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How Does Hydrostatic Work?
External Work
Hydrostatic system: A system that obeys Pascal’s law is called a hydrostatic system. Important characteristics of such a system are:
- The pressure is uniform throughout the system and acts normally outwards at everypoint on the walls.
- Any additional pressure applied at any point gets transmitted throughout the system in such a way that the pressure everywhere continues to remain uniform.
These conditions are obeyed by
1. small amounts of liquids for which the effect of gravity may be neglected and
2. gaseous systems. However, all liquids are nearly incompressible, i.e., any change in pressure does not produce an appreciable change in volume. Due to this reason, for the thermodynamic analysis of a hydrostatic system, it is convenient to take a closed gaseous system.
Hydrostatic work:
An amount of gas is enclosed by an airtight cylinder-piston arrangement where the piston can move without friction along the inner walls of the cylinder [Fig.].
Let p = pressure of the enclosed gas,
α = area of cross section of the piston.
So, the force acting on the piston = pα.
Now, suppose the piston goes through a very small displacement outwards from the position A to the position B; dis-placement AB = dx. This displacement occurs when a slight difference develops between the pressures of this system and its surroundings. A small outwards pull on the piston, a gain of a small amount of heat from the surroundings, etc., may initiate such a displacement. During this displacement, however small, the pressure p of the system may change. But at every instant, the pressure must be uniform throughout the system. A sufficiently slow motion of the piston is required to satisfy this hydrostatic condition.
Due to the small displacement dx, the gas expands from volume V to V+ dV. Clearly, dV = αdx. So, the small work done in this infinitesimal process is,
dW = force × displacement
= (pα)(dx) = p(αdx) = pdV
or concisely, dW = pdV ……….. (1)
Work done in a finite process, for which the volume changes from V1 to V2, is
W = ∫ dW = \(\int_{V_1}^{V_2} p d V\) ……. (2)
The relation (1) can be interpreted as:
i) When there is no change in volume, V = constant and dV= 0. So, dW= 0; no work is done in such a process.
ii) When the volume increases, i.e., for an expansion, dV is positive, i.e., dV > 0. So, dW > 0; positive amount of work is done in this process. The system releases some energy to its surroundings; it is termed as work done by the system.
iii) When the volume decreases, i.e., for a contraction occurring due to an inward motion of the piston. dV is negative, i.e., dV < 0. So, dW < 0; a negative work is done in this process. The system receives some energy from its surroundings; it is termed as work done on the system.
The integral in relation (2) can be evaluated for special processes :
i) Isochoric process:
V = constant and dV = 0
So, W = 0 ……… (3)
ii) Isobaric process:
p = constant.
So, W = \(\int_{V_1}^{V_2} p d V\) = p\(\int_{V_1}^{V_2} d V\) = p(V2 – V1) ….. (4)
But, when the pressure p of a system change in a process, the integral cannot be evaluated unless p can be expressed as a function of volume V. It is only possible if the equation of state of the system is known. Later in this chapter, the expressions for istothermal work and adiabatic work for an ideal gas would be evaluated, using the ideal gas equation of state, pV = RT.
There are systems in thermodynamics that are not hydrostatic. For them, the work equations (1) and (2) would be different. However, in this chapter, we shall restrict ourselves mainly to hydrostatic systems.