Question

A cylinder with adiabatic walls is closed at both ends and is divided into two compartments by a frictionless adiabatic piston. Ideal gas is filled in both (left and right) the compartments at same \(P,V,T\). Heating is started from left side until pressure changes to \( \frac{27P}{8} \). If initial volume of each compartment was \(9\) litres then the final volume in right-hand side compartment is _____ litres. (for this ideal gas \(C_P/C_V=1.5\)).

• A. \(3\)
• B. \(4\)
• C. \(14\)
• D. \(9\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question
We have a system of two compartments of an ideal gas separated by a movable, insulating piston. The left side is heated, causing it to expand and compress the right side. The entire process for the right compartment is adiabatic because the walls and piston are insulating. We need to find the final volume of the right compartment.
Step 2: Key Formula or Approach
1. The piston is frictionless and free to move. This means that at any point during the process (if done slowly) and at the end, the pressure on both sides of the piston must be equal. So, the final pressure in the right compartment is the same as the final pressure in the left compartment.
2. The right compartment undergoes an adiabatic process because it is compressed by an adiabatic piston and the cylinder walls are also adiabatic.
3. For an adiabatic process, the relationship between pressure and volume is given by \( P V^\gamma = \text{constant} \), where \( \gamma = C_p/C_v \).
Step 3: Detailed Explanation
Let the initial state of the gas in both compartments be \( (P_i, V_i, T_i) \).
We are given:
- Initial volume of each compartment, \( V_i = 9 \) litres.
- Initial pressure, \( P_i = P \).
- Final pressure in the left compartment, \( P_{L,f} = \frac{27P}{8} \).
- Ratio of specific heats, \( \gamma = C_p/C_v = 1.5 = 3/2 \).
Since the piston is frictionless and movable, the final pressure in the right compartment must be equal to the final pressure in the left compartment. \[ P_{R,f} = P_{L,f} = \frac{27P}{8} \] Now, consider the gas in the right compartment. It undergoes an adiabatic compression from its initial state \( (P_i, V_i) \) to its final state \( (P_{R,f}, V_{R,f}) \). For an adiabatic process: \[ P_i V_i^\gamma = P_{R,f} V_{R,f}^\gamma \] Substitute the known values: \[ P (9)^{3/2} = \left(\frac{27P}{8}\right) (V_{R,f})^{3/2} \] We can cancel \(P\) from both sides: \[ (9)^{3/2} = \frac{27}{8} (V_{R,f})^{3/2} \] Rearrange to solve for \( V_{R,f} \): \[ (V_{R,f})^{3/2} = \frac{8}{27} (9)^{3/2} \] \[ (V_{R,f})^{3/2} = \frac{8}{27} ( (9^{1/2})^3 ) = \frac{8}{27} (3^3) = \frac{8}{27} \times 27 = 8 \] \[ V_{R,f} = (8)^{2/3} \] \[ V_{R,f} = ( (8^{1/3})^2 ) = (2^2) = 4 \] Step 4: Final Answer
The final volume in the right-hand side compartment is 4 litres.