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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.

  • 15 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

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_{i}\right)=n R\left[T_{f}-T_{i}\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 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 =mL

                      \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-finalstate

               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

 

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Isobaric process

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