Air is contained in a frictionless piston-cylinder arrangement as shown in the figure.

The atmospheric pressure is 100 kPa and the initial pressure of air in the cylinder is 105 kPa. The area of piston is 300 cm². Heat is now added and the piston moves slowly from its initial position until it reaches the stops. The spring constant of the linear spring is 12.5 N/mm. Considering the air inside the cylinder as the system, the work interaction is ________ J. (round off to the nearest integer).

The correct answer is 544.

Analysis of the Image & System Description:
By analyzing the provided schematic diagram, we can observe the following layout inside the piston-cylinder device:
1. The piston is initially at the bottom, containing the air system.
2. There is a clearance gap of 8 cm between the piston's initial position and the free end of the linear spring.
3. The distance from the free end of the spring to the upper stops is also 8 cm.
Thus, the expansion of the air system occurs in two distinct stages:
- Stage 1: The piston rises freely by 8 cm under a constant initial pressure before making contact with the spring.
- Stage 2: The piston continues to rise by another 8 cm against both the constant external force (atmospheric pressure and piston weight) and the linearly increasing resistance of the spring until it reaches the stops.
The total displacement of the piston is:
$dx=8\text{cm}+8\text{cm}=16\text{cm}=0.16\text{m}$
And the compression of the spring is:
$\delta =8\text{cm}=80\text{mm}=0.08\text{m}$

Given Parameters:
- Initial pressure of air, $P=105\text{kPa}=105\times {10}^3{\text{N/m}}^2$
- Piston cross-sectional area, $A=300{\text{cm}}^2=300\times {10}^{-4}{\text{m}}^2$
- Spring constant of the linear spring, $k=12.5\text{N/mm}=12.5\times {10}^3\text{N/m}$

Formulation of Work Interaction:
At any point during the quasi-equilibrium movement, the pressure force of the air is balanced by the atmospheric pressure, the weight of the piston, and the spring force:
$P_{\text{air}}A=P_{\text{atm}}A+F_{\text{piston}}+F_{\text{spring}}$
Since the initial pressure of 105 kPa balances the atmospheric pressure and the piston weight before spring contact:
$P_{\text{air}}A=PA+kx$ (where $x$ is the spring compression)
Therefore, the total work interaction done by the air is:
$W=\int P_{\text{air}}dV=\int P_{\text{air}}Adx=\int (PA+kx)dx$
Evaluating this integral over the path gives:
$W=P\times A\times dx+\frac{1}{2}$
This splits the work interaction into two parts:
1. Work done against the constant initial pressure force over the entire displacement ($W_{\text{piston}}$).
2. Work done to compress the spring ($W_{\text{spring}}$).

Step-by-Step Calculation:

1. Work Done by Piston against Constant Pressure:
$W_{\text{piston}}=P\times A\times dx$
Substituting the given values:
$W_{\text{piston}}=(105\times {10}^3{\text{N/m}}^2)\times (300\times {10}^{-4}{\text{m}}^2)\times \left(0.16\text{m}\right)$
$W_{\text{piston}}=105\times 3\times 1.6=504\text{J}$

2. Work Done to Compress the Spring:
$W_{\text{spring}}=\frac{1}{2}k{\delta }^2$
Substituting the given values:
$W_{\text{spring}}=\frac{1}{2}\times (12.5\times {10}^3\text{N/m})\times {\left(0.08\text{m}\right)}^2$
$W_{\text{spring}}=0.5\times 12500\times 0.0064=40\text{J}$

3. Total Work Interaction:
$W_{\text{total}}=W_{\text{piston}}+W_{\text{spring}}$
$W_{\text{total}}=504\text{J}+40\text{J}=544\text{J}$