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# First law analysis of non-flow processes

The first law of thermodynamics can be applied to a

system to evaluate the changes in its energy when it undergoes a change of

state while interacting with its surroundings. The processes that are usually

encountered in thermodynamic analysis of

systems can be

identified as any one or a combination of the following elementary processes:

1. Constant volume (isochoric) process

2. Constant pressure (isobaric) process

3. Constant temperature

(isothermal) process.

4. Adiabatic process.

## 1. Constant volume process

Suppose a gas

enclosed in a rigid vessel is interacting with the surroundings and absorbs

energy Q as heat. Since the vessel is rigid, the work done W due to expansion

or compression is zero. Applying the first law, we get

dU = dQ or Q = U_{2} –U_{1}

That is, heat interaction is equal to the change in

internal energy of the gas. If the system contains a mass m equal of an ideal

gas, then

Q = ∆U = mC_{v} (T_{2} –T_{1})

The path followed by the gas is shown on a P-V diagram.

Now consider the fluid contained in a

rigid vessel as shown. The vessel is rigid and insulated. Shaft work is done on

the system by a paddle wheel as shown

in Fig. a. In Fig. b electric work is done on the system. Since the vessel is

rigid, the PdV work is zero. Moreover, the vessel is insulated and hence dQ = 0.

Application of the first law of thermodynamics gives

dU = dQ – dW = dQ – (dW_{pdv} +

dW_{s}) or dU = -dW_{s} or – W_{s} = ∆U = U_{2}

–U_{1}

Where dW_{pdv} is the compression /expansion

work and dW_{s} is the shaft work. That is increase in internal energy of a system at a constant

volume, which is enclosed by an adiabatic wall, is equal to the shaft work done

on the system.

**2. Constant pressure process**

Several industrial processes are carried out at constant

pressure. A few examples of constant pressure processes are: (a) reversible

heating/cooling of a gas (b) phase change (c) paddle wheel work (d) electrical work. For a constant pressure

process, the work done W is given by

W = ∫PdV = P (V

_{2}-V

_{1})

Application of the first law of thermodynamics gives

dU = dQ – dW = dQ – PdV = dQ – d(PV) or dQ =

dU + d(PV) = d(U + PV) = dH

or Q = ∆H

That is in a constant pressure process, the heat

interaction is equal to the increase in the enthalpy of the system. Now

consider the constant pressure processes in which the system is enclosed by an

adiabatic boundary. Application of the first law gives:

dU = dQ – dW = dQ – (PdV + dW

_{s})

Here, the net work done (dW) consists of two parts – the

PdV work associated with the

motion of the boundary and (-dW

_{s}), the shaft work (or electrical work) done by the surroundings. Since the system is enclosed by an adiabatic boundary, dQ = 0 the equation can be written as

-dW_{s}

= dU + d(PV) = dH

That is, the increase in the enthalpy of the system is

equal to the shaft work done on the system.

**2. Constant temperature process**

Suppose a gas enclosed in the piston

cylinder assembly is allowed to expand from P_{1} to P_{2} while the temperature is held constant. Then

application of the first law gives:

dU = dQ – dW = dQ –PdV

It is not possible to calculate work and heat

interactions unless the relationships between the thermodynamic properties of

the gas are known. Suppose the gas under consideration is an ideal gas (which follows the relation Pv = RT

and u = u(T) only) then for an

isothermal process,

dU = 0

dQ = PdV = RTdv/v or Q =W = RTln(v_{2}/v_{1})