HC Verma Concepts Of Physics Exercise Solutions Of Chapter 26 (Laws of Thermodynamics)

HC Verma Concepts of Physics Solutions - Part 2, Chapter 26 - Laws Of Thermodynamics:

EXERCISE

1. A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J/kg-K and 4200 J/kg-K respectively. Neglect any thermal expansion, (a) how much heat is transferred to the liquid—vessel system? (b) How much work has been done on this system? (c) How much is the increase in internal energy of the system?

Sol:

System: water + copper

Given: insulated; closed system; t_i = 15 ⁰c; t_e = 17 ⁰c; mass of copper, m_cu = 100 g = 0.1 kg; mass of water, m_w = 200 g = 0.2 kg; c_cu = 420 J/kg-K and c_w = 4200 J/kg-K.

(a) Due to insulated boundary, heat transfer to the system is zero.

(b) From the 1^st law,

We know, Q = ΔU + W = U₂ – U₁ + W

Or, W = U₁ – U₂              [Q = 0]

Or, W = (U₁ – U₂)_cu + (U₁ – U₂)_w

Or, W = m_cu c_cu (t_i – t_e) + m_w c_w (t_i – t_e)

Or, W = 0.1 * 420 (15 – 17) + 0.2 * 4200 (15 – 17)

Or, W = – 1764 J.

– Ve sign means work done on the system.

(c) The increase in internal energy of the system is ΔU = W = 1764 J.

2. Figure (26-E1) shows a paddle wheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200 J/K kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm. (a) how much heat is given to the liquid? (b) How much work is done on the liquid? (c) Calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.

Sol:

System: liquid + wheel + container

Given: adiabatic system; heat capacity of liquid, C_L = 4200 J/K, mass of block, m = 12 kg, change of height of block, Δh = 70 cm = 0.7 m.

(a) Heat is given to the liquid is zero, because system is adiabatic.

Q = 0

(b) Work is done on the liquid, W = change of potential energy of block

Or, W = mg (Δ h) = 12 * 10 * 0.7 = 84 J.

(c) 1^st law, Q = ΔU + W

Or, 0 = C_L (Δ t) – 84 [W = – 84 work done on the system]

Or, Δ t = 84/4200 = 0.02 ⁰c.

3. A 100 kg block is started with a speed of 2.0 m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt, (b) Consider the situation from a frame of reference moving at 2.0 m/s along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m/s. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt, (c) Find the work done in this frame by the external force holding the belt.

Sol:

Given: mass of block = 100 kg; u = 2 m/s; μ = 0.2; v = 0.

1^st law: Q = Δu + W

In this case Q = 0

Or, – Δu = W = change in K.E.

Or, Δu = – (½ mv² – ½ mu²)

Or, Δu = ½ * 100 * 2² = 200 J.

(a) The change in the internal energy of the block-belt system is 200 J.

(b) The increase in the kinetic energy of the block is 200 J.

4. Calculate the change in internal energy of a gas kept in a rigid container when 100 J of heat is supplied to it.

Sol:

Given: Q = 100 J; ΔV = 0.

1^st Law: Q = ΔU + W = ΔU + pΔV

Or, 100 = ΔU + 0 → ΔU = 100 J

So, the change in internal energy of a gas is 100 J.

5. The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc. (a) Calculate the work done by the gas. (b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?

Sol:

Given: Q = 0; v₁ = 200 cc = 2 * 10^-4 m³; v₂ = 50 cc = 0.5 * 10^-4; p₁ = 10 kPa; p₂ = 50 kPa.

(a) Work done during the process 1-2,

W = p₁ (v₂ – v₁) + ½ * (p₂ – p₁)(v₂ – v₁)

Or, W = 10 * (0.5 – 2)*10^-4 + ½ *40 *(0.5 – 2) * 10^-4

Or, W = – 0.0015 – 0.0030 = – 0.0045 kJ

Or, W = – 4.5 J

So, the work done by the gas is – 4.5 J.

(b) 1^st law: Q = ΔU + W

Or, 0 = ΔU + W

Or, ΔU = – W = 4.5 J.

6. An ideal gas is taken from an initial state i to a final state f in such a way that the ratio of the pressure to the absolute temperature remains constant. What will be the work done by the gas?

