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Chapter 16: Problem 3

A vessel contains a mixture of nitrogen of mass \(7 \mathrm{~g}\) and carbon dioxide of mass \(11 \mathrm{~g}\) at temperature \(290 \mathrm{~K}\) and pressure \(1 \mathrm{~atm}\). The density of the mixture is : (a) \(1.1 \mathrm{~g} / \mathrm{L}\) (b) \(1.2 \mathrm{~g} / \mathrm{L}\) (c) \(1.515 \mathrm{~g} / \mathrm{L}\) (d) \(1.6 \mathrm{~g} / \mathrm{L}\)

Short Answer

Expert verified
The density of the gas mixture is approximately 1.515 g/L (option c).

Step by step solution

01

Calculate Total Mass

First, identify the total mass of the gas mixture by adding the mass of nitrogen and carbon dioxide. The mass of nitrogen is given as 7 g and the mass of carbon dioxide is 11 g. So, the total mass, \( M_t \), of the mixture is:\[ M_t = 7 \, \text{g} + 11 \, \text{g} = 18 \, \text{g} \]
02

Find Volume Using Ideal Gas Law

The ideal gas law \( PV = nRT \) can be used to find the volume. First, convert the given units to standard conditions if necessary and find the number of moles \( n \). However, since we need the volume directly, let's rearrange the density formula:\[ \text{Density} = \frac{\text{Total Mass}}{\text{Volume}} \]Using the ideal gas law for standard conditions \( P = 1 \text{ atm}, T = 290 \text{ K} \), the volume \( V \) of 1 mole of an ideal gas is near 24 L at standard pressure and temperature.For simplicity and since density is directly influenced by the change in pressure and temperature, assume behavior under ideal conditions for the combined mass rather than calculate individual volumes separately since both gases are ideal.
03

Calculate Mixture Density

Assume an ideal gas occupies 22.4 L (slight deviation in calculation typically for STP as 24 L accounts a modest temperate) here for understanding density of mixture directly, we use the total mass and convert it directly acknowledging ideal gas behavior.Thus,\[ \text{For 1 atm at } \approx 290 K \: \text{ to simplify density measurements, close to:} \]\[ V = \frac{PV}{RT} \approx 24 \text{ L/mol equivalent translation for ease;} \text{use percentage values derived in gas mix.} \]\[ \text{Density } = \frac{18 \text{ g}}{12 \text{ L}} \approx 1.5 \text{ g/L} \text{ (hence computationally acceptable near: } 1.515 \text{ g/L made given choice round) }\]
04

Choose the Closest Density

From the calculation, the density is approximately 1.5 g/L. Looking at the provided options, the closest one is \(1.515 \, \text{g/L}\). Therefore, the answer is option (c).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental principle used to relate the pressure, volume, temperature, and number of moles of a gas. The equation is given by \( PV = nRT \) where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume of the gas
  • \( n \) is the number of moles of the gas
  • \( R \) is the ideal gas constant (\(8.314 \, \text{J/(mol K)}\))
  • \( T \) is the temperature of the gas in Kelvin
This law is particularly useful when dealing with gas mixtures because it allows for the calculation of the total behavior of the gases based on the sum of their individual contributions. In this exercise, by utilizing the relationship between the given conditions (1 atm and 290 K), we can determine the overall behavior of the gas mixture in the vessel.
Density Calculation
Density is a measure of the mass of a substance divided by its volume. Calculating the density of a gas mixture involves finding the total mass of the gas and dividing it by the volume it occupies.
  • Total Mass: Sum the masses of each component gas, which in this case is nitrogen (7 g) and carbon dioxide (11 g), resulting in a combined mass of 18 g.
  • Volume of Gas: Use the Ideal Gas Law or given conditions to deduce volume. At 1 atm and 290 K, 1 mole of gas typically occupies around 24 L, which helps in simplifying computations.
  • Density Formula: \( \text{Density} = \frac{\text{Total Mass}}{\text{Volume}} \)
So, the challenge is aligning conditions known for norm gas metric (standard, assumed variables) to help project the effective density of the mixture. The calculation concluded in the problem resulted in a mixture density close to 1.515 g/L.
Mole Concept
The Mole Concept is essential in understanding gas mixtures as it relates the mass of a substance to the number of molecules contained within it. Moles provide a bridge between the atomic scale and the macroscopic scale.
  • Moles and Mass: The number of moles \( n \) is calculated by dividing the mass of the substance by its molar mass: \( n = \frac{\text{mass}}{\text{molar mass}} \).
  • Application to Gases: For gases at standard conditions, the volume occupied by one mole of gas (\(22.4\) L/mol at STP or slightly higher like \(24\) L per problem鈥檚 context explained).
This concept allows comprehension of not only how many molecules or atoms are present but how these molecular quantities transform into the metrics used for calculations in density or changing from individual gas properties in a mixture.
Mixture Composition
In a gas mixture, understanding the composition means knowing which gases are present and in what quantity. This is essential for calculations related to density and other properties.
  • Component Gases: In the given exercise, the mixture includes nitrogen and carbon dioxide.
  • Mass Contribution: The mass of each gas contributes to the total mass, influencing density calculations and overall properties.
  • Behavior of Mixtures: The gas laws, especially for ideal mixtures, allow us to treat these gases separately but sum them up for understanding whole systems behavior.
Mixes correlate with their ratio of presence, so understanding combined tubing is foundational in deriving critical properties for experimentals and revised scenarios.

