Cyclic And Non Cyclic Process

In thermodynamics, processes are categorized into cyclic and non-cyclic based on how the system's state changes. A cyclic process is one where a system returns to its initial state after undergoing a series of changes, with no net change in its properties. An example is the operation of an internal combustion engine, where each cycle involves intake, compression, combustion, and exhaust. On the other hand, a non-cyclic process doesn't return the system to its original state, like charging a battery or burning fuel, where the energy is transformed, and the state changes permanently. These processes can be seen in daily life: for instance, our body’s metabolic cycles are cyclic, whereas a campfire burning wood into ash is non-cyclic. Understanding these processes helps in optimizing energy use, as cyclic processes are often more sustainable and predictable in closed systems.

Cyclic and Non-cyclic Process

In thermodynamics, cyclic and non-cyclic processes refer to how a system undergoes changes in its state.

Cyclic Process - A cyclic process consists of a series of changes that return the system back to its initial state.
Non-cyclic Process - In the non-cyclic process, the series of changes involved do not return the system back to its initial state.

Now, as we know internal energy is the point function. So when the process returns to its initial point after completing the process then the final and initial internal energy will be the same. So, the change in internal energy is zero.

In case of cyclic process as $U_{\text {final }}=U_{\text {initial }} \Rightarrow \Delta U=U_{\text {final }}-U_{\text {initial }}=0$
i.e., change in internal energy for cyclic process is zero

So, we can say that $\Delta U \propto \Delta T \Rightarrow \Delta T=0$

By applying the first law of thermodynamics to cyclic processes

$\Delta Q=\Delta U+\Delta W \Rightarrow \Delta Q=\Delta W \quad($ As $\Delta U=0)$

So, we can say that the heat given is equal to the work obtained in the cyclic process.

For the cyclic process, the initial point and final point are the same. So, the P-V graph is a closed curve and the area enclosed by the closed path gives the work done.

But, here is one assumption for the calculation of work done

If the cycle is clockwise work done is positive and if the cycle is anticlockwise work done is negative.

From the given graph and its direction, you can see that the first graph has positive work and the second graph has negative work

Now, for the Non-cyclic process, the work done is equal to the area covered between the curve and the volume axis on the P-V diagram. It does not depend on the points or state but it depends on the process of the path. We can see this in the given graph.

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Solved Examples Based on Cyclic and Non-cyclic Process

Example 1: An ideal gas is taken via the path as shown in the figure. The net work done in the whole cycle is

1) $6 P_1 V_1$
2) Zero
3) $3 P_1 V_1$
4) $-3 P_1 V_1$

Solution:

Work done in the cyclic process

It is the area under the cycle.

wherein

Work is positive if the cycle is clockwise. Work is negative if the cycle is anti-clockwise.

(for P - V diagram)

Work done is the area enclosed.

Work done $=\frac{1}{2}\left(3 V_1-V_1\right)\left(4 P_1-P_1\right)$

$
W=-3 P_1 V_1
$

Hence, the answer is the option (4).

Example 2: Two rigid boxes containing different ideal gases are placed on a table. Box $A$ contains one mole of nitrogen at temperature $T_0$, while Box $B$ contains one mole of helium at temperature ( $7 / 3$ ) $T_0$. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final temperature of the gases, $T_f$, in terms of $T_0$ is

1) $T_f=\frac{5}{2} T_0$
2) $T_f=\frac{3}{7} T_0$
3) $T_f=\frac{7}{3} T_0$
4) $T_f=\frac{3}{2} T_0$

Solution:

Since boxes are rigid

So
$
\begin{aligned}
& \Delta V=0 \\
\Rightarrow & \Delta W=0
\end{aligned}
$
As given in the question, we have to ignore the heat capacity of boxes.
So $\Delta Q=0$
So using the first law of thermodynamics, we get

$
\begin{aligned}
& \Delta U=0 \\
& \text { i.e } U_i=U_f \\
& n_1 C_{V 1} T_o+n_2 C_{V 2}\left(\frac{7}{3} T_o\right)=\left(n_1 C_{V 1}+n_2 C_{V 2}\right) T_f \\
& T_f=\frac{1 \times \frac{5 R}{2} \cdot T_o+1 \cdot\left(\frac{3 R}{2}\right) \cdot \frac{7}{3} T_o}{1 \cdot \frac{5 R}{2}+1 \cdot \frac{3 R}{2}}=\frac{6 R T_o}{4 R}=1.5 T_o
\end{aligned}
$

Hence, the answer is the option (4).

