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Polytropic Process - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
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21 Questions around this concept.
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The thermodynamic process is shown below on a PV diagram for one mole of an ideal gas.If $\mathrm{} \mathrm{V}_2=2 \mathrm{~V}_1$ then the ratio of temperature $\frac{T_2}{T_1}$ is :
An ideal gas undergoes a quasi-static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn=constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively)
An ideal gas follows a process equation $\mathrm{PT}=$ constant. The correct graph between pressure and volume of gas is
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A cylinder contain 1 mole of He gas of 200K.Pa and 27oC. The gas is compressed reversibly to a pressure of 600K.Pa following a polytropic process with exponent n=1.3. Then calculate the work required during the process.
For polytropic process $\left(P V^n=\right.$ constant $)$ n can take value as.
0.02 moles of an ideal diatomic gas with initial temperature $20^{\circ} \mathrm{C}$ is compressed from $1500 \mathrm{~cm}^3$ to $500 \mathrm{~cm}^3$. The thermodynamic process is such that $P V^2=\beta {\text {where }} \beta$ is constant. Then the value of $\beta$ is close to: (The gas constant $R=8.31 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$ )
(A) $7.5 \times 10^{-2}$ Pa.m. ${ }^6$
(B) $1.5 \times 10^{-2}$ Pa.m. ${ }^6$
(C) $3 \times 10^{-2}$ Pa.m. ${ }^6$
(D) $2.2 \times 10^{-2}$ Pa.m. ${ }^6$
An ideal gas follows a process described by $P V^2=C$ from $\left(P_1, V_1, T_1\right)_{\text {to }}\left(P_2, V_2, T_2\right)$ (C is a constant). Then
An ideal gas with heat capacity at constant volume CV undergoes a quasistatic process described by PVα in a P-V diagram, where α is a constant. The heat capacity of the gas during this process is given by
A real gas within a closed chamber at $27^{\circ} \mathrm{C}$ undergoes the cyclic process as shown in figure. The gas obeys $P V^3=\mathrm{RT}$ equation for the path $A$ to $B$. The net work done in the complete cycle is (assuming $R=8 \mathrm{~J} / \mathrm{molK}$ ):
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Try NowConcepts Covered - 1
Polytropic Process
A process $P V^N=C$ is called polytropic process. So, any process in this world related to thermodynamics can be explained by polytropic process.
For example - 1. If $\mathrm{N}=1$, then the process become isothermal.
2. If $\mathrm{N}=0$, then the process become isobaric.
3. If $\mathrm{N}=\gamma$, then the process become adiabatic
Work done by the polytropic process -
$
W_{1-2}=\int P d V
$
For a polytropic process,
$
\begin{gathered}
P V^N=P_1 V_1^N=P_2 V_2^N=C \\
P=\frac{C}{V^N}
\end{gathered}
$
Substituting in Equation, we get,
$
\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} \ldots \ldots (1)\\
P_1 V_1 & =n R T_1 \\
P_2 V_2 & =n R T_2
\end{aligned}
$
So, equation (1) can be written as -
$
W_{1-2}=\frac{n R\left(T_2-T_1\right)}{1-N}
$
And for one mole, $W_{1-2}=\frac{R\left(T_2-T_1\right)}{1-N}$
Specific heat for polytropic process -
We can write the equation of heat as - $Q=C \Delta T$
Here C = Molar specific heat -
From the first law of thermodynamics
$
\begin{aligned}
& \quad 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{aligned}
$
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