/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 The density of neon will be high... [FREE SOLUTION] | 91Ó°ÊÓ

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The density of neon will be highest at a. STP b. \(0^{\circ} \mathrm{C}, 2 \mathrm{~atm}\) c. \(273^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) d. \(273^{\circ} \mathrm{C}, 2 \mathrm{~atm}\)

Short Answer

Expert verified
The density of neon is highest at 0°C and 2 atm, which is Option B.

Step by step solution

01

Understand the Ideal Gas Law

To determine the density of neon, we use the Ideal Gas Law equation, which is \( PV = nRT \). Here, \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. Density, \( \rho \), is equal to mass divided by volume \( \rho = \frac{m}{V} \), and can also be expressed as \( \rho = \frac{PM}{RT} \) where \( M \) is the molar mass of the gas.
02

Calculate Density at Different Conditions

For each option, calculate the density using the formula \( \rho = \frac{PM}{RT} \). The molar mass \( M \) of neon is a constant at \( 20.18 \text{g/mol} \). The conditions for each option are:- **Option A** (STP): \( P = 1 \text{ atm}, T = 273.15 \text{ K} \)- **Option B**: \( P = 2 \text{ atm}, T = 273.15 \text{ K} \)- **Option C**: \( P = 1 \text{ atm}, T = 546.15 \text{ K} \) (273°C converted to Kelvin)- **Option D**: \( P = 2 \text{ atm}, T = 546.15 \text{ K} \)
03

Compare the Calculated Densities

For each option, calculate \( \rho \). The value of \( R \) (in \( \text{atm} \cdot \text{L/mol} \cdot \text{K} \)) is 0.0821:- **Option A**: \( \rho = \frac{1 \times 20.18}{0.0821 \times 273.15} \approx 0.893 \text{ g/L} \)- **Option B**: \( \rho = \frac{2 \times 20.18}{0.0821 \times 273.15} \approx 1.786 \text{ g/L} \)- **Option C**: \( \rho = \frac{1 \times 20.18}{0.0821 \times 546.15} \approx 0.446 \text{ g/L} \)- **Option D**: \( \rho = \frac{2 \times 20.18}{0.0821 \times 546.15} \approx 0.893 \text{ g/L} \)
04

Identify the Highest Density Condition

After comparing the densities from Step 3, we see that the density is highest for **Option B** at \( 0^{\circ} \text{C} \) and \( 2 \text{ atm} \), with a value of approximately \( 1.786 \text{ g/L} \). The pressure is doubled while keeping the temperature the same as STP, leading to higher density.

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Key Concepts

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

Gas Density Calculation
Calculating gas density often requires using the Ideal Gas Law, a fundamental equation in chemistry. This law is expressed as \( PV = nRT \), where:
  • \( P \) is the pressure of the gas.
  • \( V \) is the volume it occupies.
  • \( n \) is the amount in moles.
  • \( R \) is the ideal gas constant, approximately 0.0821 when using atm, L, mol, and K as units.
  • \( T \) is the temperature in Kelvin.
To find the density of a gas, we rearrange this equation to \( \rho = \frac{PM}{RT} \), where \( M \) is the molar mass of the gas. This formula shows that density \( \rho \) is directly proportional to pressure \( P \) and inversely proportional to temperature \( T \). When figuring out the density, it's important to convert all measurements to correct units — atm for pressure, Kelvin for temperature, and g/mol for molar mass. The correct application of these relationships ensures accurate density calculations.
Pressure and Temperature Effects on Gas
Pressure and temperature significantly impact gas density. Generally, as pressure increases, gas molecules are forced closer together, thus increasing the density. This relationship becomes clear in the formula \( \rho = \frac{PM}{RT} \). Higher pressure \( P \) results in a higher density \( \rho \), assuming the temperature \( T \) and molar mass \( M \) remain unchanged.On the other hand, an increase in temperature leads to a decrease in density. When gas is heated, its molecules move faster and spread out, which increases the volume and reduces density, keeping pressure constant. This is why the ideal gas law equation reflects that temperature \( T \) is inversely proportional to density when other factors are constant.Understanding these effects allows us to predict and explain changes in gas behavior under different conditions. For example, gases will expand if warmed or compressed if subjected to higher pressure. These fundamental concepts are vital for engineering, environmental science, and even weather predictions.
Molar Mass and Gas Density
Molar mass plays a crucial role in determining the density of a gas. The molar mass \( M \) refers to the mass of one mole of a substance, measured in grams per mole (g/mol). In our density equation \( \rho = \frac{PM}{RT} \), it appears that the molar mass is directly proportional to the density \( \rho \). Thus, gases with higher molar masses will have higher densities when all other variables remain constant.For example, neon gas, with a molar mass of 20.18 g/mol, will have a specific density under defined conditions. In our exercise, we used different temperature and pressure conditions to calculate the densities. This showcases how variations in molar mass, alongside changes in pressure and temperature, can affect gas density. By understanding molar mass’s influence, one can predict how a gas will behave, helping in industries ranging from pharmaceuticals to aerospace.

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Most popular questions from this chapter

If a gas expands at constant temperature a. the pressure decreases b. the kinetic energy of the molecules remains the same c. the kinetic energy of the molecules decreases d. the number of molecules of the gas increase.

Indicate the correct statement for equal volumes of \(\mathrm{N}_{2}(\mathrm{~g})\) and \(\mathrm{CO}_{2}(\mathrm{~g})\) at \(25^{\circ} \mathrm{C}\) and \(\mathrm{l} \mathrm{atm} .\) a. The rms speed remains constant for both \(\mathrm{N}_{2}\) and \(\mathrm{CO}_{2}\) b. The average translational KE per, molecule is the same for \(\mathrm{N}_{2}\) and \(\mathrm{CO}_{2}\) c. The total translational \(\mathrm{KE}\) of both \(\mathrm{N}_{2}\) and \(\mathrm{CO}_{2}\) is the same d. The density of \(\mathrm{N}_{2}\) is more that that of \(\mathrm{CO}_{2}\)

If the vapour shows ideal gas behaviour at \(1000 \mathrm{~K}\) than its average transitional kinetic energy will be? a. \(1.035 \times 10^{-10} \mathrm{~J}\) b. \(2.07 \times 10^{-30} \mathrm{~J}\) c. \(4.41 \times 10^{-20} \mathrm{~J}\) d. \(2.07 \times 10^{-20} \mathrm{~J}\)

The correct statement(s) amongst the following is/are a. A gas above its critical temperature can be liquefied by applying the minimum pressure b. Deviation of real gases from ideal behaviour becomes more pronounced as pressure is increased and temperature is decreased. c. At \(0^{\circ} \mathrm{C}, \mathrm{H}_{2}\) and He show positive deviation throughout with increasing pressure indicating thereby that the Boyle's temperatures for these two gas are below \(0^{\circ} \mathrm{C}\) d. \(\mathrm{N}_{2}(\mathrm{~g})\) shows ideal behaviour for some range of pressure \((0-100 \mathrm{~atm})\) at \(51.1^{\circ} \mathrm{C}\) indicating that \(51.1^{\circ} \mathrm{C}\) is the Boyle's temperature for \(\mathrm{N}_{2}(\mathrm{~g})\)

Whicle of the fulluwing is the incorrect statement about the Boyle's temperature \(\left(T_{B}\right)\) ? a. The value of \(\mathrm{T}_{\mathrm{B}}\) is equal to \(\mathrm{a} / \mathrm{Rb}\) b. Temperature at which second virial coefficient becomes zero c. Temperature at which first virial coefficient becomes zero d. Both a. and b.

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