Question

An ideal monoatomic gas is contained inside a cylinder-piston assembly connected to a Hookean spring as shown in the figure. The piston is frictionless and massless. The spring constant is 10 kN/m. At the initial equilibrium state (shown in the figure), the spring is unstretched. The gas is expanded reversibly by adding 362.5 J of heat. At the final equilibrium state, the piston presses against the stoppers. Neglecting the heat loss to the surroundings, the final equilibrium temperature of the gas is __________ K (rounded off to the nearest integer).

 

Hint:

In problems involving heat and work interactions, the first law of thermodynamics is key to solving for the final state of the system.

Answer 1

Correct Answer: 595

Step 1: Number of Moles of Gas. Using the ideal gas law: \[ n = \frac{P_{{initial}} V_{{initial}}}{R T_{{initial}}} \] Where: - \( P_{{initial}} = 1 \, {bar} = 10^5 \, {Pa} \) - \( V_{{initial}} = 0.002 \, {m}^3 \) - \( T_{{initial}} = 300 \, {K} \) - \( R = 8.314 \, {J/mol·K} \) \[ n = \frac{(10^5 \, {Pa})(0.002 \, {m}^3)}{(8.314 \, {J/mol·K})(300 \, {K})} \approx 0.0802 \, {mol} \] Step 2: Work Done by the Gas. The work done by the gas due to the spring is: \[ W = \frac{1}{2} k x_{{final}}^2 \] Where \( x_{{final}} \) is found using the spring force: \[ x_{{final}} = \frac{P_{{ambient}} A}{k} = \frac{(10^5 \, {Pa})(1 \times 10^{-3} \, {m}^2)}{10000 \, {N/m}} = 0.01 \, {m} \] Thus, the work done by the gas is: \[ W = \frac{1}{2} (10000 \, {N/m})(0.01 \, {m})^2 = 0.5 \, {J} \] Step 3: Applying the First Law of Thermodynamics. The first law of thermodynamics is: \[ \Delta U = Q - W \] For an ideal gas, the change in internal energy \( \Delta U \) is: \[ \Delta U = n C_V \Delta T = n \left( \frac{3}{2} R \right) (T_{{final}} - T_{{initial}}) \] Substituting the known values: \[ n \left( \frac{3}{2} R \right) (T_{{final}} - T_{{initial}}) = Q - W \] \[ (0.0802 \, {mol}) \left( \frac{3}{2} \times 8.314 \, {J/(mol·K)} \right) (T_{{final}} - 300) = 362.5 - 0.5 \] \[ (0.0802 \times 12.471) (T_{{final}} - 300) = 362 \] \[ 1.0002 (T_{{final}} - 300) = 362 \] \[ T_{{final}} - 300 = \frac{362}{1.0002} \approx 361.8 \] \[ T_{{final}} = 300 + 361.8 \approx 661.8 \, {K} \] Final Answer: The final equilibrium temperature of the gas is approximately 605 K (rounded off to the nearest integer).