Thermodynamic Variables: Extensive and Intensive

N,V, U

System separated from the Universe by Walls

Universe (Reservoir) Very big

Walls: Diathermic and Adiabatic

Mechanical and thermal interaction

Equilibrium

Quasi-Static Processes

First law of thermodynamics: Conservation of Energy.

Exchange of heat:

internal energy must increase.

Mechanical Energy:

mechanical work conservative.

$\Delta W, \Delta U$

Brings in the notion of the internal energy.

$\Delta Q$ Heat supplied to the system

$\Delta W$ Work done by the system

$\Delta U$ increase in internal energy

$\Delta Q = \Delta U + \Delta W$

Convention:

$\Delta Q$ Positive: Heat supplied to the system

$\Delta W$ Positive: Work done by the system

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

Case 1: $\Delta W=0, \Delta Q=\Delta U$

Case 2: $\Delta Q=0, \Delta U=-\Delta W$

For Ideal Gas: $U=C_VT$ + Constant

Extensive: depends on the path, $\delta Q, \delta W$.

State function: depends on the initial and final states, $d U$.

$\left(P_i V_i, T_i\right)\longrightarrow (P_f V_f, T_f)$

$T_i\longrightarrow T_f$

$\Delta U= C_V(T_f-T_i)$

Recall the notion of potential in mechanics for a conservative field.

Adiabatic Process: Work done is a state function.

Recall: Conservative Force Field.

Work done is independent of path.

Work done in a closed loop is zero.

$\delta Q, \delta W$ Depends on the path (thermodynamical process) connecting two states

$d U$ Depends on the initial and final states.

$\delta Q = 0$

Conservative $\rightarrow$ potential $\Rightarrow$ difference in potential

Ideal Gas: dU depends on initial and final temperature

Work done by the gas.

Assume Quasistatic Process.

Work done: $P A d x=P d V$

Net Work done:

$W=\int_{V_1} ^{V_2} P d V$

Area under the curve represent work done for different thermodynamic process.

Consider $n$ moles of ideal Gas:

$P V=n R T \longrightarrow P=\frac{n RT}{V}$

Work done by Isothermal Process:

$W=\int_{V_1}^{V_2} P d V=nRT\int_{V_1}^{V_2} \frac{dV}{V}=nRT\ln\frac{V_2}{V_1}$

No change in internal energy as $T$ fixed, $dU=0$

Heat supplied$\equiv$ work done by the system

Pressure is constant.

Work done by Isobaric Process:

• $W=\int_{V_1}^{V_2} P d V=P\left(V_2-V_1\right)=n R\left(T_2-T_1\right)$

$P$ constant and $V$ and $T$ change

$P V_1=n R T_1, \quad P V_2=n R T_2$

Change in internal energy: Heat absorbed goes into work done as well changing Internal energy.

$d U \neq 0$

Volume is constant, $P$ and $T$ change.

No work $\int PdV=0$.

Recall first law: The heat supplied to gas is converted to Internal energy.

This is why internal energy is essentially to conserve the total energy

Heat absorbed $\rightarrow$ increases the temperature and hence the internal energy.

These three examples also establish that work done by the gas depends on the thermodynamic processes.

$\Delta Q = \Delta U + \Delta W, \Delta W = P\Delta V$

$\Delta Q = 0 \Rightarrow \Delta U = -\Delta W$

W is path-independent.

Specific heat capacities: Ideal Gas

$C_V$ and $C_P$

Isochoric Process: Volume is constant

$C_V = (\frac{\Delta Q}{\Delta T})_V = (\frac{\Delta U}{\Delta T})_V$

Isobaric Process: pressure is constant.

$C_P = (\frac{\Delta Q}{\Delta T})_P = (\frac{\Delta U}{\Delta T})_P + P(\frac{\Delta V}{\Delta T})_P$

$\Delta Q=\Delta U+\Delta W$

$P\left(\frac{\Delta V}{\Delta T}\right)_P \Rightarrow P\left(\frac{d V}{d T}\right)_P ={\Delta V}+{P \Delta V}$

${\Delta T \rightarrow 0}\left(\frac{\Delta Q}{\Delta T}\right)_P=\left(\frac{\Delta U}{\Delta T}\right)_P+P\left(\frac{\Delta V}{\Delta T}\right)_P$

Calculate $\Delta W$ in an adiabatic process

$T$ is not constant.

We can not use $P V=$ constant

Some other relation: $P V^\gamma=$ constant

For an ideal Gas, $U$ is a function of temperature only

$\left(\frac{\Delta U}{\Delta T}\right)_V=\left(\frac{\Delta U}{\Delta T}\right)_P\rightarrow \left(\frac{d U}{d T}\right)_V =\left(\frac{d U}{d T}\right)_P$

$C_P-C_V=P\left(\frac{d V}{d T}\right)_P$

$P V=R T$

$C_P=C_V+R$

For one mole of ideal Gas: $C_P-C_V=R$

One mole of ideal gas

$\Delta Q=0 \Rightarrow \Delta U=-\Delta W=-P \Delta V$

$\Delta U=C_V \Delta T$

$P = \frac{RT}{V}$

$C_V \Delta T=-\Delta{P \Delta V} \Rightarrow C_V d T=-\frac{R T \Delta V}{V} \Rightarrow-\frac{\left(C_P-C_V\right){T \Delta V}}{V}$

$\frac{\Delta T}{T}=(1-\gamma) \frac{\Delta V}{V} \Rightarrow \frac{d T}{T}=(1-\gamma) \frac{d V}{V}$

$\gamma=(C_P / C_V)$

Integrating:

$\ln T=(1-\gamma) \ln V+\text { constant } \Rightarrow T \propto V^{(1-\gamma)}$

$C_V d T =-\frac{C_P-C_V}{V}T dV$

$\frac{d T}{T} =(1-\gamma) \frac{d V}{V}$

$\frac{d T}{T} \sim \ln T$

$P V=R T$

$T \propto V^{1-\gamma}T$ = Constant $V^{1-\gamma}$

$P V^\gamma= constant$

One mole of ideal Gas

$P_I V_I=C_I$

$P_A V_A^\gamma=C_A$

Isothermal process: $P V$ = constant $\longrightarrow C_I$

Adiabatic Process: $P V^\gamma$ = constant $\longrightarrow C_A$

$P V^\gamma$ = constant $\longrightarrow$ Quasi-static

Work done: $P V^\gamma$= constant

$W=\int_{V_1}^{V_2} P d V=\frac{R(T_1-T_2)}{\gamma-1}$

$P_1 V_1^\gamma=P_2 V_2^\gamma$

No heat absorbed.

W>0 work done by the Gas and W<0 work done on the gas.

$W= \int_{V_1}^{V_2} P d V=C_A \int_{V_1}^{V_2} \frac{d_V}{V_\gamma}$

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

$\gamma>1$, $C_p>Q_V$

$W =\left[\frac{R\left(T_1-T_2\right)}{\gamma-1}\right]$