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What is the density (in \(\mathrm{g} \mathrm{L}^{-1}\) ) of \(\mathrm{CO}_{2}\) at \(400 \mathrm{~K}\) and exerting a pressure of \(0.0821\) atm? \((\mathrm{R}=0.0821 \mathrm{~L}\) atm \(\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) \(0.01\) (b) \(0.11\) (c) \(2.5\) (d) 44

Short Answer

Expert verified
The density of \( \mathrm{CO}_2 \) is \( 0.11 \, \mathrm{g} \, \mathrm{L}^{-1} \), matching option (b).

Step by step solution

01

Understand the Formula

To find the density, we will use the Ideal Gas Law formula: \( PV = nRT \). Density can be related to the molar mass \( M \) through the equation \( \rho = \frac{PM}{RT} \), where \( \rho \) is the density, \( P \) is the pressure, \( M \) is the molar mass of \( \mathrm{CO}_2 \), \( R \) is the universal gas constant, and \( T \) is the temperature.
02

Determine Molar Mass of COâ‚‚

The molar mass \( M \) of \( \mathrm{CO}_2 \) is calculated by adding the atomic masses of its constituents: \( C (12 \, \mathrm{g/mol}) + 2 \times O (16 \, \mathrm{g/mol}) = 44 \, \mathrm{g/mol} \).
03

Applying the Formula

Substitute the given values into the density formula: \( P = 0.0821 \) atm, \( R = 0.0821 \, \mathrm{L} \, \mathrm{atm} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1} \), \( T = 400 \) K, and \( M = 44 \, \mathrm{g/mol} \). The density \( \rho \) is \( \rho = \frac{0.0821 \times 44}{0.0821 \times 400} \).
04

Simplify the Equation

Calculate the expression: \( \rho = \frac{0.0821 \times 44}{0.0821 \times 400} = \frac{44}{400} = 0.11 \, \mathrm{g/L} \).
05

Select the Correct Answer

The calculated density is \(0.11 \, \mathrm{g} \, \mathrm{L}^{-1} \), which matches option (b).

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Key Concepts

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

Density Calculation
The calculation of density for gases often involves using the Ideal Gas Law. Here, the concept of density (\( \rho \)) is explored with gases like COâ‚‚. In simpler terms, density refers to how much mass is packed into a given volume.
For gases, this relationship is sometimes not straightforward due to the properties of gases, which can vary with conditions like temperature and pressure. However, we can use the formula\[ \rho = \frac{PM}{RT} \]to calculate density.
This equation rearranges the Ideal Gas Law (\( PV = nRT \)) to express density in terms of pressure (\( P \)), molar mass (\( M \)), the universal gas constant (\( R \)), and temperature (\( T \)). This method simplifies finding the density of a gas, making it accessible with the right data, like in our COâ‚‚ example.
Molar Mass of CO2
Understanding the molar mass of COâ‚‚ is essential when performing calculations involving gas properties. The molar mass is the sum of the atomic masses of all atoms present in a single molecule of the compound.
  • Carbon (C) has an atomic mass of 12 g/mol.
  • Oxygen (O) has an atomic mass of 16 g/mol.
  • COâ‚‚ has one carbon atom and two oxygen atoms.
To calculate the molar mass of COâ‚‚, sum up these atomic masses: \[ M = 12 \, \text{(for C)} + 2 \times 16 \, \text{(for each O)} = 44 \, \text{g/mol} \]Knowing this helps in various calculations involving COâ‚‚, as it aids in determining both density and other properties related to the Ideal Gas Law.
Universal Gas Constant
The universal gas constant, represented by the symbol \( R \), is a fundamental constant in chemistry used in various equations and calculations involving gases.
Its value is generally taken as \( 0.0821 \, \mathrm{L} \, \mathrm{atm} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1} \) when working with the Ideal Gas Law.
This constant connects numerous properties of gases, enabling us to convert temperature, pressure, and volume relationships effectively.
It is pivotal in the ideal gas equation \( PV = nRT \)and its derivatives, which allow for universal applicability across various scenarios. Understanding the universal gas constant helps in comprehending how gases behave under different conditions and simplifies solving equations like those used in density calculations.
Temperature and Pressure Relationship
The relationship between temperature and pressure is a key consideration in understanding gas behavior, explained effectively through the Ideal Gas Law.
According to this relationship, when the temperature of a gas increases, the pressure also tends to increase, provided the volume remains constant.
This relationship highlights the principle that gases expand when heated, causing particles to collide more frequently with the walls of their container.
In our specific example of COâ‚‚, we see the interaction where a set temperature of 400 K at a given pressure of 0.0821 atm is used.
This interaction is simplified using the formula:\[ PV = nRT \]because it allows us to derive properties like density while understanding how changes in temperature might affect pressure and volume. Recognizing this relationship is crucial for predicting gas behavior under different environmental conditions.

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Most popular questions from this chapter

Osmotic pressure of a solution is \(0.0821 \mathrm{~atm}\) at a temperature of \(300 \mathrm{~K}\). The concentration in moles/litre will be (a) \(0.33\) (b) \(0.066\) (c) \(0.3 \times 10^{-2}\) (d) 3

\(\mathrm{KBr}\) is \(80 \%\) dissociated in aqueous solution of \(0.5 \mathrm{M}\) concentration. (Given \(K_{r}\) for water \(\left.=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\). The solution freezes at (a) \(271.326 \mathrm{~K}\) (b) \(272 \mathrm{~K}\) (c) \(270.5 \mathrm{~K}\) (d) \(268.5 \mathrm{~K}\)

If \(0.44 \mathrm{~g}\) of substance dissolved in \(22.2 \mathrm{~g}\) of benzene lowers the freezing point of benzene by \(0.567^{\circ} \mathrm{C}\), then the molecular mass of substance is, (the molal depression constant \(\left.=5.12^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\right)\) (a) \(128.4\) (b) \(156.6\) (c) \(178.9\) (d) \(232.4\)

When non-zero volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to (a) molar fraction of solute (b) molar fraction of solvent (c) concentration of the solute in grams per liter (d) concentration of the solute in gram per \(100 \mathrm{~mL}\).

Which pair of the following will not form an ideal solution? (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) (b) \(\mathrm{H}_{2} \mathrm{O}+\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}\) (c) \(\mathrm{CCl}_{4}+\mathrm{SiCl}_{4}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{14}+\mathrm{C}_{7} \mathrm{H}_{16}\)

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