Class 12 Physics First law of thermodynamics Notes

Class 12 Physics First law of thermodynamics Notes
UNIT 2
CLASS 12 PHYSICS • HEAT AND THERMODYNAMICS

First Law of Thermodynamics

Chapter 4
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Thermodynamic System

A certain region in the space used to study the thermodynamic process is called thermodynamic system.

There are three types of thermodynamic system:

a) Open System

The system which exchanges heat or matter is called open system.

b) Closed System

The system which exchanges heat only is called closed system.

c) Isolated System

The system which exchanges neither heat nor matter is called isolated system.

Thermodynamic Process

1) Isothermal Process

The process in which temperature remains constant is called isothermal process.
We have, PV = nRT
For isothermal process: PV = constant

2) Isochoric Process

The process in which volume remains constant is called isochoric process.

3) Isobaric Process

The process in which pressure remains constant is called isobaric process.

4) Adiabatic Process

The process in which pressure, volume and temperature change but amount of heat remains constant is called adiabatic process.

Work Done by Expansion

Gas in a cylinder with a movable piston moving outward by dx P, V, T Piston F dx A
Fig: Work done by expansion

Let us consider a gas kept inside a cylinder provided with movable and frictionless piston at pressure ‘P’, volume ‘V’ and temperature ‘T’. Suppose the gas expands by small volume ‘dV’ and piston moves forward by distance ‘dx’. If A is the cross-sectional area of cylinder then force on piston:

F = P.A
[∵ P = F/A]

Also, change in volume:

dV = A dx

Now, small work done:

dW = F dx
= PA dx
dW = P dV
which is external workdone.

P-V Diagram

The variation of pressure and volume in a thermodynamic process represented by diagram is called P-V diagram.
Pressure-volume diagram with shaded area under the process representing work done P V K (P₁,V₁) L (P₂,V₂) V₁ V₂ Area = workdone
P-V diagram: area under the process gives the work done
Area = P dV
= workdone

Therefore, the area of P-V diagram gives the workdone during the thermodynamic process.

Internal Energy

In case of real gases, the molecules are attached to each other by intermolecular force and molecules are in motion. So, real gas molecules exhibit both potential energy and kinetic energy. Therefore, the sum of kinetic energy and potential energy is called internal energy.

Internal energy = Kinetic energy + Potential energy

In case of ideal gas, there is no P.E. So, in case of ideal gas internal energy is K.E. only and which depends upon absolute temperature.

For ideal gas:
Internal energy = Kinetic energy

First Law of Thermodynamics

It states that if certain amount of heat is supplied to the given system then a part of heat may be used to increase the internal energy and remaining part is used for workdone.

If dQ is the amount of heat supplied to the system and dU, dW are the increase in internal energy and external workdone respectively, then according to first law of thermodynamics:

dQ = dU + dW   — (i)

Since,

dQ = dU + P dV

If dV = 0, then dW = 0

∴ dQ = dU
The scanned note then says: “It means total amount of heat supplied is used to increase external workdone only when volume is constant.” This sentence is preserved as a source issue because the equation immediately above gives dQ = dU when dV = 0.

Heat Capacity of Gas

a) Molar Heat at Constant Volume (Cv)

It is defined as the amount of heat required to rise the temperature of 1 mole of gas through 1 kelvin at constant volume. It is denoted by Cv.
Cv = dQ / (n dT)

b) Molar Heat at Constant Pressure (Cp)

It is defined as the amount of heat required to rise the temperature of 1 mole of gas through 1 kelvin at constant pressure. It is denoted by Cp.
Cp = dQ / (n dT)

Mayer’s Formula (Cp − Cv = R)

Gas in a cylinder with movable piston used in Mayer’s formula derivation P, V, T Piston
Fig: Gas in cylinder

Let us consider ‘n’ mole of gas kept in a cylinder provided with movable and frictionless piston at pressure ‘P’, volume ‘V’ and temperature ‘T’. Suppose the gas is heated at constant volume initially and dQ be the amount of heat given to the system to increase its temperature by dT, then:

dQ = nCvdT   — (i)

From first law of thermodynamics:

dQ = dU + P dV

Since at constant volume, dV = 0:

dQ = dU
dU = nCvdT   — (ii)

Again, the gas is heated at constant pressure and dQ be the amount of heat given to the system to increase the temperature by dT:

dQ = nCpdT   — (iii)

Again from first law of thermodynamics:

dQ = dU + P dV
nCpdT = nCvdT + P dV   — (iv)

From ideal gas equation:

PV = nRT

Differentiating both sides with respect to T at constant pressure:

P dV = nR dT   — (v)

Using eqn (v) in eqn (iv):

nCpdT = nCvdT + nR dT
Cp = Cv + R
Cp − Cv = R

which is Mayer’s formula.

