Problem 36 - Master Resource Book in JEE Main Physics

Question

Inside a cylinder closed at both ends is a movable piston. On one side of the piston is a mass \(m\) of a gas, and on the other side a mass \(2 m\) of the same gas. What fraction of the volume of the cylinder will be occupied by the larger mass of the gas when the piston is in equilibrium? The temperature is the same throughout. (a) \(\frac{2}{3}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{2}\) (d) \(\frac{1}{4}\)

Step-by-Step Solution

Answer

\(\frac{2}{3}\)

Step 1: Define the Ideal Gas Law

The Ideal Gas Law states that \( PV = nRT \) where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature. Since the temperature is the same throughout and the gas is the same on both sides of the piston, \( R \) and \( T \) are constant.

Step 2: Understand Equilibrium Conditions

When the piston is in equilibrium, the pressure on each side must be equal. So, we write \( P_1 V_1 = n_1RT \) and \( P_2 V_2 = n_2RT \) but \( P_1 = P_2 \), therefore, \( V_1 = \frac{n_1}{n_2}V_2 \).

Step 3: Relate Moles to Mass

Since the problem involves masses, we can use the relationship between moles and mass. Let \( M \) be the molar mass of the gas, then \( n_1 = \frac{m}{M} \) and \( n_2 = \frac{2m}{M} \). Therefore, \( \frac{n_1}{n_2} = \frac{m/M}{2m/M} = \frac{1}{2}\).

Step 4: Calculate Volume Fractions

From the equilibrium condition \( V_1 = \frac{1}{2}V_2 \), and the total volume \( V = V_1 + V_2 \), we have \( V = V_1 + 2V_1 = 3V_1 \). Thus, the volume occupied by the larger mass, \( V_2 \), is \( 2V_1 = \frac{2}{3}V \).

Step 5: Conclusion

The fraction of the volume occupied by the larger mass of the gas is \( \frac{2}{3} \). Thus, the answer is \( \frac{2}{3} \).

Key Concepts

Equilibrium Conditions

In the context of gases in an enclosed system, equilibrium conditions refer to a state where there is a balance in forces or pressures on different sides of a partition. Imagine a piston inside a cylinder containing gas on either side. For the piston to be stationary and not move, the pressure exerted by the gas on one side of the piston must be equal to the pressure exerted by the gas on the other side. In formula terms, this translates to \( P_1 = P_2 \). Since the pressure depends on three factors: the number of moles \( n \), the universal gas constant \( R \), and the temperature \( T \), in our equilibrium condition we consider these elements constant to establish \( P_1V_1 = n_1RT \) and \( P_2V_2 = n_2RT \), thus allowing us to focus on the volumes and moles related to each side. This fundamental understanding is crucial to solving many physics and chemistry problems involving balance and distribution of gases.

Moles and Mass Relationship

The relationship between moles and mass is essential in converting between these two measures in chemistry. The number of moles of a gas is the mass divided by the molar mass, \( n = \frac{m}{M} \). In our problem, this relationship helps us calculate the number of moles on each side of the piston based on the given masses. With masses of \( m \) and \( 2m \), the corresponding number of moles for each gas side will be \( n_1 = \frac{m}{M} \) and \( n_2 = \frac{2m}{M} \). These allow us to deduce the mole ratio: \( \frac{n_1}{n_2} = \frac{1}{2} \), helping us understand that the pressure balance between the gas masses can be expressed as a volume ratio.

Volume Fractions

Volume fractions deal with the part of the total volume occupied by a specific amount of substance. Once we determine that \( V_1 = \frac{1}{2}V_2 \), it implies that for every unit of volume occupied by the smaller mass \( m \), the larger mass \( 2m \) occupies twice that volume. Since the total volume \( V \) is the sum of both side volumes \( V = V_1 + V_2 \), substituting the known relationship gives \( V = V_1 + 2V_1 = 3V_1 \), meaning the total volume is divided into 3 equal parts: \( V_1 \) for the smaller mass and \( 2V_1 \) or \( \frac{2}{3}V \) for the larger mass. This fraction, \( \frac{2}{3} \), tells us the larger mass occupies two-thirds of the entire cylinder volume, illustrating the concept of how mass differences affect volume occupancy in equilibrium.