Applications of First Law of Thermodynamics

In the previous physical chemistry post named thermodynamics we learned the First law of thermodynamics. We know that we can calculate the change in internal energy by adding heat change in the process and work done by/on the system (ΔU = q + w). There are two types of processes, reversible or irreversible. Measurement of heat q can be done by calorimeter but we have to calculate the work. In this post we will learn how to calculate work.

Let’s take an example of a cylinder which contains one moleof ideal gas and is fitted with a frictionless piston. The initial volume of the gas is V_i and pressure of the gas inside is p. If the external pressure p_ex is greater than the p, then the piston will move inward till p_ex becomes equal to the p and final volume of gas will become V_f.

We can calculate the volume change by multiplying the distance travelled by the piston with the cross section of the cylinder.

            

             ΔV = V_f-V_i = l × A

How do we calculate the work done by the piston? Work can be calculated by force multiplied by distance and force can be calculated by pressure multiplied by the area.

Work done on Ideal gas by Piston

Force on the piston = pressure ×Area

             F = p_ex × A

Work = force × distance

             w = (p_ex × A) × l

             w = p_ex × ΔV = p_ex(V_f -V_i) 

In our example, V_i is greater than V_fbecause compression of gas is done by the piston. That means work is done on the system so its sign has to be positive. To get the correct sign we do a little adjustment in the work equation by adding a negative sign with ΔV.

         
            w = p_ex × (-ΔV)                                     ...............1

For the compression, w gets a positive sign but w will be negative for expansion as the work is done by the system.

What happens in reversible process? How does it differ from irreversible process? In reversible process the changes occur in infinite number of steps and there is infinite number of equilibrium stages so that system and the surrounding always remain in equilibrium. Even a minor difference can manipulate the direction of these reactions or you can say that these reactions can be reversed at any point of the reaction.

Irreversible and Reversible Compression

For a reversible reaction, the piston will take infinite number of steps to cover the distance l. In that case we will get a new set of pressure and ΔV for every single step. Then we have to get the sum of all the steps to calculate the total work done.

        
            w_rev = - Ʃ p ΔV                                     ...............2

In the case of compression, external pressure is always infinitesimally greater than the pressure of the gas and at each step; volume is decreased by an infinitesimal amount ∂V. For compression, p_ex for each step is equal to (p_in +∂p) and for expansion, p_ex = (pin -∂p) and volume is increased by ∂V.

For compression p_ex = (p_in+∂p)

For expansion p_ex = (p_in -∂p)

In generalised way we can write it as p_ex= (p_in ±∂p)

Internal pressure p_in is the pressure of gas filled in the cylinder. So we can apply the ideal gas equation.

          pV = nRT

          p = nRT/V

If the process is carried out at a constant temperature then the equation of w for isothermal process is,

         w_rev = -2.303nRT log (V_f  / V_i)

Calculations of Reversible Work Done

Applications of First Law of Thermodynamics

Now we have learnt to calculate the work done in isothermal expansion or compression for irreversible and reversible processes. Let’s calculate the internal energy change for these processes.

           ΔU = q + w

• Isothermal process: for isothermal process internal energy remain constant that means ΔU = 0. So,

            q = - w

• For isothermal irreversible process:

            q = -w = - [p_ex × (-ΔV)]

            q = p_ex × ΔV

• For isothermal reversible process:

            q = - w_rev = - [-2.303nRT log (V_f/V_i)]

            q = 2.303nRT log (V_f /V_i)

• For Isochoric process: If the process is carried out at constant volume (ΔV=0), then work done will be zero (w =0) and

            ΔU = q_v

         Where q_v is the heat supplied at constant volume.

• Free expansion: When expansion of gas occurs in vacuum (p_ex= 0) then it is called free expansion. That means no work is done during the free expansion of ideal gas neither in irreversible nor reversible process.

         ΔU = q

• Isothermal free expansion: for isothermal free expansion of ideal gas, no work is done during the process because p_ex= 0. And there is no change in internal energy since temperature is kept constant. So,

          ΔU = 0, q = 0, w = 0

• Adiabatic process: for adiabatic changes no heat transfer is allowed between the system and surrounding so q = 0.

         ΔU = q + w

         ΔU = w_ad

We have discussed changes in different conditions like in vacuum or at constant temperature but when we work in a laboratory; we carry out the reactions mostly at atmospheric pressure, which means pressure remains constant.

        ΔU = q + w

        ΔU = q_p + pΔV

Here q_p denotes that heat is supplied at a constant pressure and here new thermodynamic function comes in action which is called the enthalpy H.

            q_p = ΔH

In the next post we will discuss this new and very important function of thermodynamics and its applications. ​

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.