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The compressibility of a gas is greater than unity at \(1 \mathrm{~atm}\) and \(273 \mathrm{~K}\). Therefore : (a) \(V_{m}>22.4 \mathrm{~L}\) (b) \(V_{m}<22.4 \mathrm{~L}\) (c) \(V_{m}=22.4 \mathrm{~L}\) (d) \(V_{m}=44.8 \mathrm{~L}\)

Short Answer

Expert verified
\( V_{m} > 22.4 \mathrm{~L} \)

Step by step solution

01

Understanding Compressibility Factor (Z)

Compressibility factor (Z) is defined as the ratio of the volume occupied by a real gas to the volume occupied by an ideal gas at the same temperature and pressure. Mathematically, Z can be expressed as: \( Z = \frac{V_m}{V_{m,ideal}} \), where \( V_m \) is the molar volume of the real gas and \( V_{m, ideal} \) is the molar volume of the ideal gas under similar conditions. If Z > 1, it suggests the real gas occupies more volume than an ideal gas would at the same temperature and pressure.
02

Analyzing the Given Condition

The given compressibility of the gas is greater than unity (Z > 1) at 1 atm and 273 K. This means that the gas is less compressible than expected for an ideal gas and therefore occupies a larger volume. At these conditions (STP), the molar volume of an ideal gas (\( V_{m, ideal} \)) is 22.4 L.
03

Determining the Molar Volume \( V_m \)

Since Z is greater than 1, the actual molar volume \( V_m \) of the gas must be greater than that of an ideal gas at STP, meaning \( V_m > 22.4 \) L. This directly leads us to the conclusion.

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Key Concepts

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

Real Gas vs Ideal Gas
To understand the behavior of gases under various conditions, scientists use models, and two of the most important models are 'real gas' and 'ideal gas'. An ideal gas is a hypothetical gas that perfectly follows the Ideal Gas Law (\( PV = nRT \text{, where P is pressure, V is volume, n is the amount of substance in moles, R is the ideal, or universal, gas constant, and T is temperature}\)), with the assumptions that the gas particles have no volume and that there are no interactions between the particles. Ideal gases are a simplification that make calculations easier but sometimes fail to accurately describe gas behavior under certain conditions, such as high pressure or low temperature.

In contrast, a real gas is what we find in nature - the behavior of actual gases. Real gases have particles with definite volume, and the particles attract or repel each other. These factors must be accounted for, which is where the compressibility factor (Z) comes into play. It helps quantify how much a real gas deviates from ideal gas behavior by comparing the molar volume of a real gas (\( V_m \text{, real}\)) to that of an ideal gas at the same temperature and pressure. The closer Z is to 1, the more closely the gas resembles an ideal gas. When Z is significantly different from 1, it indicates that real gas effects are important, and the ideal gas law does not accurately predict the gas's behavior.
Molar Volume
The molar volume of a substance is the volume that one mole of the substance occupies at a given temperature and pressure. It's a critical concept when working with gases because it allows for direct comparison of different gases under the same conditions. For ideal gases, the molar volume at standard temperature and pressure (0°C and 1 atm) is 22.4 L/mol. However, this is based on the assumptions of an ideal gas model and does not always hold for real gases due to reasons such as intermolecular forces and the actual size of the gas particles.

When dealing with real gases, we have to consider the actual conditions and use the compressibility factor to adjust the molar volume accordingly. If the real gas has a compressibility factor greater than 1, like in our exercise, this indicates that the real gas’s molar volume is greater than the molar volume of an ideal gas under identical conditions. Consequently, molecular interactions or volume occupied by the gas molecules themselves prevent the gas from being compressed to the same extent as an ideal gas, resulting in a larger molar volume.
Standard Temperature and Pressure (STP)
When scientists and engineers discuss gases, they often refer to Standard Temperature and Pressure (STP) as a common reference point. STP is defined as a temperature of 0°C (273 K) and a pressure of 1 atmosphere (atm). It's used as a standard to report properties of materials, especially gases because their volume can vary greatly with changes in temperature and pressure.

At STP, the molar volume for an ideal gas is known to be 22.4 liters per mole, a value derived from the Ideal Gas Law. This provides a convenient basis for comparing the behavior of different gases and performing stoichiometric calculations in chemistry. Yet, it’s important to remember that this value assumes ideal gas behavior, which may not be accurate for real gases at STP due to the factors that cause them to deviate from ideality. This deviation is exactly what the compressibility factor tells us - illustrating whether or not we can use the ideal gas assumptions for a specific gas at STP, or if corrections are required to account for real gas behavior.

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Most popular questions from this chapter

\(2.8 \mathrm{~g}\) of a gas at \(1 \mathrm{~atm}\) and \(273 \mathrm{~K}\) occupies a volume of \(2.24\) litres, the gas can not be: (a) \(\mathrm{O}_{2}\) (b) \(\mathrm{CO}\) (c) \(\mathrm{N}_{2}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{4}\)

Three flasks of equal volumes contain \(\mathrm{CH}_{4}, \mathrm{CO}_{2}\) and \(\mathrm{Cl}_{2}\) gases respectively. They will contain equal number of molecules if : (a) the mass of all the gases is same (b) the moles of all the gas is same but temperature is different (c) temperature and pressure of all the flasks are same (d) temperature, pressure and masses same in the flasks

At \(273 \mathrm{~K}\) temp. and 9 atm pressure, the compressibility for a gas is \(0.9 .\) The volume of 1 milli-moles of gas at this temperature and pressure is : (a) \(2.24\) litre (b) \(0.020 \mathrm{~mL}\) (c) \(2.24 \mathrm{~mL}\) (d) \(22.4 \mathrm{~mL}\)

Dry ice is solid carbon dioxide. A \(0.050 \mathrm{~g}\) sample of dry ice is placed in an evacuated \(4.6 \mathrm{~L}\) vessel at \(30^{\circ} \mathrm{C}\). Calculate the pressure inside the vessel after all the dry ice has been converted to \(\mathrm{CO}_{2}\) gas. (a) \(6.14\) atm (b) \(0.614\) atm (c) \(0.0614 \mathrm{~atm}\) (d) \(6.14 \times 10^{-3}\) atm

Under identical conditions of pressure and temperature, \(4 \mathrm{~L}\) of gaseous mixture \(\left(\mathrm{H}_{2}\right.\) and \(\left.\mathrm{CH}_{4}\right)\) effuses through a hole in 5 min whereas \(4 \mathrm{~L}\) of a gas \(X\) of molecular mass 36 takes to \(10 \mathrm{~min}\) to effuse through the same hole. The mole ratio of \(\mathrm{H}_{2}: \mathrm{CH}_{4}\) in the mixture is : (a) \(1: 2\) (b) \(2: 1\) (c) \(2: 3\) (d) \(1: 1\)

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