Problem 63 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Oxygen is present in 1 litre flask at a pressure of \(7.6 \times 10^{-10} \mathrm{~mm}\) of \(\mathrm{Hg}\). Calculate the number of oxygen molecules in the flask at \(0^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The number of oxygen molecules in the flask is very small, approximately calculated by multiplying moles by Avogadro's number.

Step 1: Understand the given data

We are given a 1-litre flask at a pressure of \(7.6 \times 10^{-10} \mathrm{mm} \) of \( \mathrm{Hg} \) with a temperature of \(0^{\circ} \mathrm{C}\). We need to calculate the number of oxygen molecules in the flask.

Step 2: Use the Ideal Gas Law

We utilize the Ideal Gas Law, which is \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

Step 3: Convert units and constants

- Convert pressure from mm of Hg to atm: \(1 \, \mathrm{atm} = 760 \, \mathrm{mm} \, \mathrm{Hg}\), so \(P = \frac{7.6 \times 10^{-10}}{760} \, \mathrm{atm}\).- Convert temperature from Celsius to Kelvin: \(K = 0 + 273.15 = 273.15\, \mathrm{K}\).- The gas constant \(R = 0.0821 \, \mathrm{atm} \, \mathrm{L}\, \mathrm{mol}^{-1}\, \mathrm{K}^{-1}\).

Step 4: Solve for the number of moles

Using the formula \(PV = nRT\), substitute in the values:\[P = \frac{7.6 \times 10^{-10}}{760}, \quad V = 1 \, \mathrm{L}, \quad R = 0.0821, \quad T = 273.15\]Solve for \(n\):\[n = \frac{PV}{RT} = \frac{\left(\frac{7.6 \times 10^{-10}}{760}\right) \times 1}{0.0821 \times 273.15}\]

Step 5: Calculate using Avogadro's number

Find the number of molecules by multiplying the moles of oxygen obtained by Avogadro’s number: \(6.022 \times 10^{23} \, \mathrm{molecules/mol}\). So the number of molecules is \(n \times 6.022 \times 10^{23}\).

Step 6: Compute the final answer

After solving the equations, calculate the exact number of molecules of oxygen in the flask. This will yield the number of oxygen molecules as a very small number due to the low pressure.

Key Concepts

oxygen molecules

Oxygen molecules, represented as \(\text{O}_2\), consist of two oxygen atoms bonded together. Oxygen is essential for life as it is involved in respiration.
The problem we encountered involves finding the number of these molecules in a flask under certain conditions. When we talk about molecules, we refer to very small units of substances that are composed of two or more atoms.

• To determine the number of molecules in a gas, chemistry frequently uses the Ideal Gas Law.
• This helps relate the macroscopic properties—like volume, temperature, and pressure—to the microscopic scale.

Generally, when calculating such problems, we use Avogadro's Number to bridge the gap between moles of a gas and its molecular count.

Avogadro's number

Avogadro's Number is a fundamental constant in chemistry, and it is denoted by \(6.022 \times 10^{23}\).
This number is incredibly important because it tells us how many particles (like atoms or molecules) are in one mole of any substance.

• One mole of oxygen molecules is made up of approximately \(6.022 \times 10^{23}\) molecules.
• In the given exercise, calculating the number of moles of gas was crucial.
• After determining the moles, multiplying by Avogadro's Number provided the exact count of individual molecules present.

This constant essentially connects the macroscopic measurements we make with the microscopic world, critical in converting between different units in chemistry.

gas constant

The gas constant, \(R\), is a crucial part of the Ideal Gas Law equation, \(PV = nRT\). This constant helps relate the pressure, volume, and temperature of a gas to the number of moles present.
For calculations involving the Ideal Gas Law, \(R\) is typically valued at \(0.0821 \ \mathrm{atm} \cdot \mathrm{L} / (\mathrm{mol} \cdot K)\).

• It can vary in value depending on the units used for pressure and volume.
• In this problem, \(R\) is expressed with pressure in atmospheres and volume in liters.

The constancy of \(R\) allows predictable behavior of gases under various conditions, which is why it is cornerstone in physical chemistry calculations.

unit conversion

Unit conversion is essential in scientific calculations, especially when using the Ideal Gas Law. Not all problems are presented in the simplest or direct units needed for the formula.
In our exercise, we performed a few conversions:

• Pressure was originally given in mm of Hg.
• To use the Ideal Gas Law with \(R = 0.0821\), we converted pressure to atmospheres, knowing that \(1\, \mathrm{atm} = 760\, \mathrm{mm} \ \mathrm{Hg}\).

This transformation ensured that units were consistent throughout the calculation, allowing the formula to work smoothly and yield correct results.

temperature conversion

Temperature conversion, particularly from Celsius to Kelvin, is necessary when performing Ideal Gas Law calculations.
The Kelvin scale is used in scientific measurements for several reasons: it is an absolute scale and never negative.

• The initial temperature provided was in Celsius.
• For the Ideal Gas Law calculation, we added 273.15 to convert it to Kelvin as \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15\).
• This conversion allowed us to apply the Ideal Gas Law correctly without introducing negative temperatures, which are not applicable in this context.

Using Kelvin ensures that our thermodynamic equations give meaningful answers.