Heat Work and First law of thermodynamics | Metallurgical Thermodynamics

In thermodynamics, work performed by a system is the energy transferred by the system to its surroundings and for a closed system it is equal to area under P-V diagram when projected on the volume axis as shown below. $$\delta W = P \partial V$$ $$ \boxed{ W = \int_{1}^{2} P dV }$$

Important points :
     (1) Condition for calculating work - a) Closed System b) Quasi-static Process.
     (2) Work is a Path function.
     (3) It is an inexact differential i.e., depends on past history.
     (4) It is an boundary and transient phenomenon.
     (5) Sign Convension:
           a) Work Done by system on surroundings is taken as +Ve
           b) Work Done by surroundings on system is taken as -Ve
     (6) All process those are left to right on P-V diagram are work done by the system (+Ve) and vice-versa
     (7) All clockwise cycles on P-V diagram are power developing cycle (+Ve) and vice verse.

Work done By system for non-flow(closed) process

(1) Constant Volume (Isochoric or Isomeric) :
As for such process (Ex. Rigid Containers) \( dV = 0 \) which concludes that \[ \boxed{ W = \int_{1}^{2} P dV = 0 } \]

(2) Constant Pressure (Isobaric or Isopiestic) :
\[ \delta W = P dV \] \[ W = \int_{1}^{2} P dV \] \[ \boxed{ W = P(V_{2} - V_{1}) } \]

(3) Isothermal Process ( \( ΔT = 0 \) ) :
\[ \textrm{As We know } W = \int_{1}^{2} P dV \] \[ \textrm{As } PV=C => W = \int_{1}^{2} \frac{C}{V} dV \] \[ \textrm{On Integration : } \boxed{ W = C \ln{\frac{V_{2}}{V_{1}}} } \] \[ T = C => \frac{V_{2}}{V_{1}} = \frac{P_{2}}{P_{1}} \] \[ => W = C \ln{\frac{P_{1}}{P_{2}}} \] \[\textrm{And } C = P_{1}V_{1} = P_{2}V_{2} = nRT_{1} = nRT_{2} \]

(4) Adiabatic Process ( \( ΔQ = 0 \) ) : \( \textrm{For this process } PV^\gamma = Constant \) \[ \delta W = P dV \] \[ W = \int_{1}^{2} P dV \] \[ W = \int_{1}^{2} \frac{C}{V^\gamma} dV \] \[ \textrm{On Integration : } \boxed{ W = \frac{ P_{1}V_{1} - P_{2}V_{2} }{ \gamma - 1} } \]

As \( PV^ \gamma = C \)
\[ \textrm{ Now, } P \left( \frac{T}{P} \right)^\gamma = C => \frac{T_{2}}{T_{1}} = \left({ \frac{P_{2}}{P_{1}} } \right)^{\frac{\gamma - 1 }{ \gamma }} \textrm{ Also, } \frac{T_{2}}{T_{1}} = \left({ \frac{V_{1}}{V_{2}} } \right)^{\gamma - 1 } \]

(5) Polytropic Process : \( PV^n = C \textrm{ And } -\infty < n < \infty \)

           Generally for natural processes ( Ideal Process ) \( 1 < n < \gamma \) \[ \textrm{Work Done : } \boxed { W = \frac{ P_{1}V_{1} - P_{2}V_{2} }{ n - 1} } \]

As \( PV^n = C \)
\[ \textrm{ Now, } \frac{T_{2}}{T_{1}} = \left({ \frac{P_{2}}{P_{1}} } \right)^{\frac{n - 1 }{ n }} \textrm{ Also, } \frac{T_{2}}{T_{1}} = \left({ \frac{V_{1}}{V_{2}} } \right)^{n - 1 } \]

Gate 2008 Question : One mole of monoatomic idea gas is reversibily and isothermally expended at 100K to twice of its original volume. The work done by the gas is __________ J
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P-V Diagram for various process :

K	Process	Expansion	Compression
0	P=C Isobaric
\( \infty \)	V=C Isochoric
1	T=C Isothermal
\( \gamma \)	Adiabatic
n	Polytropic

Question : A system undergoes three process as shown in Diagram.