Sol:

Given: p/T = constant.

For ideal gas, pV = RT

Or, v = RT/p = constant

Work done, W = pΔV = 0

So, the work done by the gas is zero.

7. Figure (26-E2) shows three paths through which a gas can be taken from the state A to the state B. Calculate the work done by the gas in each of the three paths.

Sol:

Area under the path in p-v plane will give the work done.

In case path AB:

→ Work done, W = W_AB

Or, W = 10 * 10³ * (25 – 10)* 10^-6 + ½ * (30 – 10)*10³* (25 – 10)* 10^-6

Or, W = 0.15 + 0.15 = 0.30 J.

In case path ACB:

→ Work done, W = W_AC + W_CB

Or, W = 0 + 30 * 10³ * (25 – 10) * 10^-6

Or, W = 0.45 J.

In case path ADB:

→ Work done, W = W_AD + W_DB

Or, W = 10 * 10³ * (25 – 10)* 10^-6 + 0

Or, W = 0.15 J.

8. When a system is taken through the process abc shown in figure (26-E3), 80 J of heat is absorbed by the system and 30 J of work is done by it. If the system does 10 J of work during the process adc, how much heat flows into it during the process?

Sol:

Given: Q_abc = + 80 J; W_abc = + 30 J; W_adc = + 10 J.

Applying 1^st law during the process abc

→ Q = ΔU + W

Or, 80 = ΔU + 30

Or, ΔU = 50 J.

Change of internal energy in both paths are same, because the end points of two process paths are same.

Applying 1^st law during the process adc

→ Q = ΔU + W

Or, Q = 50 + 10 = 60 J.

9. 50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in figure (26-E4). Find the quantity of heat to be supplied to take it from A to B via ADB.

Sol:

Given: Q_ACB = + 50 cal = 50 * 4.2 = + 210 J.

Work through the path ACB,

→ W_ACB = W_AC + W_CB

Or, W_ACB = 50 * 10³ * (400 – 200) * 10^-6 + 0

Or, W_ACB = 10 J

Work done by the system, W = + 10 J

Applying 1^st law during the process ACB,

→ Q = ΔU + W

Or, 210 = ΔU + 10

Or, ΔU = 200 J.

Change of internal energy in both paths are same, because the end points of two process paths are same.

Work through the path ADB,

→ W_ADB = W_AD + W_DB

Or, W_ADB = 0 + 155 * 10³ * (400 – 200) * 10^-6

Or, W_ADB = 31 J

Work done by the system, W = + 31 J

Applying 1^st law during the process ADB,

→ Q = ΔU + W

Or, Q = 200 + 31 = 231 J = 55 cal.

10. Calculate the heat absorbed by a system in going through the cyclic process shown in figure (26-E5).

Sol:

For cyclic process, ΔU = 0

Therefore, Q = W

Area under curve in p-v plane will give the work done.

Work, W = area of circle

Or, W = π (300 – 100)/2 * 10³ * (300 – 100)/2 * 10^-6

Or, W = 31.4 J

Therefore, the heat absorbed by a system is 31.4 J.

11. A gas is taken through a cyclic process ABC A as shown in figure (26-E6). If 2.4 cal of heat is given in the process, what is the value of J?

Sol:

Given: Q = + 2.4 cal.

Work, W = area of the triangle

Or, W = ½ * (200 – 100) * 10³ * (700 – 500) * 10^–6

Or, W = 10 J

Work done by the system W = + 10 J.

1^st law for cyclic process, J Q = W

J is Mechanical equivalent. Where Q and W has different unit.

→ J * 2.4 = 10

Or, J = 10/2.4 = 4.17 J/cal.

12. A substance is taken through the process abc as shown in figure (26-E7). If the internal energy of the substance increases by 5000 J and a heat of 2625 cal is given to the system, calculate the value of J. 

Sol:

Given: Q = + 2625 cal, ΔU = + 5000 J.

Work, W = W_ab + W_bc

Or, W = 200 * 10³ * (0.05 – 0.02) + 0

Or, W = 6000 J

So, work done by the system W = + 6000 J.

Applying 1^st law during the process abc,

→ J Q = ΔU + W

Or, J * 2625 = 5000 + 6000

Or, J = 11000/2625 = 4.19 J/cal.

