Isobaric Process

An Isobaric process is a type of thermodynamic process in which the pressure of a system remains constant. This process is characterized by the absence of any change in pressure while other properties such as volume and temperature may vary. In simpler terms, during an isobaric process, the system's pressure remains constant. Understanding the isobaric process is crucial for analyzing various thermodynamic systems and is an important concept in both theoretical and practical applications.

1.0Definition of Isobaric Process

• It is a thermodynamic process that takes place at constant pressure of the system, but volume and temperature varies for change in state of the system.
• In an isobaric process, when the heat is transferred to the system, some work is done. However, there is also a change in the internal energy of the system. This further means that no quantities, as in the first law of thermodynamics, become zero.
•  The constant pressure is obtained when the volume is either expanded or contracted, so this means work done can be either positive or negative.
• An example of an isobaric process is the boiling of water to produce steam or the freezing of water to form ice. In these processes, a gas either expands or contracts to keep the pressure constant, resulting in work being done either by the system or on the system. 

2.0Equation of Isobaric Process

1. Process Equation 

$P=\text{ constant or }\Delta P=0$

$\frac{V}{T}=\text{ Constant }$

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

2. Work Equation 

$W=\int dW=\int PdV$, The work done is found from the equation,

$W={\int }_{V_i}^{V_f}PdV=P\left(V_f-V_i\right)$

Since the pressure is constant in an isobaric process, the integral becomes,

$W=P_i{\int }_{V_i}^{V_f}dV=P_i\left(V_f-V_i\right)=P_i\Delta V=nR\left(T_f-T_i\right)=nR\Delta T$

•  If the gas expands, $V_f>V_i$, so ΔV > 0, and the work done by the gas is positive.
•   When the gas is compressed, $V_f<V_i$, so that ΔV < 0, and the work that is done by the gas is negative.

The work done by the gas is represented on the PV diagram by the rectangular area under the isobaric path. Whether the area is positive or negative depends on whether the gas expands or compresses.

3. Ideal Gas Relationship

PV = nRT

V ∝ T (∴P = Constant)

$\frac{V}{T}=\text{ Constant }$

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where $V_1$​ and $T_1$ are the initial volume and temperature, and $V_2$ and $T_2$ are the final volume and temperature.

4. Change in Internal Energy ($\Delta U$) for an Ideal Gas:

For an ideal gas, the change in internal energy ΔU is related to the change in temperature by:

$\Delta U=nR\frac{f}{2}(\Delta T)$

$f\frac{R}{2}=C_v$

$\Delta \mathrm{U}=\mathrm{n}{C}_V\Delta T$

where n is the number of moles, $C_V$​ is the specific heat capacity at constant volume, and ΔT is the change in temperature.

5. By First Law of Thermodynamics

$\Delta Q=\Delta U+\Delta W$

$(\therefore \Delta W=P\left(V_2-V_1\right)$

$\Delta W=nR\left(T_2-T_1\right)$

$\Delta W=nR\Delta T$

$\Rightarrow \Delta U$ is the change in internal energy and $\Delta Q$ is the heat added to or removed from the system.

$\Delta \mathrm{U}=\mathrm{n}{C}_V\Delta T$

$\Delta Q=\Delta U+\Delta W$

$\Delta Q=\mathrm{n}{C}_V\Delta T+nR\Delta T$

$\Delta Q=n\Delta T\left(C_V+R\right)$

Here, $C_P=C_V+R$

$C_P$ is known as the molar heat capacity at constant pressure.

So, in Isobaric Process, $\Delta Q=\mathrm{n}{C}_P\Delta T$.

6. Molar Heat Capacity      

The amount of heat needed to increase the temperature of 1 gram mole of a gas by  1℃ at constant pressure is equivalent to the gas's specific heat at constant pressure.

$\Delta Q=n{C}_P\Delta T$

$C_P=\frac{\Delta Q}{n\Delta T}$

Also, $C_P=C_V+R$

$C_P=\frac{f}{2}R+R$

7. Derivation of Work Done in Isobaric Process

Work Done is represented by,

dW = P dV

$\int dW={\int }_{V_i}^{V_f}PdV$

$P=\text{ constant }$

$\int dW=P{\int }_{V_i}^{V_f}dV$

$W=P(Vf-Vi)=nR(Tf-Ti)$

$W=P\Delta V=nR\Delta T$

• If the gas expands, $V_f>V_i$, so ΔV > 0, and the work done by the gas is positive.
• When the gas is compressed, $V_f<V_i$, so that ΔV < 0, and the work that is done by the gas is negative.

3.0Graph of Isobaric Process

The P-V diagram for isobaric processes is a horizontal straight line(representing constant pressure) parallel to the volume axis. The direction of this line shows compression or expansion, which can also explain sign convention in work done(negative sign in compression), where area under this line shows the work done.

Isobaric Process

4.0Work Done in Isobaric process from PV graph

Work done in an isobaric process: $\int dW=P{\int }_{V_i}^{V_f}dV$

W = Area under the PV graph

In the PV diagram, the area under the isobaric curve is equal to the work done in the isobaric process(Pressure Constant).The shaded area in the diagram represents the work performed by the gas.

5.0Slope of the PV Curve

Slope of PV curve in isobaric process is equal to

${\left(\frac{dP}{dV}\right)}_{\text{Isobaric }}=0$

6.0Examples of Isobaric Process

1.  The best example can be the boiling water and how the state of water changes from one form to another. The atmospheric pressure on water remains constant, the increase in temperature of water leads to the transition of water to steam. 
2. Similar example is the transition of Ice into Water. When we keep ice at room temperature, heat flows and results in melting of ice into water, while the atmospheric pressure remains constant.
3. When the gas is heated and moves the piston such that it exerts a force equal to the sum of atmospheric pressure and the force due to gravity, the process is considered isobaric.℃
4. Many cooking processes in the kitchen are isobaric. When food is prepared in an open vessel, the pressure above it remains constant at atmospheric pressure.

7.0Solved Examples

Example 1:  What is the heat input needed to raise the temperature of 3 moles of monoatomic gas from 0⁰C to 100⁰C at Isobaric Process?

Solution:

$\mathrm{Q}=\mathrm{n}{C}_P\Delta T$

$Q=3\times \frac{5R}{2}\times (100-0)=750R$

Example 2:  A polyatomic gas with six degrees of freedom does 25 J of work when it is expanded at constant pressure. Find the heat given to the gas ?

Solution: Since the process is taking place at constant pressure, it is an Isobaric process. Work done in an Isobaric Process: $W=P\Delta V=nR\Delta T$

$W=P\Delta V=nR\Delta T=25\mathrm{J}$    …………..(1)   

$\Delta Q=\mathrm{n}{C}_P\Delta T\text{ and }{C}_P=\frac{f}{2}R+R(f=6)$

On substituting the value of f, we have C_P=4 R 

Now $\Delta Q=\mathrm{n}{C}_P\Delta T=\mathrm{n}(4R)\Delta T=4nR\Delta T=4(25)$

$(nR\Delta T=25\mathrm{J})$

$\Delta Q=100\mathrm{J}$

8.0Points to Remember in Isobaric Process

1. Constant Pressure: The defining characteristic of an isobaric process is that the pressure of the system remains constant throughout the process.

2. Pressure-Temperature Relationship (for Ideal Gases):For an ideal gas undergoing an isobaric process, the relationship between pressure and temperature is given by

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$, where $V_1$ and $T_1$ are the initial volume and temperature, and $V_2$ and $T_2$ are the final volume and temperature.

3. Change in Internal Energy: For an ideal gas, the change in internal energy ΔU can be expressed as $\Delta \mathrm{U}=\mathrm{n}{C}_V\Delta T$

4. Isobaric Specific Heat Capacity: The specific heat capacity at constant pressure $C_p$​ is used to quantify how much heat is required to change the temperature of the gas maintained at a constant pressure. 

5. Graphical Representation: On a PV diagram (Pressure vs. Volume diagram),an isobaric process is represented as a horizontal line because the pressure remains constant while volume changes.

6. Graphical Work Done: Graphically, work done in an isobaric process is obtained by calculating the area under the PV graph. If the gas expands the work done is positive and if the gas compresses work done is negative.

7. Applicability in Real Systems: There are many examples in real life where isobaric processes are taking place, like boiling of water, melting of ice and various other cooking processes in our kitchen.

9.0Sample Questions on Isochoric Process 

Q-1: A gas undergoes an isobaric process, and its initial volume is 3m3. If the temperature increases from 300K to 450K, find the final volume.

Solution: P is constant, since it is an isobaric process.

Using Isobaric process equation $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$\Rightarrow \frac{3}{300}=\frac{V_2}{450}$

$\Rightarrow V_2=3\times \frac{450}{300}=4.5{\mathrm{m}}^3$

Q-2.If the ratio of the specific heat capacities of a gas is given. Calculate the change in internal energy of 1 mole of a gas when its volume increases from  V  to  2V at constant pressure P.

Solution:

$\gamma =\frac{C_p}{C_V}$= Specific Heat Ratio

$\Delta U=n{C}_V\Delta T=\frac{nR\Delta T}{\gamma -1}$

$\Delta U=\frac{nP\Delta V}{\gamma -1}=\frac{nP(2V-V)}{\gamma -1}=\frac{nPV}{\gamma -1}$

$(\therefore \mathrm{P}\Delta \mathrm{V}=\mathrm{nR}\Delta \mathrm{T})$

For n=1:     $\Delta U=\frac{PV}{\gamma -1}$

Q-3.  What is the ratio of the work an ideal monoatomic gas does to the heat added during an isobaric process?

Solution:

$\Delta W=P\Delta V=nR\Delta T$

$\Delta Q=n{C}_P\Delta T=\frac{5}{2}nR\Delta T\left(\therefore C_P=\frac{5}{2}R\right)$

$\frac{\Delta W}{\Delta Q}=\frac{nR\Delta T}{\frac{5}{2}nR\Delta T}=\frac{2}{5}$

Q-4.A cyclic process ABCD is given for helium gas $(C_V=\frac{3}{2}R,C_P=\frac{5}{2}R$ as shown in fig. Find work done in path B to C.

Solution:

Process B $\to$ C is an isobaric process because P= Constant

1. Work Done $\mathrm{W}=\int PdV=\mathrm{P}\Delta \mathrm{V}$

$W=P\Delta V=3P(2V-V)$

W=3 P V