Equation of Adiabatic Process

From first law of thermodynamics:

dQ = dU + P dV   — (i)

For adiabatic process:

dQ = 0   [∵ heat constant]
dU + P dV = 0   — (ii)

If Cv is the molar heat capacity at constant volume for one mole of gas:

dU = CvdT   — (iii)

Using equation (iii) in equation (ii):

CvdT + P dV = 0   — (iv)

Also ideal gas equation for 1 mole of gas:

PV = RT

Differentiating both sides with respect to T:

P dV + V dP = R dT
dT = (P dV + V dP)/R   — (v)

Using equation (v) in equation (iv):

Cv(P dV + V dP)/R + P dV = 0
Cv(P dV + V dP) + RP dV = 0
(Cv + R)P dV + CvV dP = 0
[∵ Cp − Cv = R]
CpP dV + CvV dP = 0

Dividing both sides by CvPV:

(Cp/Cv) dV/V + dP/P = 0
γ dV/V + dP/P = 0   [∵ γ = Cp/Cv]

Integrating on both sides:

γ ∫dV/V + ∫dP/P = constant
γ lnV + lnP = constant
ln(PVγ) = constant
PVγ = constant   — (vi)

In Terms of Pressure & Volume

P1V1γ = P2V2γ

In Terms of Temperature & Volume

We have, PV = RT
P = RT/V

Then eqn (vi) becomes:

(RT/V)Vγ = constant
TVγ−1 = constant   — (vii)
T1V1γ−1 = T2V2γ−1

In Terms of Pressure & Temperature

We have, PV = RT
V = RT/P

Then equation (vi) becomes:

P(RT/P)γ = constant
P1−γTγ = constant   — (viii)
P11−γT1γ = P21−γT2γ

Work Done by Isothermal Process

Initial and final states of one mole of gas during isothermal process P₁, V₁, T P₂, V₂, T Isothermal
Initial and final states at constant temperature

Let us consider that 1 mole of gas is kept in a cylinder provided with movable and frictionless piston. Also let P1, V1 and T be the initial pressure, volume and temperature respectively. By keeping temperature constant volume can be changed to V2, so that pressure also changed to P2. So that small workdone:

dW = P dV   — (i)

Thus total workdone:

W = ∫V₁V₂ dW
W = ∫V₁V₂ P dV   — (ii)

For 1 mole of gas, ideal gas equation:

PV = RT
P = RT/V   — (iii)

Using eqn (iii) in eqn (ii):

W = ∫V₁V₂ (RT/V)dV
W = RT ∫V₁V₂ dV/V
W = RT[lnV]V₁V₂
W = RT(lnV2 − lnV1)
W = RT ln(V2/V1)   — (iv)

Also from isothermal process:

P1V1 = P2V2
V2/V1 = P1/P2
W = RT ln(P1/P2)   — (v)

Eqn (iv) and (v) are required expressions for isothermal process when gas is 1 mole.

For n moles:
W = nRT ln(V2/V1) = nRT ln(P1/P2)

Work Done by Adiabatic Process

Let us consider 1 mole of gas is kept in a cylinder provided with movable and frictionless piston. Also consider the gas expands adiabatically from initial volume V1 to final volume V2. The workdone is given by:

W = ∫V₁V₂ P dV   — (i)

For adiabatic process:

PVγ = k = constant
P = k/Vγ   — (ii)

Using equation (ii) in (i):

W = ∫V₁V₂ (k/Vγ) dV
= k∫V₁V₂V−γdV
= k[V1−γ/(1−γ)]V₁V₂
= k/(γ−1)[V11−γ − V21−γ]
= 1/(γ−1)[P1V1 − P2V2]   — (iii)

For one mole of gas:

PV = RT
W = 1/(γ−1)(RT1 − RT2)
W = R/(γ−1)(T1 − T2)   — (iv)

Equation (iii) and (iv) are required expressions for one mole of gas during adiabatic process.

