Physics > First Law of Thermodynamics > 4.0 Different thermodynamic processes
First Law of Thermodynamics
1.0 Introduction
2.0 Three important terms in first law of thermodynamics.
3.0 First law of thermodynamics
4.0 Different thermodynamic processes
4.1 Isobaric Process:
4.2 Isochoric process
4.3 Isothermal process
4.5 Adiabatic process
4.6 Polytropic process
5.0 Graphs
6.0 Efficiency of cyclic process
7.0 Heat engine
8.0 Refrigerator
4.6 Polytropic process
4.2 Isochoric process
4.3 Isothermal process
4.5 Adiabatic process
4.6 Polytropic process
Polytropic process is a general process given by ${P{V^\alpha } = cons\tan t}$. By putting different values of $\alpha $ we get different processes.
We consider an expansion in which temperature changes from ${T_1}$ to ${T_2}$.$$\begin{equation} \begin{aligned} W = \int\limits_{{V_1}}^{{V_2}} {PdV} = \int\limits_{{V_1}}^{{V_2}} {\frac{C}{{{V^\alpha }}}dV} = C\int\limits_{{V_1}}^{{V_2}} {\frac{{dV}}{{{V^\alpha }}} = \left[ {\frac{{C{V^{ - \alpha + 1}}}}{{1 - \alpha }}} \right]} _{{V_1}}^{{V_2}} \\ = \frac{C}{{1 - \alpha }}\left( {V_2^{1 - \alpha } - V_1^{1 - \alpha }} \right) \\ = \frac{1}{{1 - \alpha }}\left( {{P_2}V_2^\alpha V_2^{1 - \alpha } - {P_1}V_1^\alpha V_1^{1 - \alpha }} \right) \\ = \frac{{{P_2}{V_2} - {P_1}{V_1}}}{{1 - \alpha }} = \frac{{nR\left( {{T_2} - {T_1}} \right)}}{{1 - \alpha }} \\ \Delta U = n{C_V}\left( {{T_2} - {T_1}} \right) \\ Q = n\left( {{T_2} - {T_1}} \right)\left[ {{C_V} + \frac{R}{{1 - \alpha }}} \right]\left[ {\because Q = \Delta U + W} \right] \\ \Rightarrow Q = n\Delta T\left[ {{C_V} + \frac{R}{{1 - \alpha }}} \right] \\\end{aligned} \end{equation} $$
Specific heat of gas for polytropic process:
$$C = \frac{Q}{{n\Delta T}} = {C_V} + \frac{R}{{1 - \alpha }}$$
Putting different values in $\alpha $ we will get all the thermodynamic processes.
| Name of the process | $Q$ | $\Delta U$ | $W$ |
| Isothermal | $Q = W$ | $0$ | $nRT\ln \left( {\frac{{{V_f}}}{{{V_i}}}} \right) = nRT\ln \left( {\frac{{{P_i}}}{{{P_f}}}} \right)$ |
| Adiabatic | $0$ | $n{C_V}\Delta T$ | $\frac{{{P_i}{V_i} - {P_f}{V_f}}}{{1 - \alpha }} = - \Delta U$ |
| Isobaric | $n{C_P}\Delta T$ | $n{C_V}\Delta T$ | $P\left( {{V_f} - {V_i}} \right)$ |
| Isochoric | $Q = \Delta U = n{C_V}\Delta T$ | $n{C_V}\Delta T$ | $0$ |
Cyclic Process:
In cyclic process, initial and final states are same and so change in internal energy for gas in cyclic process is zero.$$ \Rightarrow \Delta U = 0$$ Thus the total work done is equal to net amount of heat supplied to gas.$$\begin{equation} \begin{aligned} \sum {\Delta U} = 0 \\ \sum Q = \sum W \\\end{aligned} \end{equation} $$
Work done gas in the cyclic process:
Work was done in cyclic process = Area enclosed by the close curve.
On $P$-$V$ graph if the process is clockwise, net work done is positive. If the process is anti-clockwise net work done is negative.
Explanation:
 |  |  |
The work done in this process is given by the sum of the area enclosed by the upper and the lower curve with the $x$ axis. Work = $W$ = ${A_1} - {A_2}$ = positive | As the cycle is in clockwise direction, the area is positive and the work done is positive. $W$ = ${A_1}$ | As the cycle is in the anti-clockwise direction, the area is negative and the work done is negative. $W$ = -${A_2}$ |
Question 15. The $P$-$V$ diagram of $0.2$ moles of a diatomic ideal gas is shown in the figure. Process $BC$ is adiabatic.The value of $\gamma $ for this gas is $1.4$.
