## Thermodynamic Processes

Subject: Physics

#### Overview

This note provides us an information on thermodynamic processesA process is said to be the thermodynamic process if there is a change in the thermodynamic parameter with respect to time. They are Isothermal Process Adiabatic Process Isochoric Process Isobaric Process
##### 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
3. Isochoric Process
4. Isobaric Process

#### Isothermal Process (a) Heat is given out. (b) Heat is absorbed. (c) The variation of pressure and volume during an 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 The variation of pressure and volume during an 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)

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

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, CV 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.

##### Things to remember

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

A thermodynamic process in which temperature of the system remains constant is called Isothermal process.

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.

A thermodynamic process in which total energy of the system remains constant is called adiabatic process.

The work done in an adiabatic process depends only upon the initial and final temperature.

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##### Videos for Thermodynamic Processes ##### Adiabatic Equation And Its Work Done 