For n moles:
W = nR/(γ−1)(T1 − T2)

Reversible and Irreversible Process

Reversible Process

The process which returns to initial state from final state exactly along the same path is called reversible process. The net workdone during reversible process is 0.
P-V diagram for reversible process using the same path in both directions (P₁,V₁) (P₂,V₂) P V
Fig: P-V diagram for reversible process

Irreversible Process

The process which does not return to initial state from final state exactly along the same path is called irreversible process. The net workdone during irreversible process is not zero.
P-V diagram for irreversible process using different forward and return paths (P₁,V₁) (P₂,V₂) P V
Fig: P-V diagram for irreversible process

Solved Numericals

Q.1 — Adiabatic compression to one-eighth volume

A gas in a cylinder is initially at temperature of 17°C & pressure 1.01×105 N/m2. If it is compressed adiabatically to one-eighth of its original volume, what would be the final temperature and pressure of gas? [γ = 1.4]

The handwritten question gives P1 = 1.01×105 N/m², while the pressure calculation line later uses 1.05×105. The working below preserves the values written in the scan.
Initial temperature (T1) = 17°C = 290 K
Initial pressure (P1) = 1.01×105 N/m2
Original volume (V1) = V
Final volume (V2) = V/8

Now,

P1V1γ = P2V2γ
1.01×105 × Vγ = P2(V/8)γ
1.01×105 = P2/8γ
1.05×105 × 81.4 = P2
P2 = 18.5×105 N/m2

Again,

T1V1γ−1 = T2V2γ−1
T2/T1 = 8γ−1
T2 = 290 × 81.4−1
T2 = 666.24 K

Q.2 — Air compressed adiabatically to half volume

Air is compressed adiabatically to half its volume. Calculate the change in its temperature. [γ = 1.4]

Initial volume (V1) = V
Final volume (V2) = V/2

Now,

T1V1γ−1 = T2V2γ−1
T1(V)γ−1 = T2(V/2)γ−1
T2/T1 = 1.319

Change of temperature in percent:

(T2/T1 − 1) × 100%
= (1.319 − 1) × 100%
= 0.319 × 100%
= 31.9%

Q.3 — Isothermal compression of five moles of ideal gas

Five moles of an ideal gas are kept at a constant temperature of 53°C while the pressure of the gas is increased from 1 atm to 3 atm. Calculate the work done by the gas.

Number of moles, n = 5
Constant temperature, T = 53°C = 326 K
Initial pressure, P1 = 1.00 atm
Final pressure, P2 = 3 atm
Work done by gas, W = ?

From isothermal process:

W = nRT ln(P1/P2)
= 5 × 8.314 × 326 × ln(1/3)
= −1.49×104 J
W = −1.49×104 J

Q.4 — One litre of air heated at constant pressure

A litre of air initially at 20°C & at 760 mm of Hg pressure is heated at constant pressure until its volume is doubled. Find (i) the temperature, (ii) external workdone by the air in expanding, (iii) the quantity of heat. Capacity at constant volume = 714 J kg−1K−1.

Initial temperature, T1 = 20°C = 293 K
Final temperature, T2 = ?
Pressure, P = 1.013×105 N/m2
Initial volume, V1 = 1 litre = 10−3 m3
Final volume, V2 = 2V1 = 2×10−3 m3

(i) T2 = ?

V1/V2 = T1/T2
10−3/(2×10−3) = 293/T2
T2 = 293×2
T2 = 586 K

(ii) dW = ?

dW = P dV
= P(V2 − V1)
= 1.013×105(2×10−3 − 10−3)
dW = 101.3 J

(iii) dQ = ?

V0/V1 = 273/293
V0 = 9.317×10−4 m3
m = V0ρ
= 9.317×10−4 × 1.293
= 1.2×10−3 kg
dU = mCvdT
= 1.2×10−3 × 714 × 293
= 251.08 J
dQ = dU + dW
dQ = 352.3 J

Q.5 — Oxygen heated at constant pressure

16 gm of oxygen having volume 0.02 m3 at temperature of 27°C & pressure of 2×105 N/m2 is heated at constant pressure until its volume increases to 0.03 m3. Calculate the external workdone & increase in internal energy of the gas if the molar heat capacity at constant volume is 0.8 J mol−1K−1 & molar mass of oxygen is 32.

V1 = 0.02 m3
V2 = 0.03 m3
T1 = 27°C = 300 K
P = 2×105 N/m2

External workdone:

dW = P dV
= 2×105(0.03 − 0.02)
= 2×105 × 0.01
dW = 2×103 J

At constant pressure:

V1/T1 = V2/T2
T2 = 0.03×300 / 0.02
T2 = 450 K

Internal energy:

dU = (m/M)CvdT
= (16/32) × 0.8 × (450 − 300)
dU = 60 J
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