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Most popular questions from this chapter

Two gases \(A\) and \(B\) are contained in the same vessel which is at temperature \(T\). The number of molecules of gas \(A\) is ' \(N^{\prime}\) and mass of each is ' \(m^{\prime}\). The number of molecules of gas ' \(B^{\prime}\) is \(2 N\), each of mass \((2 m)\). If mean square volocity of molecules of ' \(B\) ' is \(v^{2}\) and mean square velocity of \(x\) component of the velocity of ' \(A^{\prime}\) type is given by ' \(\omega^{2}\), then \(\omega^{2} / v^{2}\) is: (a) 2 (b) 1 (c) \(\frac{1}{3}\) (d) \(\frac{2}{3}\)

Two chambers, one containing ' \(m_{1}^{\prime} \mathrm{g}\) of a gas at ' \(P_{1}\) ' pressure and other containing ' \(m_{2} \mathrm{~g}\) of a gas at \({ }^{\prime} P_{2}{ }^{\prime}\) pressure, are put in communication with each other. If temperature remains constant, the common pressure reached will be : (a) \(\frac{P_{1} P_{2}\left(m_{1}+m_{2}\right)}{P_{2} m_{1}+P_{1} m_{2}}\) (b) \(\frac{m_{1} m_{2}\left(P_{1}+P_{2}\right)}{\left(P_{2} m_{1}+P_{1} m_{2}\right)}\) (c) \(\frac{P_{1} P_{2} m_{1}}{P_{2} m_{1}+P_{1} m_{2}}\) (d) \(\frac{m_{1} m_{2} P_{2}}{\left(P_{2} m_{1}+m_{2} P_{1}\right)}\)

A gas is contained in a closed vessel at \(250 \mathrm{~K}\), then the percentage increase in pressure, if the gas is heated through \(1^{\circ} \mathrm{C}\), is : (a) \(0.4 \%\) (b) \(0.6 \%\) (c) \(0.8 \%\) (d) \(1.0 \%\)

The pressure of a gas kept in an isothermal container is \(200 \mathrm{kPa}\). If half the gas is removed from it, the pressure will be: (a) \(100 \mathrm{kPa}\) (b) \(200 \mathrm{kPa}\) (c) \(400 \mathrm{kPa}\) (d) \(800 \mathrm{kPa}\)

If the temperature of 3 moles of helium gas is increased by \(2 \mathrm{~K}\), then the change in the internal energy of helium gas is: (a) \(70.0 \mathrm{~J}\) (b) \(68.2 \mathrm{~J}\) (c) \(74.8 \mathrm{~J}\) (d) \(78.2 \mathrm{~J}\)
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