13. A gas is taken along the path AB as shown in figure (26-E8). If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.

Sol:

Given: Q = – 70 cal = – 294 J.

Work, W = 200 * 10³ * (250 – 100) * 10^-6 + ½ * (500 – 200) * 10³ * (250 – 100) * 10^-6

Or, W = 30 + 22.5 = 52.5 J

Work is done on the system, therefore W = – 52.5 J

1^st law: Q = ΔU + W

Or, ΔU = Q – W

Or, ΔU = – 294 + 52.5 = – 241.5 J.

14. The internal energy of a gas is given by U = 1.5pV. It expands from 100 cm³ to 200 cm³ against a constant pressure of 1.0 * 10⁵ Pa. Calculate the heat absorbed by the gas in the process.

Sol:

Given: U = 1.5 pV; p₁ = p₂ = p = 1.0 * 10⁵ Pa; V₁ = 100 cm³; V₂ = 200 cm³.

Work, W = p (V₂ – V₁)

Or, W = 1.0 * 10⁵ * (200 – 100) * 10^-6 = 10 J

Work done by the system W = + 10 J.

Change in internal energy, ΔU = 1.5 p (V₂ – V₁)

Or, ΔU = 1.5 * 1.0 * 10⁵ * (200 – 100) * 10^-6 = 15 J

Increase in internal energy ΔU = + 15 J.

1^st law: Q = ΔU + W

Or, Q = 15 + 10 = 25 J.

15. A gas is enclosed in a cylindrical vessel fitted with a frictionless piston. The gas is slowly heated for some time. During the process, 10 J of heat is supplied and the piston is found to move out 10 cm. Find the increase in the internal energy of the gas. The area of cross-section of the cylinder = 4 cm² and the atmospheric pressure = 100 kPa.

Sol:

Given: Q = + 10 J; area of cross-section of the cylinder, A = 4 cm²; change of height, Δh = 10 cm; p = 100 kPa.

Work, W = pΔV = pAΔh

Or, W = 100 * 10³ * 4 * 10^-4 *10 * 10^-2

Or, W = 4 J

Work done by system W = + 4 J

1^st law: Q = ΔU + W

Or, ΔU = Q – W = 10 – 4 = 6 J. 

16. A gas is initially at a pressure of 100 kPa and its volume is 2.0 m³. Its pressure is kept constant and the volume is changed from 2.0 m³ to 2.5 m³. Its volume is now kept constant and the pressure is increased from 100 kPa to 200 kPa. The gas is brought back to its initial state, the pressure varying linearly with its volume, (a) whether the heat is supplied to or extracted from the gas in the complete cycle? (b) How much heat was supplied or extracted?

Sol:

In a cyclic process, Q = W

Work, W = area of the triangle

Or, W = ½ * (200 – 100) * 10³ * (2.5 – 2)

Or, W = 25000 J

W_ca = work done on the system (– ve)

W_ab = work done by the system (+ ve)

W_ca > W_ab

Therefore, work is done on the system W = – 25000 J

So, Q = – 25000 J

(a) – ve sign shows heat is extracted from the system.

(b) Q = 25000 J.

17. Consider the cyclic process ABCA, shown in figure (26-E9), performed on a sample of 2.0 mole of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC.

Sol:

Given: n = 2 mole; Q = – 1200 J; ΔU = 0 (During cyclic Process); W = W_AB + W_BC + W_CA.

W_AB = pΔV = nRΔT; W_CA = pΔV = 0 (ΔV = 0)

1^st law: Q = ΔU + W

Or, – 1200 = 0 + W_AB + W_BC + W_CA

Or, – 1200 = nRΔT + W_BC + 0

Or, – 1200 = 2 × 8.3 × 200 + W_BC 

Or, W_BC = – 400 × 8.3 – 1200 = – 4520 J.

18. Figure (26-E10) shows the variation in the internal energy U with the volume V of 2.0 mole of an ideal gas in a cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.

Sol:

Given: n = 2 moles; dV = 0 in ‘da’ and ‘bc’; T_b = 500 K; T_c = 300 K.

Work done during the process ‘da’ and ‘bc’ are zero (dV = 0).

For an ideal gas, internal energy depends on temperature only.

Process ab and cd → internal energy remains constant. So, temperature also remain constant.

