Work done in different thermodynamic Processes

Quasi-static

Ideal Gas: $P V=n R T$

Isothermal process: Temperature remains fixed. Internal energy (function of) does not change Pressure and Volume change

$P_1, V_1, T \Rightarrow P_2, V_2, T$

$P_1 V_I=C_I \rightarrow$ Isothermal process

$W=\int_{V_1}^{V_2} P d V=n R T \ln \frac{V_2}{V_1}$

Expansion $V_2>V_1 \quad$ : work done in positive: System absorbs heat and coverts to work.

Contraction $V_1>V_2$ : work done is negative (done on the system): work is done on the system and it releases heat.

$\delta Q= d u+\delta W$

$\delta u = 0$ Isothermal process

$\delta Q=\delta W$

$P V=n R T, \quad V=\frac{n R T}{P}$

$d V=-\frac{n R T}{p^2} d P$

$W =\int_{V_1}^{V_2} PdV ;$

$=-n R T \int_{P_1}^{P_2} \frac{P}{P^2}$

$=-n R T \ln \frac{P_2}{P_1}$

$P_2 V_2=P_1 V_1 \Rightarrow n R T = C_I$

Isochoric Process: Volume is kept fixed, But temperature and pressure change, Internal energy changes.

$P_1, V, T_1 \Rightarrow P_2, V, T_2$

No work done $\int P d V=0$

Heat absorbed or released = Change in internal energy

Heat absorbed $\rightarrow$ Temperature increases $\rightarrow$ Pressure increases, Internal energy increases.

Heat released $\rightarrow$ Internal energy decreases

$\delta Q = d U$ as $W = 0$.

$P \propto T$ if $V$ is constant.

Pressure is constant: Volume and temperature change

$P, V_1, T_1 \Rightarrow P, V_2, T_2$

Internal energy change: $W=P\left(V_2-V_1\right)=n R\left(T_2-T_1\right)$

Exapansion $V_2>V_1 \Rightarrow T_2>T_1$ : Work done is positive, Change in internal energy is positive. System absorbs heat

$\delta Q=d V+\delta W$

Contraction $V_1>V_2 \Rightarrow T_1>T_2:$ Work done is negative, Change in internal energy is negative. System releases heat

$V \propto T$, $P V=n R T$

Adiabatic Process: No heat exchange: Pressure, temperature and volume change $\delta Q = 0$

$P_1, V_1, T_1 \Rightarrow P_2, V_2, T_2$

$C_P-C_V=R$

$P V^\gamma$= constant

$\gamma=\frac{C_P}{C_V} ; \gamma>1$

Isothermal: $P_I V_I=C_I$

Adiabatic: $P_A V_A^\gamma=C_A$

$P_I=\frac{C_I}{V_I},\left(\frac{\partial P_I}{\partial V_I}\right)=-\left.\frac{C_I}{V_I^2}\right|_{{V_0},{P_0}}$

Isothermal: $P_I V_I=C_I$

$P_I=\frac{C_I}{V_I},\left(\frac{\partial P_I}{\partial V_I}\right)=-\left.\frac{C_I}{V_I^2}\right|_{{V_0},{P_0}}$

$= - \left.\frac{P_I V_I}{V_I^2}\right|_{V_0, P_0}$

$= - \left.\frac{P_I V_I}{V_I^2}\right|_{P_0, V_0} =-\frac{P_0}{V_0}$

Adiabatic: $P_A V_A^\gamma=C_A$

$\left(\frac{\partial P_A}{\partial V_A}\right)_{P_0, V_0}=$

$-\frac{\gamma C_A}{V_A^{\gamma+1}}|_{P_0, V_0}$

$\left(\frac{\partial P_A}{\partial V_A}\right)_{P_0, V_0}=$

$-\frac{\gamma C_A}{V_A^{\gamma+1}}|_{P_0, V_0}-\frac{\gamma P_0}{V_0} \longrightarrow$ at the point of intersection $(P_0,V_0)$

$= - \gamma(\frac{P_0}{V_0})$

$(\frac{P_0}{V_0})\longrightarrow$ slope of the isothermal

Expansion Process: $V_0 \rightarrow V_2$

Contraction process: $V_0 \rightarrow V_1$

Work done in the Isothermal process is more.

Contraction process: $\longrightarrow$ Adiabatic work done is more.

$W=\frac{1}{1-\gamma}\left[P_2 V_2-P_1 V_1\right]=n R \frac{\left(T_1-T_2\right)}{\gamma-1} ; \gamma>1$

$T_1>T_2$ Internal energy decreases: Work done is positive

$T_2>T_1$ Internal energy increases: Work done is negative

$P_1 V_1^\gamma=P_2 V_2^\gamma=C_A \Rightarrow P_1=\frac{C_A}{V_1^\gamma} ; P_2=\frac{C_A}{V_2^\gamma}$

$W=\frac{1}{1-\gamma}\left[P_2 V_2-P_1 V_1\right]=\frac{C_A}{1-\gamma}\left(\frac{1}{V_2^{\gamma-1}}-\frac{1}{V_1^{\gamma-1}}\right)$

Quasi-static and non-dissipative: No friction or viscosity

Forward process $\rightarrow P_1,V_1,T_1 \Rightarrow P_2,V_2,T_2$

Backward process $\rightarrow P_2,V_2,T_2 \Rightarrow P_1,V_1,T_1$

a. Working substance: steam in steam engines: Ideal Gas.

b. The working goes in a closed cycle:multiple thermodynamic processes.

$(P, T, V)\longrightarrow(P, V, T)$

c. Works in a closed loop between two heat reservoirs:

Hot reservoir $T_1$ and cold reservoir $T_2$ , $T_1>T_2$

In a closed loop:

Heat absorbed from the hot reservoir: $Q_1$

Heat released to thecold reservoir: $Q_2$

Work done by the system $W=Q_1-Q_2$

$W=Q_1-Q_2$

Efficiency of an engine = Work done / Heat absorbed in a complete Cycle

$\eta=\frac{Q_1-Q_2}{Q_1}=\frac{W}{Q_1}$

Maximum Value: $Q_2=0$

$\eta=1$

In a complete cycle

Hot reservoir $T_1$ and cold reservoir $T_2$

Heat absorbed from the cold reservoir: $Q_2$

Heat released to the hot reservoir: $Q_1$

Work done on the system: $W$

Coefficient of performance $=$ Heat absorbed $/$ Work done

$W=Q_1-Q_2$

Ideal : $W = 0$

$\Phi=\infty$

$\varphi=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}$