726 torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8\) torr.)

At room temperature, water is a liquid with a molar volume of \(18 \mathrm{~mL}\). At \(105^{\circ} \mathrm{C}\) and 1 atm pressure, water is a gas and has a molar volume of over \(30 \mathrm{~L}\). Explain the large difference in molar volumes.
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