20.2
Work Done During Volume Change
In mechanics, work is done on an object when the force acting on it displaces the object. In thermodynamics, work done on a system can be estimated wh…
Consider an inflated soccer ball having a volume of V. The pressure acting on the ball is the atmospheric pressure.
Suppose the ball falls off the cliff and gets punctured; the air inside the ball releases. Assuming uniform deflation, let its radius be reduced by a distance of dr.
The magnitude of the force acting on the ball depends on the atmospheric pressure and the surface area of the ball.
Therefore, the work done by the surroundings on the ball equals the negative product of the atmospheric pressure, the ball's surface area, and the change in its radius.
However, the product of surface area and dr is the change in the volume of the ball.
Therefore, during a finite change in volume from V1 to V2, work done can be expressed as the integral of the product of pressure and change in volume.
Here the negative sign indicates that the work is done on the system by the surroundings. Conversely, the work done by the system is considered positive.
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Q1: How is work calculated when a gas changes volume?
Work done during volume change is calculated using the integral of pressure multiplied by the change in volume. For a gas expanding from volume V1 to V2, the work equals the integral of pressure times dV. This formula applies to quasi-static processes where the system remains in thermal equilibrium throughout the infinitesimally small steps of expansion or compression.
Q2: What does the negative sign mean in the work formula?
The negative sign in the work equation indicates the direction of work relative to the system. When the surroundings compress the gas, work is done on the system and is considered negative. Conversely, when the gas expands and does work on the surroundings, the work is positive. This sign convention distinguishes between work input and work output.
Q3: Why is atmospheric pressure important in calculating work during volume change?
Atmospheric pressure acts on the surface of the system, such as a deflating ball or a piston. The force exerted by the surroundings depends on atmospheric pressure multiplied by the surface area. Therefore, work done by the surroundings equals the negative product of atmospheric pressure, surface area, and the change in radius or displacement.
Q4: What is a quasi-static process in thermodynamic work calculations?
A quasi-static process is one that occurs in infinitesimally small steps, keeping the system at thermal equilibrium throughout. This gradual, reversible process allows the integral formula for work to be meaningful and applicable. Without quasi-static conditions, the pressure cannot be treated as a well-defined function of volume during the process.
Q5: How does the piston area relate to work done on a gas?
The force acting on a piston equals the pressure multiplied by the piston's area. When the piston moves a distance dr, the change in volume equals the area times dr. Therefore, work done is the product of pressure and the change in volume, making piston area a key factor in determining the total work during expansion or compression.
Q6: Can you explain work done using the soccer ball example?
When a punctured soccer ball deflates, the atmospheric pressure does work on the ball as its volume decreases. The work equals the negative product of atmospheric pressure, the ball's surface area, and the change in radius. As the radius reduces uniformly, the product of surface area and radius change equals the total volume change, allowing calculation of total work done.
Q7: How does work differ between gas expansion and compression?
During gas expansion, the system does positive work on the surroundings as it pushes against external pressure. During compression, the surroundings do negative work on the system as they reduce its volume. Both processes follow the same integral formula for work, but the sign of the result indicates whether work is output from or input to the system.