Thermodynamic Processes

A process is said to be the thermodynamic process if there is a change in thermodynamic parameter with respect to time. They are

1. Isothermal Process
2. Adiabatic Process
3. Isochoric Process
4. Isobaric Process

Isothermal Process

A thermodynamic process in which temperature of the system remains constant is called Isothermal process. For an isothermal process, the process should be slow and the wall of the system should be conducting.

Equation of isothermal process

We have ideal gas equation for n mole of gas

$$PV = nRT$$

In an isothermal process, the temperature is constant. So, for given mass of gas at constant temperature NRT is constant equation (i) becomes

$$PV = \text {constant}$$

This equation of an isothermal process. If P₁V₁ be initial and P₂V₂be final pressure and volume of the system at constant temperature

$$P_1V_1 = P_2V_2$$

Application of first Law of Thermodynamics to Isothermal Process

From the first law of thermodynamics, we have

$$dQ = dU + dW$$

The temperature in an isothermal process does not change. Since the internal energy of an ideal gas depends only on the temperature, there will be no change in the internal energy of the gas. So,

$$dU = 0$$

Hence, the first law of thermodynamics isothermal process will become

$$dQ = 0 + dW$$

$$dQ = dW$$

$$= PdV$$

Thus during an isothermal process,

$$\text {Heat added to the system} = \text {work done by the system}$$

Work done in Isothermal Process

Consider n-mole of an ideal gas contain in a cylinder fitted with frictionless and weightless piston suppose the gas expand isothermally from initial state A(P₁V₁) to the final state B(PV₂V)₂) at constant temperature (T). Here P and V represents pressure and volume of the system. We have to find work done in this process.

We know that work done

$$W = \int _{v_1}^{v_2}P dV$$

and ideal gas equation for n-mole of gas is

$$PV = nRT \dots (i)$$

In isothermal process temperature T is constant. So far given mass of gas at constant temperature nRT is constant so equation (i) becomes

$$PV = K\text {constant}$$

$$P = \frac {K}{V} \dots (ii)$$

using equation (ii) in an equation (i)

$$ W = \int _{V_1}^{V_2} \frac {K}{V} dV $$

$$ = K \int _{V_1}^{V_2} \frac {dV}{V} $$

$$= K[\log V_2 - \log V_1$$

$$W = K \log \left[\frac {V_1}{V_2} \right ] $$

$$W = nRT\log \left[\frac {V_1}{V_2} \right ] $$

This is work done by the system in isothermal process.

Fro 1 mole of gas work done(W) \(=RT\log \left[\frac {V_1}{V_2} \right ] \dots (iii)\)

If P₁V₁ be initial and P₂V₂ be final pressure and volume of system at isothermal condition

$$P_1V_1 = P_2V_2$$

$$\frac {P_1}{V_1} = \frac {P_2}{V_2}$$

from equation (iii)

$$W = nRT\log \left[\frac {P_1}{P_2} \right ] $$

Adiabatic Process

A thermodynamic process in which total energy of the system remains constant is called adiabatic process. In this process, there is no exchange of heat between system and surrounding. For an adiabatic process, the wall of a system should be insulated and the process should be rapid. For example bursting of tire

First Law of Thermodynamics Applied to Adiabatic Process

Since no heat energy enters or leaves out the system in an adiabatic process, heat gained by the system, dQ = 0 and from the first law of thermodynamics,

$$dQ = dU + dW = 0$$

$$dU = -dW$$

When a gas expands adiabatically, dW is positive and therefore, dU must be negative. The internal energy of the system would decrease and the gas will be cool. The reverse is also true.

Equation of Adiabatic Process

Let us consider n-mole of an ideal gas fitted with weightless and frictionless piston. Suppose gas is expanded adiabatically so that decrease in temperature of system dt

From first law of thermodynamics

$$dQ = dV + PdV$$

From adiabatic process, there is no exchange of heat so \(dQ = 0\)

$$0 =dV + PdV \dots (i)$$

$$\text {here,} dV = n C_VdT$$

where, C_V is molar specific heat capacity of gas at constant volume

then equation (i) becomes

$$0 = n C_VdT + PdV \dots (ii)$$

Ideal gas equation for n-mole

$$PV = nRT$$

Differentiating both sides with respect to T

$$ \frac {d}{dT}PV = \frac {d}{dT} (n RT) $$

$$PdV + VdP = nRdT$$

$$dT = \frac {PdV + VdP}{nR} \dots (iii)$$

using (iii) in an equation(ii) we get

$$\frac {n C_V (PdV + VdP)} {nR} + PdV = 0 $$

$$\text {or} C_P PdV + PdV+ C_V VdP = 0 $$

$$\text {or} (C_V + R) PdV +C_VVdP = 0 [\because C_P - C_V = R] $$

$$\text {or} C_P PdV = - C_V VdP$$

$$\text {or} \frac {C_P}{C_V} \frac {dV}{V} = -\frac {dP}{P} $$

\(\frac {C_P}{C_V} =\gamma\) is the ratio of specific heat capacity of gas at constant pressure and volume

$$\gamma \frac {dV}{V} = -\frac {dP}{P}$$

Integrating both sides

$$\int \gamma \frac {dV}{V} =\int -\frac {dP}{P}$$

$$\text {or} \gamma \log V + \log P = constant $$

$$\text {or} \log V ^\gamma+ \log P = constant $$

$$\text {or} \log (V ^\gamma P) = \log C $$

$$PV^\gamma = constant$$

This is the equation for an adiabatic process relating pressure and volume of the gas.

If P₁V₁ be initial andP₂V₂ be final pressure and volume of system at an adiabatic condition then,

$$\therefore P_1V_1^\gamma = P_2V_2^\gamma$$

Work done in Adiabatic Process

Consider n-mole of a perfect gas in a cylinder having perfectly non-conducting walls and fitted with a frictionless piston. Suppose the gas expands adiabatically from the initial volume V₁ to the final volume V₂. Then work done by the gas is

$$W = \int _{v_1}^{v_2}P dV$$

During an adiabatic process \(PV^\gamma =k\)

$$\text {or} P = \frac {K}{V^\gamma} = KV^{-\gamma } $$

$$ W = \int _{v_1}^{v_2}K \frac {dV}{V^{\gamma}}$$

$$ W = K \int _{v_1}^{v_2} V^{-\gamma }dV $$

$$= K\left [\frac{v^{1- \gamma}}{1 - \gamma}\right ]_{v_1}^{v_2} $$

$$= \frac {K}{1-\gamma} [V_2^{1-\gamma} -V_1^{1-\gamma}] $$

$$= \frac {1}{\gamma-1} [KV_1^{1-\gamma} - KV_2^{1-\gamma}] $$

$$ \text {But} P_1V_1^\gamma = P_2V_2^\gamma = K $$

$$W =\frac {1}{\gamma-1} [P_1V_1^\gamma V_1^{1-\gamma} - P_2V_2^\gamma V_2^{1-\gamma}]$$

$$=\frac {1}{\gamma-1} (P_1V_1 -P_2V_2) $$

If the temperature of the initial state T₁ and that of the final state T₂, then

P₁V₁ = nRT₁ andP₂V₂= nRT₂

$$\therefore W = \frac {1}{\gamma-1} (n RT_1 - nRT_2)$$

$$= \frac {nR}{\gamma-1} (T_1 -T_2) $$

Above equation gives the work done for one of perfect gas during an adiabatic process. Hence, the work done in an adiabatic process depends only upon the initial and final temperature T₁ and T₂.