Problem 160 - A Complete Resource Book in Chemistry for JEE Main

Question

Two closed bulbs of equal volume ( \(V\) ) containing an ideal gas initially at pressure \(P_{1}\) and temperature \(T_{1}\) are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to \(T_{2}\). The final pressure \(p_{f}\) is (a) \(2 p_{t}\left(\frac{T_{1}}{T_{1}+T_{2}}\right)\) (b) \(2 p_{t}\left(\frac{T_{2}}{T_{1}+T_{2}}\right)\) (c) \(2 p_{i}\left(\frac{T_{1} T_{2}}{T_{1}+T_{2}}\right)\) (d) \(p_{t}\left(\frac{T_{1} T_{2}}{T_{1}+T_{2}}\right)\)

Step-by-Step Solution

Answer

The final pressure \(P_f\) is option (c): \(2 p_i \left(\frac{T_1 T_2}{T_1+T_2}\right)\).

Step 1: Understand Initial Conditions

Initially, each bulb has a volume \(V\) and contains an ideal gas at pressure \(P_1\) and temperature \(T_1\). Since there are two bulbs of equal volume, the total volume is \(2V\).

Step 2: Apply Ideal Gas Law to Both Bulbs

For the initial state, the number of moles \(n\) in one bulb can be given by the ideal gas equation: \[ n_1 = \frac{P_1 V}{R T_1} \]Since there are two bulbs, the total initial number of moles \(n_{total}\) is \[ n_{total} = 2n_1 = \frac{2P_1 V}{R T_1} \]

Step 3: Consider the Change in Temperature

Let the first bulb remain at \(T_1\), and raise the temperature of the second bulb to \(T_2\). The number of moles in the first bulb after change remains \(n_1 = \frac{P_1 V}{R T_1}\). For the second bulb, apply: \[ n_2 = \frac{P_2 V}{R T_2} \]

Step 4: Apply Conservation of Moles

The total number of moles should remain constant:\[ n_1 + n_2 = \frac{2P_1 V}{R T_1} \]Rearrange this equation:\[ \frac{P_f V}{R T_1} + \frac{P_f V}{R T_2} = \frac{2P_1 V}{R T_1} \]

Step 5: Solve for Final Pressure \(P_f\)

Simplify and solve for \(P_f\):\[ P_f \left( \frac{1}{T_1} + \frac{1}{T_2} \right) = \frac{2P_1}{T_1} \]\[ P_f = \frac{2P_1}{\frac{1}{T_1} + \frac{1}{T_2}} = \frac{2P_1 T_1 T_2}{T_1 + T_2} \]

Step 6: Select the Correct Answer

The expression \(P_f = \frac{2P_1 T_1 T_2}{T_1 + T_2}\) matches option (c), which states \(2 p_i \left(\frac{T_1 T_2}{T_1 + T_2}\right)\).

Key Concepts

Pressure-Volume-Temperature Relationship

The relationship between pressure, volume, and temperature is a cornerstone of the Ideal Gas Law, which is expressed as \( PV = nRT \). Here, \( P \) represents the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This law tells us how these properties change and interact with each other.In the exercise, both bulbs initially have the same pressure \( P_1 \), volume \( V \), and temperature \( T_1 \). This means the gas within each bulb initially follows the equation \( P_1V = n_1RT_1 \). When the temperature of one bulb is increased to \( T_2 \), the volume remains constant but the pressure adjusts according to the new temperature and the same initial number of moles.It's important to remember:

• Pressure of a gas increases if the temperature increases, assuming the volume remains constant.
• A drop in temperature results in decreased pressure if volume remains unchanged.

Understanding this relationship allows us to predict how the final pressure is calculated when conditions change as they did in the problem.

Conservation of Moles

When dealing with an ideal gas in a closed system, the number of moles remains constant unless there is a leak or a reaction occurs inside. That's the principle of conservation of moles. In the given exercise setup, the two bulbs are closed and connected, ensuring that no gas escapes and no additional moles are added.Initially, the total amount of gas across both bulbs calculated by \( n_{total} = \frac{2P_1 V}{R T_1} \). After changing the temperature of the second bulb to \( T_2 \), it's important to maintain the same total number of moles through the process. Thus, the new equation becomes \( \frac{P_f V}{R T_1} + \frac{P_f V}{R T_2} = \frac{2P_1 V}{R T_1} \).This holds true because no gas was lost during the temperature increase. Such understanding of mole conservation aids in correctly assessing how pressure adjusts within a thermally manipulated environment. By keeping the total count of moles constant, the final calculated pressure \( P_f \) is obtained.

Thermodynamics

The study of thermodynamics primarily focuses on how heat energy converts to different forms and how it affects matter. In this problem, the second bulb is subjected to a change in temperature, a significant component of thermal dynamics.By increasing the temperature to \( T_2 \), the internal energy of the gas increases, affecting how the pressure distributes itself within the bulbs. The final pressure \( P_f \) is then calculated based on this energy transformation and the resulting equilibrium.Key points related to thermodynamics include:

• Energy conservation means the increase in temperature of one bulb must lead to a corresponding increase in energy resulted as pressure changes within the setup.
• Using thermodynamic principles to understand how heat affects molecular movement leading to pressure changes showcases the practical applications of these laws in real-life scenarios.

The exercise demonstrates how temperature manipulation translates into measurable physical changes guided by thermodynamic laws.

Gas Laws for JEE

Preparation for exams like the Joint Entrance Examination (JEE) requires a firm grasp of gas laws and their applications. The Ideal Gas Law, along with others like Boyle's and Charles' Laws, forms the basis for many such problems. In our particular exercise, applying the Ideal Gas Law correctly helps to understand changes in physical properties when external conditions such as temperature are altered. While studying for exams, key gas laws should be remembered:

• Boyle's Law: At constant temperature, the pressure of a gas is inversely proportional to its volume.
• Charles' Law: At constant pressure, the volume of a gas is directly proportional to its temperature.
• Avogadro's Law: Equal volumes of gases at the same temperature and pressure contain equal numbers of moles.

Understanding these laws is critical in solving more complex scenarios such as those presented in the JEE examinations. They provide a foundation upon which logical problem-solving approaches are built, helping students accurately predict outcomes in gas behavior and interactions.