(a) Find the pressure and volume at points $A$,$B$ and $C$.
(b) Calculate $\Delta Q$, $\Delta W$ and $\Delta U$ for each of the three processes.
(c) Find thermal efficiency of the cycle.
Take $1 atm$ = $1.0 \times {10^5}N/{m^2}$
Solution:
(a) $${P_A} = {P_C} = 1\,atm\, = 1.01 \times {10^5}N/{m^2}$$ Process $AB$ is an isochoric process.$$\begin{equation} \begin{aligned} P \propto T \\ \frac{{{P_B}}}{{{P_A}}} = \frac{{{T_B}}}{{{T_A}}} \\ {P_B} = \left( {\frac{{{T_B}}}{{{T_A}}}} \right){P_{_A}} = \left( {\frac{{600}}{{300}}} \right)\left( {1\;atm} \right) \\ = 2\;atm \\ = 2.02 \times {10^5}\;N/{m^2} \\\end{aligned} \end{equation} $$From ideal gas equation $$\begin{equation} \begin{aligned} V = \frac{{nRT}}{P} \\ {V_A} = {V_B} = \frac{{nR{T_A}}}{{{P_A}}} \\ = \frac{{\left( {0.2} \right)\left( {8.31} \right)\left( {300} \right)}}{{\left( {1.01 \times {{10}^5}} \right)}} \approx 5.0 \times {10^{ - 3}}{m^3} \\ = 5\;liters \\\end{aligned} \end{equation} $$and$$\begin{equation} \begin{aligned} {V_C} = \frac{{nR{T_C}}}{{{P_C}}} = \frac{{\left( {0.2} \right)\left( {8.31} \right)\left( {455} \right)}}{{\left( {1.01 \times {{10}^5}} \right)}} \\ = 7.6 \times {10^{ - 3}}\,{m^3} \\ = 7.6\;liters \\\end{aligned} \end{equation} $$
| State | $P$ | $V$ |
| $A$ | $1 atm$ | 5 $L$ |
| $B$ | $2 atm$ | 5 $L$ |
| $C$ | $1 atm$ | 7.6 $L$ |
Process $A-B$ : Is an isochoric process. Hence,$$\begin{equation} \begin{aligned} \Delta {W_{AB}} = 0 \\ \Delta {Q_{AB}} = \Delta {U_{AB}} = n{C_V}\Delta T = n\left( {\frac{5}{2}R} \right)\left( {{T_B} - {T_A}} \right) \\ = \left( {0.2} \right)\left( {\frac{5}{2}} \right)\left( {8.31} \right)\left( {600 - 300} \right) \\ \approx 1246\;J \\\end{aligned} \end{equation} $$Process $BC$ is an adiabatic process. Hence,$$\begin{equation} \begin{aligned} \Delta {Q_{BC}} = 0 \\ \Delta {W_{BC}} = - \Delta {U_{BC}} \\ \Delta {U_{BC}} = n{C_V}\Delta T = n{C_V}\left( {{T_C} - {T_B}} \right) \\ = \left( {0.2} \right)\left( {\frac{5}{2}R} \right)\left( {455 - 600} \right) \\ = \left( {0.2} \right)\left( {\frac{5}{2}} \right)\left( {8.31} \right)\left( { - 145} \right)J \\ \approx - 602\;J \\ \Delta {W_{BC}} = - \Delta {U_{BC}} = 602\;J \\\end{aligned} \end{equation} $$Process $CA$ is an isobaric process. Hence,$$\begin{equation} \begin{aligned} \Delta {Q_{CA}} = n{C_P}\Delta T = n\left( {\frac{7}{2}R} \right)\left( {{T_A} - {T_C}} \right) \\ = \left( {0.2} \right)\left( {\frac{7}{2}} \right)\left( {8.31} \right)\left( {300 - 455} \right) \\ \approx - 902\;J \\ \Delta {U_{CA}} = n{C_V}\Delta T \\ = \frac{{\Delta {Q_{CA}}}}{\gamma }\quad \quad \left( {as\;\,\gamma = \frac{{{C_P}}}{{{C_V}}}} \right) \\ = - \frac{{902}}{{1.4}} \approx - 644\;J \\ \Delta {W_{CA}} = \Delta {Q_{CA}} - \Delta {U_{CA}} = - 258\;J \\\end{aligned} \end{equation} $$
| Process | $$\Delta Q\,\left( {\,in\;J} \right)$$ | $$\Delta W\,\left( {\,in\;J} \right)$$ | $$\Delta U\,\left( {\,in\;J} \right)$$ |
| $AB$ | $1246$ | $0$ | $1246$ |
| $BC $ | $0$ | $602$ | $-602$ |
| $CA$ | $-902$ | $-258$ | $-644$ |
| Total | $344$ | $344$ | $0$ |
(c) Efficiency of cycle$$\begin{equation} \begin{aligned} \eta = \frac{{{W_{Total}}}}{{\left| {{Q_{ + ve}}} \right|}} \times 100 \\ = \frac{{344}}{{1246}} \times 100 \\ = 27.6\% \\\end{aligned} \end{equation} $$
Question 16. One mole of a monoatomic ideal gas is taken through the cycle shown in the figure.
