Question

At \( T(K) \), a vessel contains \( V \) litres of an ideal gas. The vessel was partitioned into three equal parts. The volume (in L) and temperature (in K) in each part are respectively

• A. \( \frac{V}{3} \, , \frac{T}{3} \)
• B. \( \frac{V}{3} \, , T \)
• C. \( 3V \, , T \)
• D. \( \frac{V}{3} \, , 3T \)

Hint:

In such partitioning problems, remember that the total volume is divided equally, while the temperature remains constant for each part of the ideal gas if the system is in thermal equilibrium.

Answer 1

The Correct Option is B

Step 1: The problem states that a vessel of ideal gas is partitioned into three equal parts, each with the same ideal gas. The volume of the total gas is \( V \), and the volume of each part is \( \frac{V}{3} \), as the gas is divided into three equal parts. \[ \text{Volume of each part} = \frac{V}{3} \] Step 2: Now, considering the temperature \( T(K) \) in each part, as the vessel is partitioned equally, the temperature of each part remains the same, which is \( T \), because temperature is not affected by partitioning when each part is in equilibrium. \[ \text{Temperature in each part} = T \] Step 3: Thus, the volume in each part is \( \frac{V}{3} \) and the temperature remains \( T \) in each part. Therefore, the correct answer is: \[ \boxed{\frac{V}{3}, T} \]