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Heat (Q)

Heat is the form of energy generated due to difference in the temperature difference between substances or systems. As a form of energy, heat is conserved, i.e., it cannot be created or destroyed.
Heat is directly proportional to mass(m) i.e., Q ∝ m
Also, Heat generated in the presence of thermal gradient(ΔT) i.e., Q ∝ ΔT
\[ \boxed{ \delta Q = mCdT } \] Where , C = Specific Heat Capacity \( \frac{kJ}{kg-K} \textrm{ or } \frac{kJ}{mol-K} \)

Specific Heat Capacity(C) :
It is the amount of energy of energy required to change the temperature of unit mass of a substance through unit degree temperature difference.
Important 'Specific heat' relation (For ideal gas) :
      (1) \(C_{p} > C_{v} \)
      (2) \(C_{p} - C_{v} = R \) (R can be characteristic or Universal, depends on unit of Cp and Cv)
      (3) \( \frac{C_{p}}{C_{v}} = \gamma \)
          For Monoatomic gas \( (He, Ar) , \gamma \) = 1.67
          For Diatomic gas \((H_{2}, N_{2}) , \gamma \) = 1.4
          For Triatomic gas \( (CO_{2}) , \gamma \) = 1.33
      (4) \(T\uparrow \Rightarrow C_{p}\uparrow , C_{v}\uparrow , \gamma\downarrow \)

First law of thermodynamics (Joules Law)

First law of thermodynamics is an application of conservation of energy principle of heat and thermodynamic processes such as : \[ \boxed{ dU = \delta Q - \delta W } \] But for a closed syatem , undergoing a cycle, the net heat transfer is equal to net work transfer. It is valid for all type of cycles ( Reversible as well as irreversible ), So for cyclic proces \( dU = 0 \) \[ \Rightarrow \sum Q = \sum W \] Internal Energy (U) :
Internal energy states as the energy associated with the random and disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects; In shot notes, it refers to invisible microscopic energy due to atomic and molecular vibration ( Kinetic energy ) and interatomic distance change during heating (Potential energy)

Enthalpy (H) :
Enthalpy is an energy possesed by system due to its properties i.e., sum of the internal energy and the product of pressure and volume of a thermodynamic system. \[ \boxed {H = U +PV} \] On differentiating,
\( dH = dU + PdV + VdP \)

Case 1 : When P is constant
      \( dH = dU + PdV \)
      As from first law of thermodynamics \( \delta Q = dU + PdV ,\)
      \( \Rightarrow dH = dQ \) And also , \( dQ = mC_{p}dT \) \[ dH = mC_{p}dT \]       So, Total Heat interaction \( Q = \int{dH} \) \[ \boxed{ Q = \int{mC_{p}dT} } \] Case 2 : When V is constant
\[ \boxed{ dU = mC_{v}dT } \] Gate 2007 Question :When one mole of Cu is quenched from 1000K to 300K, the amount of heat released is __________ kJ
Given \( C_{p} = 22.68 + 6.3 \times 10^{-3} T \frac{J}{mol-K}\)
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Free Expension ( irreversible isothermal process)

The Image shown above is an special case where in an isolated system ( i.e., \( \delta Q=0 \) ), one side is filled with gas and other kept as vacuum. The wall between them moved infinitely slowly such that work done can be considered zero (i.e., \( \delta W = 0 \))
As from 1st law \( \delta Q = dU + \delta W \Rightarrow dU = 0 \)
\( \Rightarrow U_{i} = U_{f} \)
\( As U = mC_{v}dT \Rightarrow C_{v} T_{i}= C_{v} T_{f} \)
\(\Rightarrow \boxed{ T_{i} = T_{f} } \) The catch is that we did not started as Isothermal Process but still \(T_{i} = T_{f}\). It Means free expansion process are isothermal process.