Polytropic Process - Practice Questions & MCQ

Solve by difficulty

MEDIUM

One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure.

The change in internal energy of the gas during the transition is:

$-20 kJ$

$20 J$

$-12 kJ$

$20 kJ$

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Concepts Covered - 1

Polytropic Process

A process $PV^{N} = C$ is called polytropic process. So, any process in this world related to thermodynamics can be explained by polytropic process.

For example - 1. If N = 1 , then the process become isothermal.

2. If N=0, then the process become isobaric.

3. If N = $\gamma$ , then the process become adiabatic

Work done by polytropic process -

$W_{1-2}=\int P d V$

$\begin{array}{l}{\text { For a polytropic process, }} \\ {\qquad \begin{aligned} P V^{N} &=P_{1} V_{1}^{N}=P_{2} V_{2}^{N}=C \\ & P=\frac{C}{V^{N}} \end{aligned}}\end{array}$

$\begin{array}{l}{\text { Subsiting in Equation } \text { , we get, }} \\ \\ {\qquad \begin{aligned} \int P d V &=\int \frac{C d V}{V^{N}}=C \int V^{-N} d v \\ \\ &=\left[V^{1-N}\right]_{1}^{2}=\left(V_{2}^{1-N}-V_{1}^{1-N}\right) \\ \\ W_{1-2} &=\frac{P_{2} V_{2}-P_1 V_{1}}{1-N} \text { or } \frac{P_1 V_{1}-P_{2} V_{2}}{N-1}. . . . . . (1) \\ \\ P_{1} V_{1} &=n R T_{1} \\ \\ P_{2} V_{2} &=n R T_{2} \end{aligned}}\end{array}$

So, equation (1) can be written as -

$\begin{aligned} W_{1-2} &=\frac{n R\left(T_{2}-T_{1}\right)}{1-N} \\ \text { And for one mole, } & W_{1-2}=\frac{R\left(T_{2}-T_{1}\right)}{1-N} \end{aligned}$

Specific heat for polytropic process -

We can write equation of heat as - $Q=C \Delta T$

Here C = Molar specific heat -

$\begin{array}{l}{\text { From the first law of thermodynamics }} \\ \\ {\qquad Q=\Delta U+W} \\ \\ {\text { or } C \Delta T=C_{v} \Delta T-\frac{R \Delta T}{(N-1)}} \\ \\ {\therefore \quad C=C_{v}-\frac{R}{(N-1)}=\frac{R}{(\gamma-1)}-\frac{R}{(N-1)}}\end{array}$