Thermodynamics

 Thermodynamics system

$\text{Thermodynamics system:}$$\to$It is an assembly of an extremely large number of particles (atoms or molecules)$\\$having a certain pressure, volume and temperature.$\\$$\to$It is of two types:$\\$1)Isolated System: (no heat can enter or leave out the system)$\\$2)Closed system: (There can be heat exchange between the system and surrounding)$\\$

$\textbf{Thermal Equilibrium:}\\$A thermodynamics system is in thermal equilibrium if the temperature of all parts of it is same.

Zeroth law of Thermodynamics

$\text{Law:}$When two bodies A and B are in thermal equilibrium with third body C then the body A and B$\\$also must be in thermal equilibrium with each other.

Work Done

Points:$\\$

$\to$Work is said done when volume of gas changes.$\\$

$\to$Work done is positive if expansion takes place and is negative if compression takes place.$\\$

Work (W) = PdV$\\$

$\to$Area under PV curve between volume axis is equal to work done.$\\$

$\to$For a closed cycle, area of closed loop gives work done.$\\$

$\to$When P remains constant throughout the expansion, the work done by the gas is$\\$

W=P(V2-V1)$\\$

$\textbf{Internal Energy of a Gas}\\$The sum of energy due to molecular motion (KE) and due to molecular configuration (PE)$\\$is called internal energy of gas.$\\$:. Internal energy (U) = PE + KE$\\$For ideal gas intermolecular force of attraction is neglected so PE=0,$\\$so internal energy of ideal gas is KE which is only the function of temperature.

First law of thermodynamics

$\text{Intro:}$When heat energy is given to a system then some part of heat energy supplied is used to$\\$change the internal energy of system and rest of energy is used to do external work.$\\$∆Q=∆U+∆W$\\$

$\textbf{Note:}\\$For cyclic process, the change in internal energy of the system is zero because the system is$\\$brought back to the initial condition. Therefore, dU=0 and from the first law of thermodynamics,$\\$dQ= du + PdV =0+dW= dW$\\$

$\textbf{Molar Heat Capacities:}\\$1)Molar heat capacity at constant pressure ($C_p$)$\\$$\to$Heat required to rise the temperature of one mole of gas through 1 degree C at constant pressure.$\\$Its unit is J/(molK)$\\$$\to$Heat required (dQ)=n$C_p$dT$\\$2)Molar heat capacity at constant volume($C_v$):$\\$$\to$Heat required to rise the temperature of mole of gas through 1 degree C at constant volume.$\\$Its unit is J/(molK)$\\$$\to$Heat required (dU) = n$C_v$dT$\\$

$\textbf{Mayer`s Formula:}\\$$\to$ $C_p-C_v$=R

Specific heat capacities

$\text{Specific heat capacity at constant pressure :}$($c_p$)$\\$$\to$Heat required to rise the temperature of unit mass of gas through 1 degree C temperature$\\$at constant pressure. $\to$Heat required (dQ)=m$c_p$dT$\\$$\to C_p$=M$c_p \\$

$\textbf{Specific heat capacity at constant volume :}$($c_v$)$\\$$\to$Heat required to rise the temperature of unit mass gas through 1 degree C temperature$\\$at constant volume. $\to$Its unit is J/(kg K)$\\$$\to$Heat required (du)=m$c_v$dT$\\$$\to C_v$=M$c_v \\$

$\textbf{Note:}\\$$\to$Heat required to rise certain temperature at constant pressure is always greater than heat$\\$required to rise same temperature at constant volume. So gas has two types of heat capacities$\\$$\to$ i.e. $C_p$>$C_v \\$$\to$Because in constant pressure, internal energy and work done both is done.

Thermodynamical process

$\text{Isochoric Process:}$$\to$Volume remains constant$\\$$\to$Work done (dw)=0$\\$$\to$ Heat supplied = change in internal energy: dQ=dU$\\$$\to$nCvdT=dU$\\$

$\textbf{ Isobaric Process:}\\$$\to$Pressure remains constant$\\$$\to$dQ=CvdT+ PdV$\\$

$\textbf{Isothermal Process:}\\$$\to$Temperature remains constant. i.e. dT=0$\\$$\to$For this process cylinder with conducting wall is used and ideal gas filled inside is allowed to$\\$expand or is compressed very slowly.$\\$$\to$Eg: Melting process and boiling process$\\$$\to$Specific heat capacity during isothermal process is infinity$\\$$\to$Change in internal energy(du)= 0$\\$$\to$Gas Equation: $P_1 V_1$=$P_2 V_2 \\$$\to$Slope of curve (dP/dV)=-P/V$\\$$\to$Work Done (w) =nRT ln($V_2$/$V_1$)$\\$=$P_1 V_1$ ln($V_2$/$V_1$)=$P_1 V_1$ ln($P_1$/$P_2$)$\\$

$\textbf{Adiabatic Process: }\\$$\to$The process in which exchange of heat energy is zero i.e. dQ=0$\\$$\to$So, dW=-dU i.e. work is done by gas on the expense of internal energy so cooling is observed after$\\$adiabatic expansion$\\$$\to$Fast process in which wall of cylinder is perfectly insulator$\\$$\to$Specific Heat capacity of gas is 0.$\\$$\to$Eg: Propagation of sound wave, sudden bursting of tire, the compression stroke in an internal$\\$combustion engine.$\\$$\to$Slope of curve (dP/dV)=-γP/V$\\$$\to$Gas equation is$\\$$P_1 V_1^γ$=$P_2 V_2^γ \\$$T_1 V_1^{γ-1}$=$T_2 V_2^{γ-1} \\$$T_1^γ$/$P_1^{γ-1}$ =$T_2^γ$/$P_2^{γ-1}$