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To establish a pressure of 2.00 atm in a 2.24 L cylinder containing \(1.60 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C},\) (a) add \(1.60 \mathrm{g} \mathrm{O}_{2} ;(\mathrm{b})\) add \(0.60 \mathrm{g} \mathrm{He}(\mathrm{g}) ;(\mathrm{c})\) add \(2.00 \mathrm{g} \mathrm{He}(\mathrm{g})\) (d) release \(0.80 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\)

Short Answer

Expert verified
Each action should be performed as follows: (a) Add 1.60 g of O2. (b) Add 0.694 g of He. (c) Add 2.00 g of He. (d) Release 0.846 g of O2.

Step by step solution

01

Convert temperature to Kelvin

Temperature in the Ideal Gas Law must be in Kelvin. Convert 0 degrees Celsius to Kelvin using the formula K = °C + 273.15. Hence, the temperature is 273.15 K.
02

Calculate number of moles to achieve desired pressure

Rearrange the Ideal Gas Law to solve for n, the number of moles. n=PV/RT. Substitute P=2.00 atm, V=2.24 L, R=0.0821 L.atm/(mol.K) (Ideal Gas Law constant), T=273.15 K into the formula. This gives the required moles of gas.
03

Calculate mass for given substances

The number of moles obtained in step 2 is used to compute the mass of each gas using their molar masses. (a) For O2 the molar mass is 32.00 g/mol and for He it's 4.00 g/mol. (b, c) Compute the masses for 0.60g of He and 2.00g of He. (d) Deduct mass of 0.8g O2 from the initial 1.6g O2.
04

Add or release required amount of each gas

Add or release the required amount of each gas found in step 3 to the cylinder to achieve the pressure of 2.00 atm.

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Key Concepts

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

Moles Calculation
Understanding the concept of moles is crucial when dealing with gases and their reactions. The mole is a unit that represents a specific number of particles, typically atoms or molecules. In chemistry, the number of moles helps understand quantities at the atomic scale. You can calculate the number of moles using the Ideal Gas Law, represented as \( n = \frac{PV}{RT} \). Here:
  • \( n \) is the number of moles of the gas.
  • \( P \) represents the pressure in atmospheres.
  • \( V \) is the volume in Liters.
  • \( R \) is the Ideal Gas Constant (\(0.0821\, \text{L.atm/(mol.K)}\)).
  • \( T \) is the temperature in Kelvin.
By rearranging the formula, you can solve for \( n \) using the known variables such as pressure, volume, and temperature. This calculation is pivotal for understanding how much gas is needed or produced in a given reaction.
Gas Pressure
Gas pressure is the force that gas molecules exert per unit area when they collide with the walls of their container. It is measured in different units including atmospheres (atm). The Ideal Gas Law, \( PV = nRT \), connects pressure with other factors like number of moles, volume, and temperature. An increase in the number of gas molecules in a constant volume results in increased pressure because more particles are colliding with the container walls.It's also interesting to note how the addition or release of gas can affect pressure. For example, adding helium to an existing oxygen gas will increase the total pressure. This concept is crucial when adjusting the gas until you reach a desired pressure like the 2.00 atm required in the cylinder example.
Temperature Conversion
In the context of gas laws, it is vital to work with the absolute temperature scale, which is Kelvin. The Kelvin scale is directly connected to the energy of particles, as it starts from absolute zero - a theoretical point where particle motion ceases.Converting Celsius to Kelvin is straightforward. Simply add 273.15 to the Celsius measurement. For example, \(0^\circ \text{C}\) converts to \(273.15\, \text{K}\).This step is essential as the Ideal Gas Law requires temperature to be in Kelvin, allowing for accurate calculations of volume, pressure, and number of moles. Thus, always remember:
  • Celsius to Kelvin: K = °C + 273.15
Molar Mass
The molar mass of a substance is the mass of one mole of its molecules or atoms, expressed in grams per mole (g/mol). It allows us to convert between the mass of a substance and the number of moles. For gases, the molar mass plays a crucial role in understanding how to achieve certain pressures with given amounts. Different elements have different molar masses, for example:
  • Oxygen (\(O_2\)) has a molar mass of \(32.00\, \text{g/mol}\).
  • Helium (He) has a molar mass of \(4.00\, \text{g/mol}\).
When you know the mass of your gas, you can calculate the number of moles by dividing the mass by the molar mass. This is a key step to determine how much you need to add or remove to reach a specific state, such as achieving the desired 2.00 atm pressure in the exercise example.

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Most popular questions from this chapter

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(24^{\circ} \mathrm{C}\) The volume of gas is 1.16 L. In a subsequent experiment, it is determined that the mass of \(\mathrm{O}_{2}\) present is 1.46 g. What must have been the barometric pressure at the time the gas was collected? (Vapor pressure of water \(=22.4 \text { Torr. })\)

A 3.05 g sample of \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s})\) is introduced into an evacuated 2.18 L flask and then heated to \(250^{\circ} \mathrm{C}\).What is the total gas pressure, in atmospheres, in the flask at \(250^{\circ} \mathrm{C}\) when the \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) has completely decomposed? $$\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$

The heat required to sustain animals while they hibernate comes from the biochemical combustion of fatty acids, such as arachidonic acid, \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) What volume of air, measured at \(298 \mathrm{K}\) and \(1.00 \mathrm{atm}\) is required to burn \(2.00 \mathrm{kg} \mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2} ?\) Air is approximately \(78.1 \% \mathrm{N}_{2}\) and \(20.9 \% \mathrm{O}_{2},\) by volume. Other gases make up the remaining \(1.0 \%\)

In the reaction of \(\mathrm{CO}_{2}(\mathrm{g})\) and solid sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) solid sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) and oxy- gen gas are formed. This reaction is used in submarines and space vehicles to remove expired \(\mathrm{CO}_{2}(\mathrm{g})\) and to generate some of the \(\mathrm{O}_{2}(\mathrm{g})\) required for breathing. Assume that the volume of gases exchanged in the lungs equals \(4.0 \mathrm{L} / \mathrm{min},\) the \(\mathrm{CO}_{2}\) content of expired air is \(3.8 \% \mathrm{CO}_{2}\) by volume, and the gases are at \(25^{\circ} \mathrm{C}\) and \(735 \mathrm{mmHg}\). If the \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) in the above reaction are measured at the same temperature and pressure, (a) how many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) are produced per minute and \((\mathrm{b})\) at what rate is the \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) consumed, in grams per hour?

Calculate \(u_{\mathrm{rms}},\) in meters per second, for \(\mathrm{Cl}_{2}(\mathrm{g})\) molecules at \(30^{\circ} \mathrm{C}\)

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