Adiabatic Process MCQ - Practice Questions with Answers

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 the adiabatic process.

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

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

Now in adiabatic -

$$
0=\Delta U+\Delta W
$$

So, $\Delta U=-\Delta W$ for the adiabatic process

Now, let us take two cases, the first is for expansion in which the work done is positive and the second one is compression in which the work done is negative - 

If $\Delta W=$ positive then $\Delta U$ becomes negative so temperature decreases ie., adiabatic expansion produces cooling.
If $\Delta W=$ negative then $\Delta U$ becomes positive so temperature increases ie., adiabatic compression produces heating.

Equations of Adiabatic process - 

1. $P V^\gamma=$ constant; where $\gamma=\frac{C_P}{C_V}$ -- -- Relating Pressure and volume
2. $T V^{\gamma-1}=$ constant $\Rightarrow T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$ or $T \propto V^{1-\gamma}---$ - -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 the adiabatic curve on the 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 $=$

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

Also, we have studied that the slope of the isothermal curve on the PV diagram is = So, we can say that the - $\quad$ (Slope $)_{\text {adiabatic }}=\gamma \times(\text { Slope })_{\text {isother mal }}$, or $\frac{(\text { Slope })_{\text {adiabatic }}}{(\text { Slope })_{\text {isother mal }}}>1$

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

                                  or,        

Specific heat in adiabatic process - The specific heat of a 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 the Specific heat for an adiabatic process is zero.

Work done in adiabatic process -

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

$\mathrm{Sa}_1$ if $\gamma$ is increasing then 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 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,

$$
\begin{aligned}
& \quad E_\phi=\text { Elasticity in adiabatic process } \\
& \text { and } \\
& E_\theta=\text { Elasticity in isothermal process }
\end{aligned}
$$