Question

0.72. Derive the expression for heat exchanged in
case of an isobaric process.
Ans​

Answer 1

Explanation:

this is the answer which is required

Answer 2

The expression for heat exchanged in the case of an isobaric process is given below:

Let the initial volume of the gas be Vi, the temperature of the gas be Ti and the Final volume and temperature of the gas be Vf, Tf.

We know that in an isobaric process, the final pressure = initial pressure

⇒ ∆P = 0

Let the work done in the process be W

So W = PΔV

= P (Vf - Vi)                 -(1)

Using the ideal gas equation, pV= nRT

Substituting in equation 1,

⇒ W = nR (Tf - Ti)            -(2)

The Change in internal energy is given by ∆U

∆U = nCv ∆T

⇒∆U = nCv (Tf - Ti)        - (3)

Here, Cv is the specific heat at constant Volume

Applying the First Law of Thermodynamics to the isobaric process,

Q = ∆U + W

From equations 2 and 3

Q = nCv (Tf -Ti) + nR (Tf - Ti)

or Q = n Cv + nR (Tf - Ti)        -(4)

or Q = n (Cv + R) (Tf - Ti)

∴ Q = n Cp (Tf - Ti)

= nCpΔT