Work Done During Volume Changes In Physics, Important Topics For JEE 2025
Shrivastav 9 Jun, 2024
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Work Done During Volume Changes : A simple example of a thermodynamic system is a quantity of gas enclosed in a cylinder with a movable piston. Internal-combustion engines, steam engines, and compressors in refrigerators and air conditioners all use some version of such a system. In the next several sections we’ll use the gas-in-cylinder system to explore several kinds of thermodynamic processes.
We’ll use a microscopic viewpoint, based on the kinetic and potential energies of individual molecules in a material, to develop intuition about thermodynamic quantities. But it is important to understand that the central principles of thermodynamics can be treated in a completely macroscopic way, without reference to microscopic models. Indeed, part of the great power and generality of thermodynamics is that it does not depend on details of the structure of matter.Work Done During Volume Changes In Physics
- A gas in a cylinder with a movable piston is a simple example of a thermodynamic system. Figure shows a gas confined to a cylinder that has a movable piston at one end. If the gas expands against the piston, it exerts a force and does work on the piston. If the piston compresses the gas as it is moved inward, work is done on the gas. The work associated with such volume changes can be determined as follows.
- Let the gas pressure on the piston face be p . Then, the force on the piston due to the gas is pA , where A is the area of the face.
When the piston is pushed outward an infinitesimal distance dx , the work done by the gas is
dW = Fdx = pA dx
which, since the change in volume of the gas is d V = Adx , becomes
For a finite change in volume from V i to V f , this equation is then integrated between V i to V f to find the network
Now, there are five methods of finding work done by a gas.
Method 1. This is used when p - V equations is known to us. Suppose p as a function of V is known to us.
p = f ( V )
then work done can be found by
Method 2. The work done by a gas is also equal to the area under p - V graph. Following different cases are possible:
V is increasing
W AB = Shaded area
V is decreasing
W AB = –Shaded area
Cyclic process
Work Done In An Isothermal Process On An Ideal Gas
Suppose an ideal gas has initial pressure, volume and temperature as p 1 , V 1 and T respectively. In a process, the temperature is kept constant and its pressure and volume are changed from p 1 , V 1 to p 2 , V 2 . The work done by the gas is
As
Thus,
Work Done In An Isobaric Process
Suppose the pressure of a system remains constant at a value p and the volume changes from V 1 to V 2 .
The work done by the system is
Work Done In An Isochoric Process
An isochoric process the volume remains constant and no work is done by the system.
Work Done In Adiabatic Process
Further,
Signs For Heat And Work In Thermodynamics
We describe the energy relationships in any thermodynamic process in terms of the quantity of heat Q added to the system and the work W done by the system. Both Q and W may be positive, negative, or zero. A positive value of Q represents heat flow into the system, with a corresponding input of energy to it; negative Q represents heat flow out of the system. A positive value of W represents work done by the system against its surroundings, such as work done by an expanding gas, and hence corresponds to energy leaving the system. Negative W , such as work done during compression of a gas in which work is done on the gas by its surroundings, represents energy entering the system.
Another way to understanding the concept is that when a thermodynamic system changes from an initial state to a final state, it passes through a series of intermediate states. We call this series of states a path. There are always infinitely many possibilities for these intermediate states. When all are equilibrium states, the path can be plotted on a pV -diagram. Point 1 represents an initial state with pressure p 1 and volume V 1 , and point 2 represents a final state with pressure p 2 and volume V 2 . To pass from state 1 to state 2, we could keep the pressure constant at p 1 while the system expands to volume V , then reduce the pressure to p 2 (probably by decreasing the temperature) while keeping the volume constant at V 2 (to point 2). The work done by the system during this process is the area under the line 1 → 3; no work is done during the constant volume process 3 → 2. Or the system might traverse the path 1 → 4→ 2; then the work is the area under the line 4 → 2, since no work is done during the constant-volume process 1 → 4.
The smooth curve from 1 to 2 is another possibility and the work for this path is different from that for either of the other paths. We conclude that the work done by the system depends not only on the initial and final states, but also on the intermediate states—that is, on the path. Furthermore, we can take the system through a series of states forming a closed loop, such as 1 → 3 → 2 → 4 → 1. In this case the final state is the same as the initial state, but the total work done by the system is not zero.
Examples Of Work Done During Volume Changes
Q.1 : In a given process work done on a gas is 40 J and increase in its internal energy is 10 J. Find heat given or taken to/from the gas in this process.
Sol.
Given,
Work done on the gas is 40 J. Therefore, work done by the gas used in the equation,
Q
=
W
+
Q
=
W
+
We have,
Q = –40 + 10 = –30J
Here, negative sign indicates that heat is taken out from the gas.Q.2. Temperature of two moles of a monoatomic gas is increased by 600 K in a given process. Find change in internal energy of the gas.
Sol. Using the equation,
Q.3 : Calculate the work done by a gas as it is taken from the state a to b , b to c and c to a as shown in figure
Sol. The work done by the gas in the process a to b is the area of abode. This is
W ab = (120 kPa) 9250 cc)
= 120 × 10 3 × 250 × 10 –6 J = 30 J
In the process b to c the volume remains constant and the work done is zero.
In the process c to a the gas is compressed. The volume is decreased and the work done by the gas is negative. The magnitude is equal to the area of caed. This area is cab + baed.
= 10 J + 30 J = 40 J
Thus, the work done in the process c to a is –40 J.
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