Calculate the entropy changes of an Ideal Gas in different processes?

There are three types of processes in which entropy changes of an ideal gas. These three processes are:

1. Isothermal process
2. Isobaric process
3. Isochoric process

1. Isothermal process:
The process in which there is no change in temperature is known as Isothermal process. Entropy changes from S1 to S2 when gas absorbs heat during expansion. The heat taken by the gas is given by the area under the curve which represents the work done during expansion.
In other words, Q = W.
But Q = S2∫ S1 Tds = T(S2 – S1)
And W = P1 V1 ln (V2 / V1)
= R T1 ln (V2 / V1)

Therefore, T (S2 – S1) = R T1 ln (V2 / V1)
S2 – S1 = R ln (V2 / V1)
∆ST = R ln (V2 / V1)
= – R ln (P2 / P1)
= R ln (P1 / P2)

Hence, ∆ST = R ln (V2 / V1)
= R ln (P1 / P2)
Hence, isothermal expansion of an ideal gas is accompanied by increase in entropy.

2. Isobaric Processes:
The process in which there is no change in pressure is known as Isobaric process. Consider an ideal gas at constant pressure and its temperature changes from T1 to T2 and entropy changes from S1 to S2.
Then, Q = Cp (T2 – T1)
Differentiating to find small increase in heat, dQ of the ideal gas when temperature rises is dT.

dQ = Cp.dT
dividing both sides by T, we will get:
dO / T = Cp. dT / T
dS = Cp. dT / T
Integrating both sides, we will get:
S2∫ S1 dS = Cp (T2∫ T1) dT / T
S2 – S1 = Cp ln (T2 / T1)
∆SP = Cp ln (T2 / T1)

Hence, increase in temperature of an ideal gas at constant pressure is accompanied by increase in entropy.

3. Isochoric Processes:
The process in which there is no change in volume is known as Isochoric process. Consider an ideal gas at constant volume and its temperature changes from T1 to T2 and entropy changes from S1 to S2.
Then, Q = Cv (T2 – T1)
Differentiating to find small increase in heat, dQ of the ideal gas when temperature rises is dT.
dQ = Cv .dT
dividing both sides by T, we will get:
dO / T = Cv. dT / T
dS = Cv. dT / T
Integrating both sides, we will get:
S2∫ S1 dS = Cv (T2∫ T1) dT / T
S2 – S1 = Cv ln (T2 / T1)
∆SP = Cv ln (T2 / T1)

Hence, increase in temperature of an ideal gas at constant volume is accompanied by increase in entropy of an ideal gas.

Category: Second Law of Thermodynamics

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