Adiabatic Process - Practice Questions & MCQ

Adiabatic process - 

When a thermodynamic system undergoes a process, such that there is no exchange of heat takes place between the system and surroundings, this process is known as adiabatic process.

In this process $\mathrm{P}, \mathrm{V}$ and T changes but $\Delta Q=0$.
From first law of thermodynamics -

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

Now in adiabatic -

$
0=\Delta U+\Delta W
$

So, $\Delta U=-\Delta W$ for adiabatic process
Now, let us take two cases, first is for expansion in which the work done is positive and second one is compression in which the work done is negative -
If $\Delta W=$ positive then $\Delta U$ become negative so temperature decreases ie., adiabatic expansion produce cooling.
If $\Delta W=$ negative then $\Delta U$ become positive so temperature increases ie., adiabatic compression produce heating.

Equations of Adiabatic process -
1. $P V^\gamma=$ constant; where $\gamma=\frac{C_P}{C_V} \quad \ldots$ Relating Pressure and volume
2. $T V^{\gamma-1}=$ constant $\Rightarrow T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$ or $T \propto V^{1-\gamma} \ldots$ - Relating Temperature and volume
3. $\frac{T^\gamma}{P^{\gamma-1}}=$ const. $\Rightarrow T_1^\gamma P_1^{1-\gamma}=T_2^\gamma P_2^{1-\gamma}$ or $T \propto P^{\frac{\gamma-1}{\gamma}}$ or $P \propto T^{\frac{\gamma}{\gamma-1}}$

For the slope of adiabatic curve on PV curve, we have to differentiate the adiabatic relation -
As, $P V^\gamma=$ constant
So,

$
\begin{aligned}
& d P V^\gamma+P \gamma V^{\gamma-1} d V=0 \\
& \frac{d P}{d V}=-\gamma \frac{P V^{\gamma-1}}{V^\gamma}=-\gamma\left(\frac{P}{V}\right)
\end{aligned}
$
 

So we can say that in the given graph, the slope =

$
\tan \left(180^{\circ}-\phi\right)=-\gamma\left(\frac{P}{V}\right)
$

Also, we have studied that the slope of the isothermal curve on PV diagram is $=\frac{-P}{V}$

So, we can say that the -

$
\text { (Slope) }_{\text {adiabatic }}=\gamma \times(\text { Slope })_{\text {isother mal }}, \text { or } \frac{(\text { Slope })_{\text {adiabatic }}}{(\text { Slope })_{\text {isother mal }}}>1
$
 

With the help of graph we can see that the adiabatic curve is more steeper than the isothermal curve- 

                                  or,        

Specific heat in the adiabatic process - Specific heat of gas during adiabatic change is zero. Mathematically  -

$
C=\frac{Q}{m \Delta T}=\frac{0}{m \Delta T}=0 \quad[\text { As } Q=0]
$

Note- Even though heat is not supplied or taken out during the process but still, the temperature change is taking place. So we can say that Specific adiabatic process is zero.

Work done in the adiabatic process -

$
W=\int_{V_i}^{V_f} P d V=\int_{V_i}^{V_f} \frac{K}{V^\gamma} d V \Rightarrow W=\frac{\left[P_i V_i-P_f V_f\right]}{(\gamma-1)}=\frac{\mu R\left(T_i-T_f\right)}{(\gamma-1)}
$

So, if $\gamma$ is increasing then the work done will be decreasing. As we know that -

$
\because \gamma_{\text {mono }}>\gamma_{\text {diatomic }}>\gamma_{\text {triatomic }} \Rightarrow W_{\text {mono }}<W_{\text {diatomic }}<W_{\text {triatomic }}
$

Elasticity in the adiabatic process - As, $P V^\gamma=$ constant

Differentiating both sides of the above equation, we get $-d P V^\gamma+P \gamma V^{\gamma-1} d V=0$

$
\gamma P=\frac{d P}{-d V / V}=\frac{\text { Stress }}{\text { Strain }}=E_\phi \Rightarrow E_\phi=\gamma P
$

i.e. adiabatic elasticity is $\gamma$ times that of pressure

$
E_\theta=P \Rightarrow \frac{E_\phi}{E_\theta}=\gamma=\frac{C_P}{C_V}
$

i.e. the ratio of two elasticity of gases is equal to the ratio of two specific heats.

                                                      Where,
$E_\phi=$ Elasticity in adiabatic process
and
$E_\theta=$ Elasticity in isothermal process