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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

b) Work Done by surroundings on system is taken as

(6) All process those are left to right on P-V diagram are work done by the system

(7) All clockwise cycles on P-V diagram are power developing cycle

(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 \( PV^n = C \)

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

Show Answer

Answer : W = 5762.82J

As for Isothermal process \( W = C \ln{\frac{V_{2}}{V_{1}}} \)

As given \( V_{f} = 2V_{i} \) , Mole n = 1, T = 1000K And R=8.324 J/mol-K

=> \( W = nRT \ln \frac{V_{2}}{V_{1}} \)

=> \( W = 1 \times 8.314 \times 1000 \ln \frac{V_{2}}{V_{1}} = 1 \times 8.314 \times 1000 \ln{2} \)

=> \( W = 5762.82J \)

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 |

1 - 2 : Isobaric Process (P = C)

2 - 3 : Polytropic Process (n = 1.4)

3 - 1 : Isothermal Process (PV = C)

Find 1) \( V_{2}\) = __________ \( m^3 \)

2) Wnet = ____________KJ

Show Answer

As 3 - 1 Isothermal Process i.e., PV = C => \( P_{1}V_{1} = P_{3}V_{3} \)

\( => 400 \times 1 = 100 \times V_{3}\)

\( V_{3} = 4 m^3 \)

For 2 - 3 Polytropic Process i.e., \( PV^n = C => P_{3}V_{3}^n = P_{2}V_{2}^n \)

\( => 100 \times 4^1.4 = 400 \times V_{2}^n \)

\( V_{2} = 1.43 m^3 \)

Wnet = \( W_{1-2} + W_{2-3} + W_{3-1}\)

Wnet = \( P_{2}(V_{3} - V_{1}) + \frac{ P_{2}V_{2} - P_{3}V_{3} }{ n - 1} + C \ln{\frac{V_{3}}{V_{1}}} \)

Wnet = \( 192.00 + 480 - P_{3}V_{3} \ln{\frac{V_{3}}{V_{1}}} \)

Wnet = 125.88KJ

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 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,

\( dH = dU + PdV + VdP \)

\( dH = dU + PdV \)

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 \( Q = \int{dH} \) \[ \boxed{ Q = \int{mC_{p}dT} } \]

\[ \boxed{ dU = mC_{v}dT } \]

Given \( C_{p} = 22.68 + 6.3 \times 10^{-3} T \frac{J}{mol-K}\)

Show Answer

As we know, \( Q = \int{mC_{p}dT} \)

\( Q = \int_{1000}^{300} ( 22.68 + 6.3 \times 10^{-3} T) dT \)

\( Q = -18.74 kJ \)

But as asked Heat released \( \Rightarrow Ans = 18.74 kJ \)

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.