Organic Chemistry Chapters for NEET 2025 - Organic Chemistry Weightage

Isobaric Process MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Isobaric process is considered one the most difficult concept.

  • 24 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.

NEET 2024: Cutoff (OBC, SC, ST & General Category)

NEET 2024 Admission Guidance: Personalised | Study Abroad

NEET 2025: SyllabusMost Scoring concepts NEET PYQ's (2015-24)

NEET PYQ's & Solutions: Physics | ChemistryBiology

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

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}}=\text { constant }

  • P-V Indicator diagram for an isobaric process

            Its PV graph has slope=0 (i.e \frac{dP}{dV}=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_{f}\right)=n R\left[T_{f}-T_{t}\right]=nR \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

So \\\Delta Q=n \frac{R}{(\gamma-1)} \Delta T+nR \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=nC_p\Delta T

           So \\\Delta Q =nC_p\Delta T

  • Examples of the isobaric process-

               1. Conversion of water into vapor 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 =mL

                      \begin{aligned} \Delta Q &=\Delta U_{P}+P\left[V_{f}-V_{t}\right] \\ \Delta U_{P} &=\Delta Q-P\left[V_{f}-V_{t}\right] \\ \Delta U_{P} &=m L-P\left[V_{f}-V_{i}\right] \end{aligned}

               2. Conversion of ice into water

                From FLOT     \Delta Q= \Delta U+\Delta W   and using, \\\Delta Q =mL

               we get   mL= \Delta U_{P}+\Delta U_{K}+\Delta W

                             mL= \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, I.e, when ice convert into water then change in volume, is negligible]

Hence \Delta U_{P}= mL

Study it with Videos

Isobaric process

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

Books

Reference Books

Isobaric process

Physics Part II Textbook for Class XI

Page No. : 312

Line : 55

E-books & Sample Papers

Get Answer to all your questions

Back to top