200 Marks in JEE Mains Percentile 2026 - Expected Percentile and Rank

Isobaric Process - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Isobaric process is considered one the most difficult concept.

  • 34 Questions around this concept.

Solve by difficulty

When heat is given to a system in an isobaric process, then 

If C_{P} and C_{V} denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then

One mole of ideal monoatomic gas (\gamma =5/3) is mixed with one mole of diatomic gas (\gamma =7/5) . What is \gamma for the mixture? \gamma denotes the ratio of specific heat at constant pressure, to that at constant volume.

1 mole of a gas with \gamma =7/5 is mixed with 1 mole of a gas with \gamma =5/3 , then the value of \gamma  for the resulting mixture is

Choose the correct P-V graph for the Isobaric process

 

work done in the isobaric process for ideal gas can be given as 

A sample of gas expands from Vto V2. In which of the following, the work done will be greatest? 

NIELIT University(Govt. of India Institution) Admissions

Campuses in Ropar, Agartala, Aizawl, Ajmer, Aurangabad, Calicut, Imphal, Itanagar, Kohima, Gorakhpur, Patna & Srinagar

Vishwa Vishwani BS Computer Science Admissions 2025

27 Years of Academic Excellence | State of Art Infrastructure

Concepts Covered - 1

Isobaric process

Isobaric Process- A Thermodynamic process in which pressure remains constant is known as the isobaric process.

In this process, V and T change keeping P constant.I.e Charle’s law is obeyed in this process

   Key points in the Isobaric Process-

  • Its Equation of state is given as $\frac{V}{T}=$ constant
    I.e $\frac{V_1}{T_1}=\frac{V_2}{T_2}=$ constant
    - P-V Indicator diagram for an isobaric process

    Its PV graph has slope $=0$ (i.e $\frac{d P}{d V}=0$,

        

The above Graph represents an isobaric expansion.

 The P-V diagram for this process is a line is parallel to the volume line.

 

  • Specific heat of gas during the isobaric process is given by

     

$
C_P=\left(\frac{f}{2}+1\right) R
$

- The bulk modulus of elasticity during the isobaric process is given by

$
K=\frac{\Delta P}{-\Delta V / V}=0
$

- Work done in the isobaric process-

$
\Delta W=\int_{V_i}^{V_f} P d V=P \int_{V_i}^{V_f} d V=P\left[V_f-V_i\right]
$


Or we can write $\Delta W=P\left(V_f-V_i\right)=n R\left[T_f-T_i\right]=n R \Delta T$
- Internal energy in an isobaric process

$
\Delta U=n C_V \Delta T=n \frac{R}{(\gamma-1)} \Delta T
$

- Heat in an isobaric process

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

$
\begin{aligned}
& \quad \Delta Q=n \frac{R}{(\gamma-1)} \Delta T+n R \Delta T=n R \Delta T\left[\frac{1}{\gamma-1}+1\right] \\
& \Rightarrow \Delta Q=n R \Delta T \frac{\gamma}{\gamma-1}=n\left(\frac{\gamma}{\gamma-1}\right) R \Delta T=n C_p \Delta T \\
& \text { So } \Delta Q=n C_p \Delta T
\end{aligned}
$
 

  • Examples of the isobaric process-

               1. Conversion of water into vapour phase (boiling process)   

             

From the first law of thermodynamics

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

since $\Delta U_K=0$ [as there is no change in temperature] and using, $\Delta Q=m L$

$
\begin{aligned}
\Delta Q & =\Delta U_P+P\left[V_f-V_i\right] \\
\Delta U_P & =\Delta Q-P\left[V_f-V_i\right] \\
\Delta U_P & =m L-P\left[V_f-V_i\right]
\end{aligned}
$


Here, i- initial state and f-final state
2. Conversion of ice into water

From FLOT $\quad \Delta Q=\Delta U+\Delta W$ and using, $\Delta Q=m L$
we get $m L=\Delta U_P+\Delta U_K+\Delta W$

$
m L=\Delta U_P+\Delta U_K+P\left(V_f-V_i\right)
$

since $\Delta U_K=0$ [as there is no change in temperature]
and $\Delta W=0$ [As $V_f-V_i$ is negligible, l.e, when ice convert into water then change in volume, is negligible] Hence $\Delta U_P=m L$

 

Study it with Videos

Isobaric process

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions