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It is observed that \(\mathrm{H}_{2}\) and He gases always show positive deviation from ideal behaviour i.e., \(Z>1\). This is because (a) the value of \(a\) is very large due to high attractive forces (b) the weak intermolecular forces of attraction due to which \(a\) is very small and \(a / V^{2}\) is negligible (c) the value of \(b\) is very large due to large size of the molecules (d) both \(a\) and \(b\) are very small and negligible.

Short Answer

Expert verified
The correct reason why \(\mathrm{H}_{2}\) and He gases show a positive deviation from ideal behavior (\(Z>1\)) is option (d) both 'a' and 'b' are very small and negligible.

Step by step solution

01

Understanding the Compressibility Factor

The compressibility factor, denoted by the symbol 'Z', is a measure of how much the real behavior of a gas deviates from ideal behavior. It is defined as the ratio of the volume of the gas at a given pressure and temperature to the volume predicted by the ideal gas law for the same conditions. The equation is given by \(Z = \frac{PV}{nRT}\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. If \(Z > 1\), it indicates that the gas occupies more volume than expected, suggesting a deviation from ideal behavior.
02

Assessing the Reasons for Positive Deviation

In the case of \(\mathrm{H}_{2}\) and He gases, the positive deviation (\(Z > 1\)) is observed because the actual volume is greater than the volume predicted by the ideal gas law. This can occur for two main reasons related to the van der Waals equation: (a) the 'a' term, which accounts for the attractive forces between molecules, and (b) the 'b' term, which represents the volume occupied by the molecules themselves. For \(\mathrm{H}_{2}\) and He gases, whose molecules are small and exhibit weak intermolecular forces, both 'a' and 'b' would be small. If 'a' was large, that would mean strong attractive forces, which would lead to a negative deviation.
03

Eliminating Incorrect Options

Given the properties of \(\mathrm{H}_{2}\) and He gases, we can systematically eliminate options that don't fit the behavior causing \(Z>1\):(a) This option suggests high attractive forces, which would actually result in negative deviation, so it is incorrect.(c) A large value of 'b' would mean a significant volume correction due to the size of the molecules, which is not characteristic of these small gases, so this is also incorrect.
04

Identifying the Correct Reason

Since we know that \(\mathrm{H}_{2}\) and He gases are small and exhibit weak intermolecular forces, we can conclude that:(b) Options (a) and (c) have been ruled out, leaving us with option (b), which states that 'a' is very small due to weak intermolecular forces and 'a / V^{2}' is negligible, fitting the observed behavior.(d) Both 'a' and 'b' being very small might also contribute to the positive deviation because they are negligible, making this a correct statement as well.
05

Choosing the Best Answer

Between options (b) and (d), we choose the option that best explains why \(Z>1\) specifically for \(\mathrm{H}_{2}\) and He gases. While both could be considered correct, (d) is the more comprehensive answer as it addresses the negligible nature of both 'a' and 'b', which together contribute to the positive deviation from ideal behavior.

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Key Concepts

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

Understanding the Ideal Gas Law
The Ideal Gas Law is a fundamental principle in chemistry and physics, often represented as the equation \(PV = nRT\), which connects the pressure (P), volume (V), number of moles (n), ideal gas constant (R), and temperature (T) of an ideal gas. In this equation, it is assumed that gas particles have no volume and do not attract or repel each other. This provides a useful approximation for how gases behave under many conditions, especially when at high temperature and low pressure.

However, there are instances where real gases do not follow the ideal gas law, as is the case with \(\mathrm{H}_2\) and He gases, which exhibit a positive deviation and have a compressibility factor (Z) greater than 1. This deviation stems from the failure of the ideal assumptions to accurately capture the behavior of these gases in reality, specifically their intermolecular forces and finite size.
The Van der Waals Equation and Real Gases
The Van der Waals Equation is an important refinement of the Ideal Gas Law that accounts for the real conditions of gases. It includes factors to correct for the volume of the gas particles themselves, not considered in the ideal gas law. The equation can be represented as \[\left(P + \frac{n^2a}{V^2}\right)\left(V - nb\right) = nRT\], where 'a' accounts for attractive forces between particles, and 'b' represents the volume occupied by the particles.

For gases like \(\mathrm{H}_2\) and He, which are known for their small sizes and weak intermolecular forces, the constants 'a' and 'b' are relatively insignificant. This characteristic leads to their real behavior deviating from the ideal, with a larger volume than anticipated, thus resulting in a compressibility factor (Z) that is greater than the value of 1 predicted by the Ideal Gas Law.
Real Gas Deviation
Real Gas Deviation refers to the difference between the behavior of real gases and the predictions of the Ideal Gas Law. Variation from ideality is quantified using the compressibility factor (Z), which equals the actual volume divided by the volume predicted by the ideal equation. For real gases, this deviation occurs because of the presence of intermolecular forces and the finite size of gas molecules.

When dealing with very small molecules such as \(\mathrm{H}_2\) and He, which show positive deviation, we see that they behave less ideally due to their weak intermolecular forces being overwhelmed by the kinetic energy of the particles, causing them to occupy a greater volume. This greater-than-expected volume leads to a compressibility factor greater than 1. Understanding real gas deviation is crucial in accurately predicting and manipulating the behavior of gases in real-world applications.

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Most popular questions from this chapter

If the assumption that there is no force of attraction between the molecules of a gas is correct, what will be the consequences? (a) All gases will be ideal gases. (b) The gases will never liquefy when cooled and compressed. (c) Gases will have definite volume. (d) Gases will occupy a definite space.

What is the relationship between thermal energy and intermolecular interaction energy of a substance in three states in terms of \(X\) and \(Y ?\) Gas \(\longrightarrow\) Liquid \(\longrightarrow\) Solid (a) \(X\)-Thermal energy, \(Y\)-Intermolecular interactions (b) \(X\)-Thermal energy, \(Y\)-Thermal energy (c) \(X\)-Intermolecular interactions, \(Y\)-Thermal energy (d) \(X\)-Intermolecular interactions, \(Y\)-Intermolecular interactions

At 1 atmospheric pressure and \(0^{\circ} \mathrm{C}\), certain mass of a gas measures \(0.4\) L. Keeping the pressure constant, if the temperature is increased to \(273^{\circ} \mathrm{C}\), what will be its volume? (a) \(0.8 \mathrm{~L}\) (b) \(22.4 \mathrm{~L}\) (c) \(54.6 \mathrm{~L}\) (d) \(0.4 \mathrm{~L}\)

Liquids are similar to gases because (a) both possess the property of flowing and take the volume of the containers (b) both diffuse and take the shape of the containers (c) both are readily compressible and diffuse (d) both are capable of infinite expansion.

Taking into account the pressure and volume corrections, the gas equation can be written as (a) \(\left(P+\frac{a^{2}}{V^{2}}\right)(V-b)=n R T\) (b) \(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T\) (c) \(\left(P+\frac{a V}{n R T}\right)\left(\frac{V-b}{n R T}\right)=R T\) (d) \(\left(P+\frac{Z}{V^{2}}\right)(V-Z b)=n R T\)

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