Zeroth & First Law of Thermodynamics | Kinetic Theory & Thermodynamics - Physics PDF Download

Zeroth Law of Thermodynamics

If two systems say, A and B are separately in thermal equilibrium with a third system say C, then they must also be in thermal equilibrium with one another.

It is analogous to the transitive property in math (if A = C and B = C , then A = B ).

Another way of stating the zeroth law is that every object has a certain temperature, and

when two objects are in thermal equilibrium, their temperatures are equal.

First Law of Thermodynamics

If ∂Q amount of heat is added to the system and if system will do ∂W amount of work, then change in internal energy is dU , which is given by dU = ∂Q - ∂W . First Law of thermodynamics is law of conservation of energy in which Heat added to system can be divided into mechanical work done by system and increase in internal energy of system .

(obviously, heat exchange and work done is dependent on path and internal energy is state function.)

One can also represent first law as ∂Q = dU + ∂W

Different Types of Thermodynamical Process and use of First Law of Thermodynamics.

Isochoric Process

When volume remains constant during the process the process is said to be isochoric process.

One can also show Isochoric process in T - V Diagram

For Ideal Gas Isochoric Process can be shown in P -T diagram which is actually straight line appear to pass through origin which is shown in figure

Work done in isochoric process

dW = PdV

Since, dV = 0 ⇒ dW = 0

In an isochoric process work done during the process is zero.

From first law of thermodynamics,

dU = ∂Q - ∂W

Since, ∂W = 0 ⇒ dU = ∂Q = nC_vdT

In isochoric process, change in internal energy is equal to heat exchange, So in isochoric process the Heat exchanged is perfect (Exact) differential, So in isochoric process Heat exchanged is Path independent.

Isobaric Process

When pressure of the system remain constant during the process, the process is set to be isobaric process.

The Isobaric process shown in P - V diagram

One can also show Isobaric process in T - P Diagram

For Ideal Gas Isochoric Process can be shown in P -T diagram which is actually straight line appear to pass through origin which is shown in figure

Work done during the process is given by

where, ΔT is change in temperature during the process.

∂Q = nC_vΔT [since, C_v + R = C_P for the ideal gas]

Heat exchange during the process for n mole of gas is equal to nC_vΔT , where C_P is specific heat capacity at constant pressure.

∂Q = nC_vΔT where C_P =

∂Q =  ⇒  ⇒∂Q =

Isothermal Process

When temperature remains constant during the process, then the process is said to be isothermal process.

Isothermal process in P - V diagram shown in figure

Isothermal process in T - V Diagram shown in figure

Isothermal process in V - T Diagram shown in figure

In isothermal process, change in internal energy is zero because ΔT = 0

during the isothermal process, heat exchange

Adiabatic Process

When there is not any heat exchange during the process, then process is said to be adiabatic process.

Adiabatic process is defined by,

PV^γ = constant, where γ = C_p/C_v

For Ideal gas, PV = RT for one mole of gas.

Slope of Adiabatic process in P - V diagram

PV^γ = C ⇒  = 0

 < 0 where γ, P,V > 0

Slope of Adiabatic process in T-V diagram

So, TV^γ-1 = constant

TV^γ-1 = C

 = 0

, where γ >1, T > 0, V > 0

Slope of adiabatic process in T-P diagram

P^1-γT^γ = constant, where γ = C_p/C_v

Slope of adiabatic process on P -T diagram is

Work done during adiabatic process

∂W = PdV

For n mole Ideal gas

W =

For Adiabatic process ∂Q = dU + ∂W

Since, ∂Q = 0 ⇒dU = - ∂W

For adiabatic process ⇒ ∂W = -dU so work done is perfect (exact) differential .in adiabatic process is work done is path independent.

So, in adiabatic process change in internal energy is equal to negative of work done i.e.

dU =

Example 1: n mole of certain ideal gas at temperature T₀ were cooled isochorically, so that the gas pressure reduced n times. Then as a result at the isobaric process the gas expanded till its temperature get back to initial value. Find the total amount of heat absorbed by the gas in the process.

Let at state A , the pressure, volume, temperature is P₀,V₀,T₀,n is number of  mole. 

According to question,

In process A to B

In process B to C

Total heat absorbed by gas,

=  =

Example 2: A sample of an ideal gas is taken through the cyclic process “abcd ” as shown in figure. It absorbed 50Jof heat during part ab, no heat during bc and reject 70J of heat during ca, 40J of work is done on the gas during part bc .

(a) Find the internal energy of the gas at b and c if it is 1500 J at a .

(b) Calculate the work done by the gas during the part ca.

(a) During process a→b , V = constant

ΔQ = dU = U_{b -}U_{a = 50 J}

_{If at ‘ a ’, Ua = 1500J, so at ‘ b’, Ub= 1550J}

_{Work done during process, b→c = -40 J}

_{ΔU = Q - W = 0 - (-40) = 40J}

U_c - U_b = 40

U_c = 1550 + 40 = 1590

U_b = 1550J ⇒U_c = 1590(no heat absorbed at bc)

(b) During process, c_→a

_{ΔU = 1500 - 1590 = - 90J}

_{ΔQ = -70J}

_{ΔW = ΔQ - ΔU = -70 + 90 = 20J}

Example 3: A sample of an Ideal gas has pressure P₀, volume V₀ and temperature T₀ . It is isothermally expanded to twice its original volume. It is then compressed at constant pressure to have the original volume V₀ . Finally, the gas is heated at constant volume to get the original temperature.

