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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

There is a standard value of temperature and pressure at which the molar volume of a gas is \(22.4 \mathrm{~L}\). The correct values are (a) \(273 \mathrm{~K}, 1 \mathrm{~atm}\) (b) \(300 \mathrm{~K}, 760 \mathrm{~mm}\) (c) \(25^{\circ} \mathrm{C}, 760 \mathrm{~mm}\) (d) \(373 \mathrm{~K}, 1 \mathrm{~atm}\)

A graph is plotted between pressure and volume at different temperatures. On the basis of the graph what changes will you observe in the volume if (i) the pressure is increased at constant temperature. (ii) the temperature is decreased at constant pressure. (a) volume increases in both the cases (b) volume decreases in both the cases (c) volume increases in (i) and decreases in (ii) (d) volume decreases in (i) and increases in (ii).

In a flask of volume \(V\) litres, \(0.2\) mol of axygen, \(0.4\) mol of nitrogen, \(0.1\) mol of ammonia and \(0.3 \mathrm{~mol}\) of helium are endosed at \(27^{\circ} \mathrm{C}\). If the total pressure exerted by these non-reacting gases is one atmosphere, the partial pressure exerted by nitrogen is (a) \(0.1\) atmosphere (d) \(0.4\) atmosphere (c) \(0.3\) atmosphere

A closed container contains equal number of moles of two gases \(X\) and \(Y\) at a total pressure of \(710 \mathrm{~mm}\) of \(\mathrm{Hg}\). If gas \(X\) is removed from the mixture, the pressure will (a) become double (b) become half (c) remain same (d) become one-fourth.

If 4 moles of an ideal gas at 300 K occupy volume of \(89.6 \mathrm{~L}\), then pressure of the gas will be (a) \(2 \mathrm{~atm}\) (b) \(1 \mathrm{~atm}\) (c) \(1.099 \mathrm{~atm}\) (d) \(2.910 \mathrm{~atm}\)

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