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 S_{1} to S_{2} 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 = S_{2}∫ S_{1} Tds = T(S_{2} – S_{1})

And W = P_{1} V_{1} ln (V_{2} / V_{1})

= R T_{1} ln (V_{2} / V_{1})

Therefore, T (S_{2} – S_{1}) = R T_{1} ln (V_{2} / V_{1})

S_{2} – S_{1} = R ln (V_{2} / V_{1})

∆S_{T} = R ln (V_{2} / V_{1})

= – R ln (P_{2} / P_{1})

= R ln (P_{1} / P_{2})

Hence, ∆S_{T} = R ln (V_{2} / V_{1})

= R ln (P_{1} / P_{2})

*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 T_{1} to T_{2} and entropy changes from S_{1} to S_{2}.

Then, Q = C_{p} (T_{2} – T_{1})

Differentiating to find small increase in heat, dQ of the ideal gas when temperature rises is dT.

dQ = C_{p}.dT

dividing both sides by T, we will get:

dO / T = C_{p}. dT / T

dS = C_{p}. dT / T

Integrating both sides, we will get:

S_{2}∫ S_{1} dS = C_{p} (T_{2}∫ T_{1}) dT / T

S_{2} – S_{1} = C_{p} ln (T_{2} / T_{1})

∆S_{P} = C_{p} ln (T_{2} / T_{1})

*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 T_{1} to T_{2} and entropy changes from S_{1} to S_{2}.

Then, Q = C_{v} (T_{2} – T_{1})

Differentiating to find small increase in heat, dQ of the ideal gas when temperature rises is dT.

dQ = C_{v }.dT

dividing both sides by T, we will get:

dO / T = C_{v}. dT / T

dS = C_{v}. dT / T

Integrating both sides, we will get:

S_{2}∫ S_{1} dS = C_{v} (T_{2}∫ T_{1}) dT / T

S_{2} – S_{1} = C_{v} ln (T_{2} / T_{1})

∆S_{P} = C_{v } ln (T_{2} / T_{1})

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