(a) Show the process in VT diagram.

(b) Calculate the heat absorbed in the process.

(a) 

(b) Hence the process is cyclic then change in internal energy in cycle is zero. So heat supplied in the cycle is equal to work done.

Work done during isothermal process, A→B is,

Since, process b→c is isobaric, so work done during the process is PdV .

P_aV_a = P_bV_b

P₀V₀ = P_b2V₀ ⇒ P_b = P₀/2

W_b→c = P₀/2(V₀ - 2V₀) = - P₀V₀/2

Since, W_c→a is isochoric, so W_c→a = 0

Thus, total work done is given by,

W =

Since, the process is cyclic i.e.,

dU = 0 ⇒ dQ = dW =

Example 4: One mole of an ideal monatomic  with gas is taken round the cyclic process ABCA as shown in figure. Calculate

(a) The work done by the gas.

(b) The heat rejected by the gas in the path CA and  the heat

absorbed by the gas in the path AB ;

(c) The net heat absorbed by the gas in the path BC ;

(d) The maximum temperature attained by the gas during the cycle.

Number of mole of ideal monatomic gas = n =1 mole

For monatomic gas: , So γ = C_p/C_v = 5/3

ABCA is cyclic process, so change in internal energy is zero during the cycle

A→B : An isochoric process at constant volume V₀

(a) Work done by the gas: dW , which is equivalent to Area of triangle ABC

dW =

(b) The process C → A: is an isobaric compression at constant pressure P₀

Heat rejected alone 

So magnitude of heat rejected in process C to A is 5/2P₀V₀ negative sign justify the heat rejection during the process

The process A→B is an isochoric process at constant volume V₀

Heat rejected alone

So magnitude of heat rejected in process A to B is 3P₀V₀ positive sign justify the heat

absorbed during the process

(c) In a cyclic process, dU = 0 , Net heat absorbed along BC :

dQ = dW =  = dW ⇒  = P₀V₀

 the magnitude of heat exchanged is P₀V₀/2 negative sign justify the rejection of heat in the process.

(d) The maximum temperature of gas: T_max : 

BC is a straight line. The maximum temperature of the gas T_max will lie on BC at some

point.

For a straight line y = mx + C

For BC, slope m = , P = mV + C ⇒ P =  will satisfy any of given coordinate on line, so 3P₀ = ⇒ C = 5P₀

 From equation of state P =  and n =1

For T to be maximum, dT/dV = 0, d²T/dV² = -ve.

Hence,

T_max =  ⇒ T_max =

⇒ T_max =

Example 5: The rectangular box shown in figure has a partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of a mono-atomic ideal gas (γ = 5/3) at a pressure P₀ volume V₀ and temperature T₀. The chamber on the left is slowly heated by an electric heater. The walls of the box and the partition are thermally insulted. Heat loss through the lead wires of the heater is negligible. The gas in the left chamber expands pushing the partition until the final pressure in both chambers becomes 243P₀/32 .

Determine:

(i) the final temperature of the gas in each chamber and

(ii) The work done by the gas in the right chamber.

(i) For left chamber, the process of heating is slow. Initially, the parameter is P_0,V₀,T₀ finally, the parameters are

or, T₁ =                                      (i)

Adiabatic compression occurs in the right chamber. The initial quantities are P₀,V₀, γ,T₀ The final qualities are

or V₂ =                                           (ii)

But V₁ + V₂ = 2V₀ given V₁ + = 2V₀

or V₁ =

From (i) and (iii),

T₁ =                                 (iii)

or T₁ =                              (iv)

Again for the right chamber, use P - T equation for an adiabatic operation.

T^γ ∞ P^γ-1

so,  ⇒

or

or  or

or T₂ = 9/4 T₀ or T₂ = 2.25T₀

(ii) Work done by the gas in right chamber.

Adiabatic work done =

W =

or W =

or W =  =

or W = -15.58T₀ Joule

Work done is negative as the volume in the right chamber decreases.

Example 6: Find the molar heat capacity of an ideal gas in a polytrophic process

pVⁿ = const if the adiabatic exponent of the gas is equal to γ At what values of the

polytrophic constant n will the heat capacity of the gas be negative?

We know from first law of thermodynamics, dQ = dU + dW

Let us assume C is specific heat related for process so, dQ = μCdT

C_v is specific Heat capacity at constant volume internal energy is path independent so  dQ = μCdT which will same for every process for Ideal gas

So first law of thermodynamics can be written as μCdT = μC_vdT + PdV

The equation of process is given by PVⁿ = constant

Differentiate equation

Also, the equation of state for Ideal gas is given by PV = μRT

PdV + VdP = μRdT

PdV + nPdV = μRdT ⇒ PdV =

 ⇒

C =

If C < 0 then

Example: An ideal gas has an adiabatic exponent γ . In some process its molar heat capacity varies as C =α /T , where α is a constant.

(a) What is the work performed by one mole of gas during its heating from the temperature T₀ to the temperature η times higher.

(b) Find the that the equation of the process in the variables T,V .

(c) Prove that the equation of the process in the variables P,V is given by PV^γ .exp = constant

(a) C = α/T

The heat exchanged during process is where n =1 is number of mole 

From first law of thermodynamics dQ = dU + dW ⇒ dQ = _nC_vdT + PdV

dW =  ⇒ W = ⇒ W =

(b) From equation of state P = RT/V

From first law of thermodynamics, dQ =

 where β is constant

 = constant

(c)  = constant

 = constant

 = constant

 = constant