Consider a weightless, frictionless piston with a 2 kg mass placed on it as shown in the figure. At equilibrium in position 1, the cylinder contains 0.1 kg of air. The piston cross-sectional area is 0.01 m2. The ambient pressure in the surroundings outside the piston-cylinder arrangement is 0 bar (absolute). When the mass above the piston is removed instantaneously, it moves up and hits the stop at position 2, which is 0.1 m above the initial position.
Assuming \( g = 9.81 \, {m/s}^2 \), the thermodynamic work done by the system during this process is ________ J (answer in integer).

Step 1: Understanding the work done.
The problem involves a weightless, frictionless piston with a given mass placed on it, which moves up when the mass is removed. The thermodynamic work in this case is done by the system as the air expands and pushes the piston upward. The work done is given by: \[ W = P \Delta V \] where:
\( P \) is the pressure exerted by the gas,
\( \Delta V \) is the change in volume.
Step 2: Pressure inside the cylinder.
Since the ambient pressure outside the piston-cylinder arrangement is 0 bar (absolute), the pressure inside the cylinder is due to the weight of the piston and the air above it. However, since the problem states that the pressure outside the cylinder is 0 bar (absolute), and there is no external force acting to compress the gas, the pressure inside the cylinder becomes 0.
Step 3: Work calculation.
Since the pressure inside the cylinder is 0, the work done during the expansion is: \[ W = P \Delta V = 0 \times \Delta V = 0. \]
Final Answer: The thermodynamic work done by the system is \( \boxed{0} \, {J} \).
A closed system is undergoing a reversible process 1–P–2 from state 1 to 2, as shown in the figure, where X and Y are thermodynamic properties. An irreversible process 2–Q–1 brings the system back from 2 to 1. The net change in entropy of the system and surroundings during the above-mentioned cycle are _______ respectively.

A closed system undergoing a thermodynamic cycle consisting of two reversible isothermal and two reversible adiabatic processes is shown in the following figure. If \(\delta Q\) is the infinitesimal heat transfer and T is the instantaneous temperature, then the value of the contour integral \( \oint \frac{\delta Q}{T} \) is 