# Work, First Law of Thermodynamics

####
**Nature of Heat and Work**

When a change in the state
of a system occurs, energy is transferred to or from the surroundings. This
energy may be transferred as heat or mechanical work.

**Work = force**×

**distance**

###
**Units of Work**

**❒**In CGS system the unit of work is

**erg**

**❒**

**erg:**is the work done when a resistance of 1 dyne is moved through a distance of 1 centimeter.

**❒**Since the erg is so small, a bigger unit, the joule (J) is now used.

1
joule = 10

^{7}ergs
1
erg = 10

^{–7}J**❒**We often use kilojoule (kJ) for large quantities of work:

1 kJ
= 1000 J

###
**Units of Heat**

**❒**The unit of heat, which was used for many years, is calorie (cal).

**❒**

**A calorie:**is quantity of heat required to raise the temperature of 1 gram of water by 1º C in the vicinity of 15ºC.

**❒**Since heat and work are interrelated, SI unit of heat is the joule (J).

1
joule = 0.2390 calories

1
calorie = 4.184 J

1
kcal = 4.184 kJ

###
**Sign Convention of Heat**

**❒**The symbol of heat is q.

**❒**If the heat flows from the surroundings into the system to raise the energy of the system, it is taken to be positive, +q.

**❒**If heat flows from the system into the surroundings, lowering the energy of the system, it is taken to be negative, –q.

###
**Sign Convention of Work**

**❒**The symbol of work is w.

**❒**If work is done on a system by the surroundings and the energy of the system is thus increased, it is taken to be positive, +w.

**❒**If work is done by the system on the surroundings and energy of the system is decreased, it is taken to be negative, –w.

####
**Pressure–Volume Work**

**❒**In physics, mechanical work is defined as force multiplied by the distance through which the force acts.

**❒**In elementary thermodynamics the only type of work generally considered is the work done in expansion (or compression) of a gas. This is known as pressure-volume work or PV work or expansion work.

**❒**Consider a gas contained in a cylinder fitted with a frictionless piston. The pressure (force per unit area) of the gas, P, exerts a force on the piston. This can be balanced by applying an equal but opposite pressure from outside on the piston. Let it be designated as P

_{ext}.

**❒**It is important to remember that it is the external pressure, P

_{ext}. and not the internal pressure of the gas itself which is used in evaluating work. This is true whether it be expansion or contraction.

**❒**If the gas expands at constant pressure, the piston would move, say through a distance

*l*. We know that:

work
= force × distance (by definition)

w =
f ×

*l*…....(1)
Since pressure is force per unit area,

f =
P

_{ext}× A …....(2)
where A is the cross-section area of the piston.

From (1) and (2), we have

w =
P

_{ext}× A ×*l*
= P

_{ext}× ΔV
where ΔV is the increase in volume of the gas

**❒**Since the system (gas) is doing work on the surroundings (piston), it bears negative sign. Thus,

w =
–P

_{ext}× ΔV**❒**Proceeding as above the work done in compression of a gas can also be calculated. In that case the piston will move down and sign of the work will be positive.

w =
P

_{ext}× ΔV**❒**As already stated, work may be expressed in dynes-centimetres, ergs, or joules. PV work can as well be expressed as the product of pressure and volume units e.g., in litre or atmospheres.

**❒**It may be noted that the work done by a system is not a state function. This is true of the mechanical work of expansion. We shall show presently that the work is related to the process carried out rather than to the internal and final states. This will be evident from a consideration of the reversible expansion and an irreversible process.

###
**Solved Problem**

**Calculate the pressure-volume work done when a system containing a gas expands from 1.0 litre to 2.0 litres against a constant external pressure of 10 atmospheres. Express the answer in calories and joules.**

**Solution:**

####
**Isothermal Reversible Expansion Work of an Ideal Gas**

**❒**Consider an ideal gas confined in a cylinder with a frictionless piston. Suppose it expands in a reversible manner from volume V

_{1}to V

_{2}at a constant temperature. The pressure of the gas is successively reduced from P

_{1}to P

_{2}.

**❒**The reversible expansion of the gas takes place in a finite number of infinitesimally small intermediate steps. To start with the external pressure, P

_{ext}, is arranged equal to the internal pressure of the gas, P

_{gas}, and the piston remains stationary. If P

_{ext}is decreased by an infinitesimal amount dP the gas expands reversibly and the piston moves through a distance d

*l*.

**❒**Since dP is so small, for all practical purposes,

P

_{ext}= P_{gas}= P
The work done by the gas in one infinitesimal step dw, can be
expressed as:

(A = cross-sectional area of piston)

dw =
P × A × d

*l*
= P
× dV

where dV is the increase in volume

**❒**The total amount of work done by the isothermal reversible expansion of the ideal gas from V

_{1}to V

_{2}is, therefore:

**❒**Isothermal compression work of an ideal gas may be derived similarly and it has exactly the same value with the sign changed. Here the pressure on the piston, P

_{ext}, is increased by dP which reduces the volume of the gas.

####
**Isothermal Irreversible Expansion Work of an Ideal Gas**

Suppose we have an ideal gas contained in a cylinder with a piston.
This time the process of expansion of the gas is performed irreversibly i.e., by
instantaneously dropping the external pressure, P

_{ext}, to the final pressure P_{2}.
The work done by the system is now against the pressure P

_{2}throughout the whole expansion and is given by the following expression:####
**Maximum Work done in Reversible Expansion**

**❒**The isothermal expansion of an ideal gas may be carried either by the reversible process or irreversible process as stated above.

**❒**The reversible expansion is shown in the following figure in which the pressure is falling as the volume increases. The reversible work done by the gas is given by the expression:

which is represented by the shaded area.

**❒**If the expansion is performed irreversibly by suddenly reducing the external pressure to the final pressure P

_{2}, the irreversible work is given by:

– w

_{in}= P_{2}(V_{2}– V_{1})
which is shown by the shaded area in Fig (b).

**❒**In both the processes, the state of the system has changed from A to B but the work done is much less in the irreversible expansion than in the reversible expansion. Thus mechanical work is not a state function as it depends on the path by which the process is performed rather than on the initial and final states. It is a path function.

**❒**It is also important to note that the work done in the reversible expansion of a gas is the maximum work that can be done by a system (gas) in expansion between the same initial (A) and final state (B). This is proved as follows :

We know that the work always depends on the external pressure, P

_{ext}; the larger the P_{ext}the more work is done by the gas. But the P_{ext}on the gas cannot be more than the pressure of the gas, P_{gas}or a compression will take place. Thus the largest value P_{ext}can have without a compression taking place is equal to P_{gas}. But an expansion that occurs under these conditions is the reversible expansion. Thus, maximum work is done in the reversible expansion of a gas.####
**Solved Problem**

**Problem (1): One mole of an ideal gas at 25ºC is allowed to expand reversibly at constant temperature from a volume of 10 litres to 20 litres. Calculate the work done by the gas in joules and calories.**

**Solution:**

**Problem (2): Find the work done when one mole of the gas is expanded reversibly and isothermally from 5 atm to 1 atm at 25ºC.**

**Solution:**

*Reference:**Essentials of Physical Chemistry /Arun Bahl, B.S Bahl and G.D. Tuli / multicolour edition.*

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