physics-class-11-unit-12-chapter-02-first-law-work-donein-different-thermodynamic-processes-l-6-9-dvgtdj3uezc

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> Recap </primary>
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- 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

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<footer style = "font-size: 18px"> $\rightarrow$ Thermodynamics $\rightarrow$ <span style = "color:blue"> Recap </span> $\rightarrow$ First Law of Thermodynamics $\rightarrow$ First Law of Thermodynamics </footer>

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> First Law of Thermodynamics </primary>
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- 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.

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<footer style = "font-size: 18px">Thermodynamics $\rightarrow$ Recap $\rightarrow$ <span style = "color:blue"> First Law of Thermodynamics </span> $\rightarrow$ First Law of Thermodynamics $\rightarrow$ Internal Energy </footer>

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> First Law of Thermodynamics </primary>
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- $\Delta Q$ 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: $\Delta Q=0, \Delta U=-\Delta W$
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<footer style = "font-size: 18px">Recap $\rightarrow$ First Law of Thermodynamics $\rightarrow$ <span style = "color:blue"> First Law of Thermodynamics </span> $\rightarrow$ Internal Energy $\rightarrow$ Work done </footer>

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> Internal Energy </primary>
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- 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.

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<footer style = "font-size: 18px">First Law of Thermodynamics $\rightarrow$ First Law of Thermodynamics $\rightarrow$ <span style = "color:blue"> Internal Energy </span> $\rightarrow$ Work done $\rightarrow$ State Function </footer>

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> Work done </primary>
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- 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.
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<footer style = "font-size: 18px">State Function $\rightarrow$ State Function $\rightarrow$ <span style = "color:blue"> Work done </span> $\rightarrow$ Isothermal Process $\rightarrow$ Isobaric Process </footer>

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> Isothermal Process </primary>
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- 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

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<footer style = "font-size: 18px">State Function $\rightarrow$ Work done $\rightarrow$ <span style = "color:blue"> Isothermal Process </span> $\rightarrow$ Isobaric Process $\rightarrow$ Isochoric process </footer>

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> Isobaric Process </primary>
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- 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.

- $d U \neq 0$
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<footer style = "font-size: 18px">Work done $\rightarrow$ Isothermal Process $\rightarrow$ <span style = "color:blue"> Isobaric Process </span> $\rightarrow$ Isochoric process $\rightarrow$ Adiabatic process </footer>

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> Isochoric process </primary>
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- 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.
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<footer style = "font-size: 18px">Isothermal Process $\rightarrow$ Isobaric Process $\rightarrow$ <span style = "color:blue"> Isochoric process </span> $\rightarrow$ Adiabatic process $\rightarrow$ Specific Heat </footer>

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> Specific Heat </primary>
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- 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} $

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

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<footer style = "font-size: 18px">Isochoric process $\rightarrow$ Adiabatic process $\rightarrow$ <span style = "color:blue"> Specific Heat </span> $\rightarrow$ Adiabatic process $\rightarrow$ Adiabatic process </footer>

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### <primary style="font-size:24px"> Adiabatic process </primary>
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- 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 $

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<footer style = "font-size: 18px">Specific Heat $\rightarrow$ Adiabatic process $\rightarrow$ <span style = "color:blue"> Adiabatic process </span> $\rightarrow$ Work Done in an Adiabatic Process $\rightarrow$ Work Done in an Adiabatic Process </footer>

### <Success style="font-size:25px"> First Law Work Donein Different Thermodynamic Processes L-6 </Success>
### <primary style="font-size:24px"> Work Done in an Adiabatic Process </primary>
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- One mole of ideal gas

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

- $\Delta U=C_V \Delta T$

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

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

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

- $\gamma=(C_P / C_V)$

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<footer style = "font-size: 18px">Adiabatic process $\rightarrow$ Adiabatic process $\rightarrow$ <span style = "color:blue"> Work Done in an Adiabatic Process </span> $\rightarrow$ Work Done in an Adiabatic Process $\rightarrow$ Work Done in an Adiabatic Process </footer>

- 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 $

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

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

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<footer style = "font-size: 18px">Adiabatic process $\rightarrow$ Work Done in an Adiabatic Process $\rightarrow$ <span style = "color:blue"> Work Done in an Adiabatic Process </span> $\rightarrow$ Work Done in an Adiabatic Process $\rightarrow$ Work Done in an Adiabatic Process </footer>

- $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

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<footer style = "font-size: 18px">Work Done in an Adiabatic Process $\rightarrow$ Work Done in an Adiabatic Process $\rightarrow$ <span style = "color:blue"> Work Done in an Adiabatic Process </span> $\rightarrow$ Work Done in an Adiabatic Process $\rightarrow$ Thank You </footer>

- 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 $, $C_p>Q_V$

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

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