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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.

  • 32 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? 

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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 $\Delta U_P=m L$

 

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Isobaric process

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Isobaric process

Physics Part II Textbook for Class XI

Page No. : 312

Line : 55

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