Question

An isolated system consists of two perfectly sealed cuboidal compartments \( A \) and \( B \) separated by a movable rigid wall of cross-sectional area \( 0.1 \, \text{m}^2 \). Initially, the movable wall is held in place by latches \( L_1 \) and \( L_2 \) such that the volume of compartment \( A \) is \( 0.1 \, \text{m}^3 \). Compartment \( A \) contains a monatomic ideal gas at \( 5 \, \text{bar} \) and \( 400 \, \text{K} \). Compartment \( B \) is perfectly evacuated and contains a massless Hookean spring of force constant \( 0.3 \, \text{N/m} \) at its equilibrium length (stored elastic energy is zero). The latches \( L_1 \) and \( L_2 \) are released, the wall moves to the right by \( 0.2 \, \text{m} \), where it is held at the new position by latches \( L_3 \) and \( L_4 \). Assume all the walls and latches are massless. The final equilibrium temperature, in \( K \), of the gas in compartment \( A \), rounded off to 1 decimal place, is: \includegraphics[width=0.4\linewidth]{q56 CE.PNG}

Hint:

For isolated systems involving springs and gas expansion, use the first law of thermodynamics and account for both mechanical and thermal energies.

Answer 1

For an isolated system: \[ \Delta U = \Delta Q + \Delta W. \] \[ \text{Since it is an isolated system, } \Delta Q = 0. \] For a Monatomic Ideal Gas: \[ \Delta U = \frac{0.3}{0.2} nR \Delta T. \] Using the ideal gas law: \[ PV = nRT. \] Substitute \( P = 5 \times 10^5 \, \text{Pa}, V = 0.1 \, \text{m}^3, R = 400 \, \text{J/mol·K} \): \[ n = \frac{5 \times 10^5 \times 0.1}{400} = 125. \] As the compartment moves to the right, the spring exerts a force towards the left, indicating that work is done on the system. Force and Pressure Relations: \[ F = -Rx, \quad P = \frac{F}{A} = -\frac{Rx}{A}. \] Work done (\(W\)) is given by: \[ W = \int P \, dV. \] Substituting: \[ W = \int -\frac{Rx}{A} \, dV. \] Since \( dV = A \, dx \): \[ W = -\int \frac{R}{A} x \, A \, dx = -\frac{R}{A} \int x \, dx. \] \[ W = -\frac{R}{2} x^2 \Big|_{x_1}^{x_2}. \] Substituting the limits and values: \[ W = -\frac{0.3}{2} \times (0.2)^2. \] \[ W = -0.03 \, \text{J}. \] Final Temperature Calculation: Using the energy equation: \[ \Delta U = W. \] Substitute: \[ \frac{3}{2} nR (T_f - 400) = -\frac{0.3}{2} \times (0.2)^2. \] Simplify: \[ \frac{3}{2} \times 125 \times 400 (T_f - 400) = -0.03. \] \[ \frac{3}{2} \times 5 \times 10^5 \times 0.1 \frac{(T_f - 400)}{R \times 400} = -0.03. \] Solve for \( T_f \): \[ T_f = 399.9 \, \text{K}. \] Final Answer: The final temperature is: \[ T_f = 399.9 \, \text{K}. \]