Problem 141 - A Complete Resource Book in Chemistry for JEE Main

Question

A gas at \(350 \mathrm{~K} 15\) atm has a molar volume 12 percent smaller than that calculated from the perfect gas law. Compressibility factor under these conditions can be expressed in scientific notation as \(88 \times 10^{-\mathrm{x}}\). The value of \(x\) is

Step-by-Step Solution

Answer

The value of \( x \) is 2.

Step 1: Understand the Perfect Gas Law

Recall that the perfect gas law is given by the equation \( PV = nRT \), 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. Here, temperature \( T = 350 \mathrm{~K} \) and pressure \( P = 15 \mathrm{~atm} \).

Step 2: Calculate Ideal Molar Volume

For 1 mole, the ideal molar volume \( V_{ideal} \) can be calculated using \( V_{ideal} = \frac{RT}{P} \). \( R \) in atm. L/mol. K is \( 0.0821 \), so, \( V_{ideal} = \frac{0.0821 \times 350}{15} \approx 1.915 \mathrm{~L/mol} \).

Step 3: Calculate Actual Molar Volume

The problem states the real molar volume is 12% smaller than ideal. Therefore, \( V_{actual} = V_{ideal} \times (1 - 0.12) = 1.915 \times 0.88 \approx 1.685 \mathrm{~L/mol} \).

Step 4: Calculate Compressibility Factor (Z)

The compressibility factor \( Z \) is the ratio \( Z = \frac{V_{actual}}{V_{ideal}} \). Therefore, \( Z = \frac{1.685}{1.915} \approx 0.88 \).

Step 5: Convert Z into Scientific Notation

We already have \( Z = 0.88 \), which can be written as \( 88 \times 10^{-2} \). Thus, the value of \( x \) in scientific notation is \( 2 \).

Key Concepts

Perfect Gas Law

The Perfect Gas Law is a fundamental equation that helps describe the behavior of ideal gases. It is expressed as \( PV = nRT \), where:

• \( P \) represents the pressure of the gas,
• \( V \) stands for the volume,
• \( n \) is the number of moles of gas,
• \( R \) is the ideal gas constant (\(0.0821\) L atm/mol K), and
• \( T \) denotes the temperature in Kelvin.

Br> This equation assumes that the gas behaves perfectly, meaning it follows the assumptions of having negligible volume and no intermolecular forces. While this is a useful approximation, in reality, not all gases behave perfectly. Deviations occur under high pressure and low temperature. In the given exercise, the gas deviates from the perfect gas law, having a molar volume 12% smaller than what would be expected in an ideal scenario. Therefore, we must use the compressibility factor to quantify this deviation.

Molar Volume

Molar Volume is the volume occupied by one mole of a substance. For an ideal gas, it is calculated using the Perfect Gas Law formula rearranged for volume:\[V_{ideal} = \frac{RT}{P}\]
In the context of gases, it describes the space one mole of gas occupies under specified conditions of pressure and temperature. For the given exercise, the ideal molar volume is calculated to be approximately \(1.915\) L/mol at a temperature of \(350 \mathrm{~K}\) and a pressure of \(15 \mathrm{~atm}\). However, the real gas deviates from this calculation, suggesting the actual molar volume is only \(1.685\) L/mol. This deviation of 12% indicates real gases can take up less space compared to the ideal calculation, due to interactions between gas molecules.
Understanding molar volume helps in distinguishing between real and ideal gas behaviors, an important step in applications involving gas reactions or conditions where deviations are significant.

Scientific Notation

Scientific Notation is a way to express numbers that are too big or small to be conveniently written in decimal form. It is commonly used in science and engineering to simplify working with very large or very small numbers.Here’s how it works: A number is expressed as a product of a number between 1 and 10, and a power of 10. For example, \( 0.88 \) can be expressed as \( 88 \times 10^{-2} \) in scientific notation.
In the exercise, we find the compressibility factor \( Z \), calculated as \( 0.88 \). It's then represented in scientific notation as \( 88 \times 10^{-2} \), indicating that the decimal has been moved two places to the right, which is why the exponent \(-2\) is negative.
The main advantage of scientific notation is that it simplifies calculations and makes it easier to read and write very large or very small numbers. It is especially useful in chemistry and physics, where one often deals with quantities that span many orders of magnitude.