Example 3: When a system is taken from $i$ to state $f$ along the path $i a f$, it is found that $\mathrm{Q}=50$ cal and $\mathrm{W}=20$ cal. Along the path $i b f \mathrm{Q}=36$ cal. $W$ along the path $i b f$ is (in calories)

1) 6

2) 14

3) 16

4) 66

Solution:

Change in internal energy for a cyclic process
$
\Delta U=0
$

wherein
Since in a cyclic process, the initial and final state is the same.

$
U_f=U_i
$
Change in internal energy is the same for both paths.

$
\begin{aligned}
& \Delta U_{i a f}=\Delta U_{i b f} \\
& =>(50-20) \mathrm{Cal}=\left(36 \mathrm{Cal}-W_{i b f}\right) \\
& =>W_{i b f}=6 \mathrm{Cal}
\end{aligned}
$

Hence, the answer is the option (1).

Example 4: An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_1$ and contains ideal gas at pressure $P_1$ and temperature $T_1$. The other chamber has volume $V_2$ and contains ideal gas at pressure $P_2$ and temperature $T_2$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

1) $\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_1+P_2 V_2 T_2}$
2) $\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}$
3) $\frac{P_1 V_1 T_1+P_2 V_2 T_2}{P_1 V_1+P_2 V_2}$
4) $\frac{P_1 V_1 T_2+P_2 V_2 T_1}{P_1 V_1+P_2 V_2}$

Solution:

Since in a cyclic process, the initial and final state is the same.

$\begin{aligned} & U_f=U_i \\ & n_1 C_{V 1} T_1+n_2 C_{V 2} T_2=\left(n_1+n_2\right) C_{V f} T \\ & C_{V 1}=C_{V 2}-C_{V f} \\ & \therefore T=\frac{n_1 C_{V 1} T_1+n_2 C_{V 2} T_2}{\left(n_1+n_2\right) C_v} \\ & =\frac{n_1 T_1+n_2 T_2}{n_1+n_2} \\ & n_1=\frac{P_1 V_1}{R T_1}, n_2=\frac{P_2 V_2}{R T_2} \\ & T=\frac{\left(P_1 V_1+P_2 V_2\right) T_1 T_2}{P_1 V_1 T_2+P_2 V_2 T_1}\end{aligned}$

Hence, the answer is the option (2).

Example 5: For a cyclic process which of the following is not true

1) change in internal energy is zero

2) The temperature of the system remains constant

3) Heat supplied is equal to the work done by the system

4) If the cycle is clockwise work done is negative

Solution:

Cyclic Process

A cyclic process consists of a number of changes that return the system back to its initial state.

in cyclic process $\Delta U=0$
In the case of a cyclic process

$
U_f=U_i \Rightarrow \Delta U=0
$

$\Delta U \propto \Delta T$ so $\Delta T=0$

$
\Delta Q=\Delta V+\Delta W
$
From FLOT $\Delta Q=\Delta W$

For the clockwise cyclic process work done is positive

Hence, the answer is the option (4).

Summary

In thermodynamics, a cyclic process returns a system to its initial state, with no net change in internal energy, meaning the heat supplied equals the work done. The work done is represented by the area enclosed on a P-V diagram, and its sign depends on the direction of the cycle (positive for clockwise, negative for counterclockwise). A non-cyclic process, however, does not return the system to its initial state, and the work done depends on the path taken rather than specific states. Understanding these concepts is crucial in applying the first law of thermodynamics.