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Chapter 5: Problem 49

The steel reaction vessel of a bomb calorimeter, which has a volume of \(75.0 \mathrm{~mL}\), is charged with oxygen gas to a pressure of 145 atm at \(22^{\circ} \mathrm{C}\). Calculate the moles of oxygen in the reaction vessel.

Short Answer

Expert verified
There are approximately 4.53 moles of oxygen gas in the reaction vessel.

Step by step solution

01

Write down the Ideal Gas Law formula.

The Ideal Gas Law equation relates the pressure, volume, moles, and temperature of any gas. It can be written as: \[PV = nRT\] Where P represents the pressure (in atm), V represents the volume (in L), n represents the number of moles, R is the ideal gas constant (0.0821 \( L\, atm / K\, mol\)), and T is the temperature (in Kelvin).
02

Convert the given volume from mL to L.

The Ideal Gas Law is applicable in units of Liters (L) for volume, so we need to convert the given volume in milliliters (mL) to liters. We can do that by using the conversion factor: 1 L = 1000 mL So, to convert the volume from mL to L: \(V = 75.0 \, mL \cdot \frac{1 \, L}{1000 \, mL} = 0.075 \, L\)
03

Convert the given temperature from Celsius to Kelvin.

The Ideal Gas Law uses temperature in Kelvin. We can convert the given temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature: \(T = 22\, ^{\circ}C + 273.15 \, K = 295.15 \, K\)
04

Solve for moles of gas (n).

We know the pressure (145 atm), volume (0.075 L), and temperature (295.15 K). We need to solve for n (moles) using the Ideal Gas Law: \[PV = nRT\] \[n = \frac{PV}{RT}\] Substitute the values into the formula: \[n = \frac{(145 \, atm)(0.075 \, L)}{(0.0821 \, L\,atm/K\,mol)(295.15 \, K)}\]
05

Calculate the moles of oxygen gas.

Perform the calculations to find the number of moles of oxygen gas: \[n \approx 4.53\, moles\] So, there are approximately 4.53 moles of oxygen gas in the reaction vessel.

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Key Concepts

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

PV=nRT Equation
Understanding the Ideal Gas Law, often encapsulated in the equation PV=nRT, is crucial for studying gas behavior under various conditions of pressure (P), volume (V), and temperature (T). This equation offers a straightforward way to calculate the amount of gas, in moles (n), given the other conditions.

The 'R' in the equation stands for the ideal gas constant, which has a value of 0.0821 L atm/K mol. This constant provides the necessary conversion factor to work with standard atmospheric pressure, volume in liters, and temperature in Kelvin. Given this formula, one can predict how a gas will respond to changes in its environment, making it an indispensable tool in fields like chemistry and physics.
Calculating Moles of Gas
To find out how much gas is present in a system, we calculate the number of moles using the Ideal Gas Law. Moles (n) are a measure of substance amount, providing a way to translate between microscopic particles and quantities we can interact with in the lab.

The calculation for moles in the context of the Ideal Gas Law is straightforward once you have the pressure, volume, and temperature. By rearranging the PV=nRT equation to solve for 'n', we get:
\[\begin{equation} n = \frac{PV}{RT} \end{equation}\]

This allows us to plug in our known values for pressure, volume, and temperature (after converting to the correct units) and solve for the number of moles of gas.
Gas Pressure and Volume Relationship
Within the PV=nRT equation, pressure and volume have an inverse relationship when the amount of gas and temperature are held constant, following Boyle's Law. This means if you increase the volume of a gas's container, the pressure decreases, and vice versa, as long as the temperature and number of moles remain unchanged.

Understanding this relationship is crucial when calculating the volume or pressure of a gas using the Ideal Gas Law. Keeping one variable constant, you can predict the behavior of another, allowing for practical applications such as in calculating how pressurized gases will behave under different volumes, which is essential in tasks like filling oxygen tanks or designing engines.
Temperature Conversion to Kelvin
Temperature is a key factor in the behavior of gases and must be measured in Kelvin when using the Ideal Gas Law. This is because Kelvin is the SI unit for thermodynamic temperature and starts at absolute zero, the point at which particles have minimal thermal motion.

Temperature conversion to Kelvin from Celsius is simple: add 273.15 to your Celsius temperature to get Kelvin. This step is essential since failing to convert temperature into Kelvin can lead to incorrect results when applying the PV=nRT equation. Always double-check your temperature units to ensure accurate calculations!

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Most popular questions from this chapter

A glass vessel contains \(28 \mathrm{~g}\) nitrogen gas. Assuming ideal behavior, which of the processes listed below would double the pressure exerted on the walls of the vessel? a. Adding enough mercury to fill one-half the container. b. Raising the temperature of the container from \(30 .{ }^{\circ} \mathrm{C}\) to \(60 .{ }^{\circ} \mathrm{C}\). c. Raising the temperature of the container from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). d. Adding 28 g nitrogen gas.

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.)

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3} .\) This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In 1894 , A. Combes (Comptes Rendus 1894, p. 1221\()\) reacted beryllium with the anion \(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows: If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3}\). Show how Combes's data help to confirm that beryllium is a divalent metal.

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?

We-state that the ideal gas law tends to hold best at low pressures and high temperatures. Show how the van der Waals equation simplifies to the ideal gas law under these conditions.
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