The unit of \(a\) in van der Waals equation, \(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T .\) is (a) \(\mathrm{atm} \mathrm{L}^{2} \mathrm{~mol}^{-2}\) (b) atm \(\mathrm{L} \mathrm{mol}^{-2}\) (c) \(\operatorname{atm} \mathrm{L} \mathrm{mol}^{-1}\) (d) \(\mathrm{atm} \mathrm{L}^{2} \mathrm{~mol}^{-1}\)

Short Answer

Expert verified
The unit of 'a' in the van der Waals equation is atm L^2 mol^{-2}, which is option (a).

Step by step solution

01

Identify the correct equation

Firstly, understand that the van der Waals equation corrects the ideal gas law for the volume occupied by gas molecules and the attraction between them. The equation is given by: \((P+\frac{an^2}{V^2})(V-nb) = nRT\), where the term \(\frac{an^2}{V^2}\) corrects for the attractive forces between molecules.
02

Analyze the equation

Assuming that all the terms must balance out dimensionally in the equation, we understand that the units of pressure (P) and the term \(\frac{an^2}{V^2}\) must be the same, since they are being added together.
03

Determine units for 'a'

The pressure P has the units of atm. To match these units, the term \(\frac{an^2}{V^2}\) must also have the units of atm. Thus, solving for the units of a, we have: \[\text{units of a} = \text{units of P} \times \left(\frac{V^2}{n^2}\right)\]Since V is volume, it has the unit L (liters), and n is the amount of substance, which has the unit mol (moles). \[\text{units of a} = \text{atm} \times \left(\frac{L^2}{\text{mol}^2}\right) = \text{atm L}^2 \text{mol}^{-2}\]
04

Choose the correct option

Having determined the units of a as atm L^2 mol^{-2}, we can see that the correct answer matches option (a).

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Key Concepts

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

Intermolecular Forces
When we delve into the realm of chemistry, specifically gas behavior, intermolecular forces play a pivotal role in understanding how gas particles interact with one another. These forces are essentially the attractions or repulsions that occur between molecules, influencing the physical properties of substances, such as boiling points, melting points, and their state of matter under given conditions.

For gases, which tend to have much weaker intermolecular forces compared to solids or liquids, these forces become significant when the gas particles are close together. This could be under conditions of high pressure, low temperature, or both. The presence of these forces means that actual gases do not behave perfectly as predicted by the ideal gas law. To account for the attractive force between molecules, which effectively reduces the pressure exerted by a gas, we introduce the correction factor 'a' in the van der Waals equation. This factor 'a' represents the magnitude of the intermolecular attractive forces at play within a volume of gas, which is crucial for real gases that do not follow ideal behavior.
Real Gases
The concept of real gases is integral when discussing gases that don’t adhere strictly to the ideal gas law. Real gases, unlike the hypothetical ideal gases, are influenced by two major factors: the volume of the gas particles themselves and the intermolecular forces among these particles.

The van der Waals equation is tailored to encompass the behaviors of real gases by incorporating variables 'a' and 'b', which correct for intermolecular attractions and the finite volume of gas molecules, respectively. While real gases might behave nearly ideally at high temperatures and low pressures, where intermolecular forces are less significant, it is the deviations at lower temperatures and higher pressures that the van der Waals equation aims to address. Understanding the real gas law allows us to predict the behavior of gases more accurately in various chemical and physical processes, making it an essential tool for chemists and engineers alike.
Gas Laws
Gas laws are fundamental principles that describe how the pressure, volume, and temperature of a gas relate to each other. In their simplest form, these laws assume that gases are composed of particles that move randomly and do not interact. Among these is the ideal gas law, represented as 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.

However, the ideal gas law falls short in describing the behavior of real gases under certain conditions. This is where the van der Waals equation comes into play, offering a more refined model by adjusting for intermolecular forces and the volume occupied by gas molecules. Through this equation, students can comprehend the more complex nature of gas behavior in real-world scenarios, especially during transitions from ideal to non-ideal conditions. The van der Waals equation showcases the importance of understanding gas laws not just for academic pursuits but also for their practical applications in fields such as meteorology, engineering, and environmental science.

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Most popular questions from this chapter

In the corrections made to ideal gas equation for real gases, the reductions in pressure due to forces of attractions between the molecules is directly peoportional to (a) \(\frac{n}{V}\) (b) \(\frac{n^{2}}{V^{2}}\) (c) \(\boldsymbol{V}-\boldsymbol{n} b\) (d) \(n b\)

Read the following statements and identify the incorrect statement. (a) Volume of one mole of a gas at critical temperature is called molar volume. (b) Pressure of a gas at critical temperature is called critical pressure. (c) The critical temperature, pressure and volume are called critical constants. (d) Critical temperature is the highest temperature at which a gas can exist as liquid, above this temperature it is a gas.

A bubble of air is underwater at temperature \(15^{\circ} \mathrm{C}\) and pressure \(1.5\) bar. If the bubble rises to the surface where the temperature is \(25^{\circ} \mathrm{C}\) and the pressure is \(1.0\) bar, what will happen to the volume of the bubble? (a) Volume will become greater by a factor of \(1.6\), (b) Volume will become greater by a factor of \(1.1 .\) (c) Volume will become smaller by a factor of \(0.70\) (d) Volume will become greater by a factor of \(2.5\).

Boilingpoint of hydrogen fluoride is highest amongst HF, HCl, HBr and HI. Which type of intermolecular forces are present in hydrogen fluoride? (a) \(\mathrm{H}-\mathrm{F}\) has highest van der Waals' forces and dipole moment. (b) \(\mathrm{H}-\mathrm{F}\) has highest London forces. (c) H-F has highest dipole moment hence has dipole-dipole, London forces and hydrogen bonding. (d) H-F has strong intermolecular interactions like dipole-induced dipole.

It is easier to liquefy ammonia than oxygen because (a) it is easier to compress oxygen than \(\mathrm{NH}_{3}\) (b) \(\mathrm{NH}_{3}\) has a very low critical temperature as compared to \(\mathrm{O}_{2}\) (c) \(\mathrm{O}_{2}\) has a higher value of van der Waals constant \(a\) and higher critical temperature than \(\mathrm{NH}_{3}\) (d) \(\mathrm{NH}_{3}\) has a higher value of van der Waals constant \(a\) and higher critical temperature than oxygen.

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