$A-B$ Adiabatic expansion
$B-C$ Cooling at constant volume
$C-D$ Adiabatic compression.
$D-A$ Heating at constant volume.
The pressure and temperature at $A$,$B$ are denoted by ${P_A}$, ${T_A}$ ; ${P_B}$, ${T_B}$ etc. respectively.
Given ${T_A}$ = 1000K, ${P_B} = \left( {\frac{2}{3}} \right){P_A}$. Calculate
(a) the work done by the gas in the process $A - B$
(b) the heat lost by the gas in the process $B - C$ and
Given ${\left( {\frac{2}{3}} \right)^{0.4}} = 0.85\;and\;R = 8.314\;J/mol - K$
Solution: As for adiabatic change $$\begin{equation} \begin{aligned} P{V^\gamma }\, = \,constant\\ P{\left( {\frac{{nRT}}{P}} \right)^\gamma } = constant\quad \left[ {as\;PV = nRT} \right] \\ \frac{{{T^\gamma }}}{{{P^{\gamma - 1}}}} = cons\operatorname{tant} \;so\;{\left( {\frac{{{T_B}}}{{{T_A}}}} \right)^\gamma } = {\left( {\frac{{{P_B}}}{{{P_A}}}} \right)^{\gamma - 1}}\quad where\quad \gamma = \frac{5}{3} \\ {T_B} = {T_A}{\left( {\frac{2}{3}} \right)^{1 - \frac{1}{\gamma }}} = 1000{\left( {\frac{2}{3}} \right)^{\frac{2}{5}}} = 850K \\ so,\quad {W_{AB}} = \frac{{nR\left[ {{T_F} - {T_I}} \right]}}{{\left[ {1 - \gamma } \right]}} = \frac{{1 \times 8.31\left[ {1000 - 850} \right]}}{{\left[ {\left( {\frac{5}{3}} \right) - 1} \right]}} \\ {W_{AB}} = \left( {\frac{3}{2}} \right) \times 8.31 \times 150 = 1869.75J \\\end{aligned} \end{equation} $$
(b) For $B-C$, $V$ = constant so $\Delta W = 0$
So, from first law of thermodynamics $$\begin{equation} \begin{aligned} \Delta Q = \Delta U + \Delta W = n{C_V}\Delta T + 0 \\ \Delta Q = 1 \times \left( {\frac{3}{2}R} \right)\left( {{T_C} - 850} \right)\quad as\,{C_V} = \frac{3}{2}R \\\end{aligned} \end{equation} $$Now, along path $BC$, $V$=constant; $$\begin{equation} \begin{aligned} P \propto T \\ \frac{{{P_C}}}{{{P_B}}} = \frac{{{T_C}}}{{{T_B}}} \\ {T_C} = \frac{{\left( {\frac{1}{3}} \right){P_A}}}{{\left( {\frac{2}{3}} \right){P_A}}} \times {T_B} = \frac{{{T_B}}}{2} = \frac{{850}}{2} = 425\,K \\ so,\;\Delta Q = 1 \times \frac{3}{2} \times 8.31\left( {425 - 850} \right) = - 5297.625J \\\end{aligned} \end{equation} $$[Negative heat means, heat is lost by the system]
