Problem 6 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An open vessel at \(27^{\circ} \mathrm{C}\) is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is : (a) \(500^{\circ} \mathrm{C}\) (b) \(500 \mathrm{~K}\) (c) \(750{ }^{\circ} \mathrm{C}\) (d) \(750 \mathrm{~K}\)

Step-by-Step Solution

Answer

The temperature is 500 K (option b).

Step 1: Understand the Problem

We are given a vessel containing air, which we will treat as an ideal gas. Initially, the temperature is 27°C. We need to find the final temperature at which two-fifths of the air has escaped. The vessel is open, so the pressure is constant, and we need to consider the volumetric ratios of the air.

Step 2: Identify Known Values and Formulas

Initial temperature \(T_1\) is 27°C, which is equivalent to \(27 + 273 = 300\, \mathrm{K}\). We need to find the final temperature \(T_2\). Since two-fifths of the air has escaped, three-fifths remains, i.e., \(n_2 = \frac{3}{5} n_1\). The pressure and volume are constant, so we'll use the relation \(\frac{n_1 T_1}{P} = \frac{n_2 T_2}{P}\).

Step 3: Set Up the Equation

Using the relationship \(n_1 T_1 = n_2 T_2\), substitute \(n_2 = \frac{3}{5} n_1\) to get \(n_1 \cdot 300 = \left(\frac{3}{5} n_1\right) T_2\). Cancel out \(n_1\) from both sides.

Step 4: Solve for Final Temperature

Simplify the equation \(300 = \left(\frac{3}{5}\right) T_2\) to find \(T_2\). Multiply both sides by \(\frac{5}{3}\) to isolate \(T_2\):\[T_2 = 300 \times \frac{5}{3} = 500\, \mathrm{K}.\]

Step 5: Conversion and Conclusion

The final temperature in Kelvin is \(500\, \mathrm{K}\). No conversion to Celsius is necessary. Match this result with the answer choices provided, and we see that option (b) \(500 \mathrm{~K}\) is correct.

Key Concepts

Open System

When we talk about an "+open system+", in the context of gases, it simply means that the system is not closed off from its surroundings. In other words, gases or molecules can move in or out of the system without restriction. This is crucial when considering how gases behave in various environments.

In an open system, like the one in this exercise, the pressure is typically constant because the system is in equilibrium with the atmospheric pressure. This equilibrium allows us to focus on other aspects of the problem, such as temperature and volume, without worrying about pressure changes. Remember, the volume of the vessel is constant in this situation, but the amount of gas in it is changing.

• Open systems can interact with their environment.
• Pressure remains constant when the system is open and in equilibrium with the surroundings.
• Gas molecules in an open system can escape, leading to changes in quantity but not pressure or volume.

Temperature Conversion

Understanding "+temperature conversion+" is vital, especially when calculating scenarios involving gases. We frequently switch between Celsius and Kelvin in scientific problems. The reason for the emphasis on Kelvin is that Kelvin is the standard unit of temperature in the scientific world, especially when dealing with the Ideal Gas Law.

To convert Celsius to Kelvin, simply add 273 to the Celsius temperature. This is because 0 degrees Celsius is equivalent to 273.15 Kelvin. For this exercise, starting with 27°C, we add 273 to obtain 300 K as the initial temperature.

• Celsius to Kelvin conversion: Add 273 to the Celsius value.
• Kelvin is the preferred scale for scientific calculations, including those in the Ideal Gas Law.
• Keep conversions in mind while solving problems with the Ideal Gas Law.

Fraction of Molecules Escaping

In this problem, the "+fraction of molecules escaping+" is an important concept as it affects how we calculate the final state of the system. Initially, when the air molecules are trapped in the vessel, they occupy a certain volume at a given temperature and pressure. But as heat is applied, some molecules gain enough energy to escape.

Here, exactly two-fifths of the air molecules have escaped, meaning three-fifths remain. This ratio directly influences the calculation for the final temperature. It showcases a practical application of the conservation of moles balance in the context of gases.

• The fraction that escapes is an indicator of energy changes within the system.
• Only 60% of the molecules remain after two-fifths have left the vessel.
• This leftover fraction is crucial in determining the new equilibrium state, particularly the final temperature.