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Isobaric process is considered one the most difficult concept.
32 Questions around this concept.
When heat is given to a system in an isobaric process, then
If and denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then
One mole of ideal monoatomic gas is mixed with one mole of diatomic gas . What is for the mixture? denotes the ratio of specific heat at constant pressure, to that at constant volume.
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1 mole of a gas with is mixed with 1 mole of a gas with , then the value of 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 V1 to V2. In which of the following, the work done will be greatest?
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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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