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ChemistryEdu Logo Thermodynamics | Polytropic Process#

Polytropic Process#

  • Polytropic process is a process which follows the equation:
\[PV^x = k\]

where, x = polytropic index and k = constant

Special Cases:

  1. If x = 0 ⇒ P = constant ⇒ Isobaric process
  2. If x = 1 ⇒ PV = constant ⇒ Isothermal process
  3. If x = γ ⇒ PVγ = constant ⇒ Adiabatic process

Work done in a polytropic process#

For a reversible polytropic process, work done can be calculated as:

\[W = - \int\limits_{V_1}^{V_2} P_{ext}dV\]
\[W = - \int\limits_{V_1}^{V_2} P_{gas}dV\]
\[W = - \int\limits_{V_1}^{V_2} {k \over V^x}dV\]
\[W = {-k(V_2^{1-x} - V_1^{1-x}) \over 1-x}\]

Calculation of molar heat capacity#

  • Molar heat capacity is defined as heat capacity per unit mole.
\[C_m = {dq \over ndT}\]
\[C_m = {dU - dW \over ndT}\]
\[C_m = {dU - (-PdV) \over ndT}\]
\[C_m = {dU \over ndT} + {PdV \over ndT}\]
\[C_m = {nC_vdT \over ndT} + {PdV \over ndT}\]
\[C_m = C_v + {PdV \over ndT}\]
  • Let's calculate dV/dT so that it can be used in above expression.
\[PV^x = k\]
\[ {T \over V}V^x = k\]
\[TV^{x-1} = k\]

Differentiating both sides:

\[V^{x-1}dT + T(x-1)V^{x-2}dV = 0\]
\[ {dV \over dT} = {V \over T(1-x)}\]

Putting in the expression of Cm:

\[C_m = C_v + {PV \over nT(1-x)}\]

Since, PV = nRT (ideal gas equation), we can write:

\[C_m = C_v + {nRT \over nT(1-x)}\]
\[C_m = C_v + {R \over (1-x)}\]

Questions based on Polytropic Process#

For a monoatomic ideal gas, the ratio of pressure and square of volume is 3. How much heat should be supplied to increase the temperature of 5 moles of the gas by 50oC?

\[ {P \over V^2} = 3\]
\[ PV^{-2} = constant\]

This is a polytropic process with x = -2.

Molar heat capacity is given by:

\[C_m = C_v + {R \over (1-x)}\]
\[C_m = C_v + {R \over (1-(-2))}\]
\[C_m = {3R \over 2} + {R \over 3}\]
\[C_m = {11R \over 6}\]

Also, molar heat capacity is given by:

\[C_m = {dq \over ndT}\]
\[dq = nC_mdT\]
\[q = \int\limits_{T_1}^{T_2} nC_mdT\]
\[q = nC_m(T_2 - T_1)\]
\[q = nC_mΔT\]
\[q = 5 \times {11R \over 6} \times 50\]
\[q = {1375R \over 3}\]
\[q = {1375 \times 8.314 \over 3}\]
\[q = 3810.58\ J\]

The molar heat capacity of a gas (at room temperature) for which pressure and volume are equal is 7R/2. What will be the rotational degree of freedom for this gas? Also, predict the atomicity of gas.

\[Given:\ P = V\]
\[PV^{-1} = constant\]

This is a polytropic process with x = -1.

Molar heat capacity is given by:

\[C_m = C_v + {R \over (1-x)}\]
\[⇒{7R \over 2} = C_v + {R \over (1-(-1))}\]
\[⇒{7R \over 2} = C_v + {R \over 2}\]
\[⇒C_v = 3R\]

Thus, it is a triatomic gas.

\[C_v = {fR \over 2}\]
\[⇒3R = {fR \over 2}\]
\[⇒f = 6\]

Thus, total degree of freedom = 6

Let ftr = Translational degree of freedom and fr = Rotational degree of freedom. Then, At room temperature:

\[f_{tr} + f_r = f\]
\[3 + f_r = 6\]
\[f_r = 3\]

Thus, rotational degree of freedom is 3.

Calculate the molar heat capacity of monoatomic gas for which the ratio of pressure and volume is 1.

\[Given:\ {P \over V} = 1\]
\[PV^{-1} = constant\]

This is a polytropic process with x = -1.

Molar heat capacity is given by:

\[C_m = C_v + {R \over (1-x)}\]

Since, Cv = 3R/2 for a monoatomic gas, we can write:

\[C_m = {3R \over 2} + {R \over (1-(-1))} \]
\[C_m = {3R \over 2} + {R \over 2}\]
\[C_m = 2R\]