Question

The compressibility factor for nitrogen at \(330 \mathrm{~K}\) and \(800 \mathrm{~atm}\) is \(1.90\) and at \(570 \mathrm{~K}\) and 200 atm is \(1.10 .\) A certain mass of \(\mathrm{N}_{2}\) occupies a volume of \(1 \mathrm{dm}^{3}\) at \(330 \mathrm{~K}\) and \(800 \mathrm{~atm} .\) Calculate volume occupied by same quantity of \(\mathrm{N}_{2}\) gas at \(570 \mathrm{~K}\) and \(200 \mathrm{~atm}\) : (a) \(1 \mathrm{~L}\) (b) \(2 \mathrm{~L}\) (c) \(3 \mathrm{~L}\) (d) \(4 \mathrm{~L}\)

Short answer

2 L

Step-by-step solution

Understand the Compressibility Factor

The compressibility factor (Z) is a measure of how much the real behavior of a gas deviates from the ideal behavior. It is defined as Z = PV/RT, where P is pressure, V is volume, R is the gas constant, and T is temperature.

Convert units and apply initial conditions

First, ensure all units are consistent. Since 1 dm^3 is equivalent to 1 L (liter), the initial volume is 1 L. Apply the initial conditions to find the moles of N2 using the compressibility factor (Z1 = 1.90) for temperature T1 = 330 K and pressure P1 = 800 atm.

Calculate initial moles of N2

Use the compressibility factor to calculate the initial moles n of N2, using the formula n = (P1 * V) / (Z1 * R * T1), where R is the ideal gas constant with a value of 0.0821 L atm / (mol K).

Apply final conditions to solve for final volume

Now use the final conditions with compressibility factor Z2 = 1.10, temperature T2 = 570 K, and pressure P2 = 200 atm along with the moles n calculated in Step 3, to find the final volume V2 using the formula V2 = (Z2 * n * R * T2) / P2.

Compare results and choose correct option

Finally, compare the calculated volume with the options given to determine the correct answer.

Key concepts

Physical Chemistry JEE

When preparing for competitive exams like the Joint Entrance Examination (JEE), which is a gateway to prestigious engineering colleges in India, understanding the nuances of physical chemistry is crucial. Among the fundamental concepts, the study of gases holds a significant place.

In the context of the exercise concerning the compressibility factor, students preparing for the JEE must be well-versed in real gas equations and how real gases deviate from ideal behavior. The compressibility factor is a perfect example to illustrate this deviation. Moreover, solving related problems helps in developing analytical skills required to tackle the physical chemistry section of the JEE. Students must pay close attention to the units and conversion factors, which can often be a source of error in calculations.

Ideal Gas Constant

The ideal gas constant, represented as 'R', is a fundamental parameter in the equation of state for an ideal gas. Its value is the same for all gases and amounts to approximately 0.0821 Latm / (molK) or 8.314 J/(molK) in SI units.

Understanding the role of the ideal gas constant is essential when transitioning to problems involving real gases. In the given exercise, the use of the ideal gas constant connects the number of moles, temperature, and pressure to the volume a gas occupies.

Importance in Calculations

In calculations, it is pivotal to use the correct units for R to avoid mismatches, as seen in the transition from an initial state to a final state of nitrogen in the problem. Remember always to check the consistency of units in any given problem to ensure that the calculations are correct.

Real Gas Behavior

Real gases do not always follow the ideal gas law due to intermolecular forces and the volume occupied by the gas molecules themselves. The ideal gas law serves as a foundation for understanding gases but often requires adjustment for real-world scenarios.

This leads to the introduction of the compressibility factor (Z), which is the ratio of the product of pressure and volume to the product of the ideal gas constant and temperature. It thus provides a correction for the ideal gas equation to account for real gas behavior.

Significance of Compressibility Factor (Z)

A Z value of 1 indicates ideal behavior, while values diverging from 1 signal real gas behavior that deviates from ideality. This concept is crucial in the problem provided, as nitrogen is experiencing high pressures and temperatures, leading to a compressibility factor significantly different from 1. Such exercises play a pivotal part in comprehending thermodynamics and are often included in JEE chemistry curricula to test the students' grasp of real gas behavior.