Work done, W = ∫pdV = ∫ (nRT/V) dV = nRT ln (V₂/V₁)

∴ W_ab = 2 * 8.3 * 500 * ln (2V₀/V₀)

Or, W_ab = 5753 J

And, W_cd = 2 * 8.3 * 300 * ln (V₀/2V₀)

Or, W_cd = – 3452 J

Work done in cyclic process, 

→ W = W_ab + W_bc + W_cd + W_da

Or, W = 5753 + 0 – 3452 + 0 = 2301 J

For a cyclic process, Q = W = 2301 J

So, the heat absorbed by the gas during the process is 2301 J.

19. Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J/kg-K and its densities at 0°C and 4°C are 999.9 kg/m³ and 1000 kg/m³ respectively. Atmospheric pressure = 10⁵ Pa.

Sol:

Given: mass of water, m = 2kg; change of temperature, ΔT = 4 °C or K; specific heat capacity, C_w = 4200 J/kg-K; ρ @ 0⁰c = 999.9 kg/m³; ρ @ 4⁰c = 999.9 kg/m³; p = 10⁵ Pa.

Specific volume, v₁ = 1/ρ = 1/999.9 = 0.001001 m³/kg

Specific volume, v₂ = 1/ρ = 1/1000 = 0.001 m³/kg

Heat absorbed by water, 

→ Q = m C_w ΔT

Or, Q = 2 * 4200 * 4 = 33600 J

Work done, W = pΔV

Or, W = m p Δv = m p (v₂ – v₁)

Or, W = 2 * 10⁵ * (0.001 – 0.001001)

Or, W = – 0.02 J

1^st Law: Q = ΔU + W

Or, ΔU = Q – W = (33600 + 0.2) J.

20. Calculate the increase in the internal energy of 10 g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg/m³. Specific heat capacity of water = 4200 J/kg-°C and the latent heat of vaporization of water = 2.25 * 10⁶ J/kg.

Sol:

Given: mass of water, m = 10 g = 0.01 kg; p = 100 kPa = 10⁵ Pa; C_w = 4200 J/kg-°C; latent heat of vaporization, l = 2.25 * 10⁶ J/kg.

Density of water, ρ_w = 1000 kg/m³, Specific vol. v_w = 0.001 m³/kg

Density of steam, ρ_s = 0.6 kg/m³, Specific vol. v_s = 1.667 m³/kg

Heat absorbed, Q = m C_w ΔT + ml

Or, Q = 0.01 * 4200 * (100 – 0) + 0.01 * 2.25 * 10⁶

Or, Q = 4200 + 22500 = 26700 J.

Work done, W = pΔV

Or, W = m p Δv = m p (v_s – v_w)

Or, W = 0.01 * 10⁵ * (1.667 – 0.001)

Or, W = 1666 J

1^st Law: Q = ΔU + W

Or, ΔU = Q – W

Or, ΔU = 26700 – 1666

Or, ΔU = 25034 = 2.5 * 10⁴ J.

21. Figure (26-E11) shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p₁, T₁ on the left and p₂, T₂ on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides, (a) how much work has been done by the gas on the left part? (b) Find the final pressures on the two sides, (c) Find the final equilibrium temperature, (d) how much heat has flown from the gas on the right to the gas on the left? 

Sol:                           

Given: internal energy, U = 1.5 nRT; ΔU = 1.5 nRΔT; Q = 0 (adiabatic).

(a) Since the wall cannot be moved thus change of volume is zero.

Hence, work done by the gas on the left part, W = 0.

1^st Law: Q = ΔU + W → ΔU = 0

Let, the final temperature be T

So, ΔU = ΔU₁ + ΔU₂ = 0

Or, 1.5 n₁R (T – T₁) + 1.5 n₂R (T – T₂) = 0

Or, T = (n₁T₁ – n₂T₂)/ (n₁ + n₂)

Or, T = [(p₁V/2RT₁)*T₁ – (p₂V/2RT₂)*T₂]/ [(p₁V/2RT₁) – (p₂V/2RT₂)]

Or, T = (p₁ + p₂) T₁T₂/ (p₁T₂ + p₂T₁)

Or, T = (p₁ + p₂) T₁T₂/λ   (where, λ = p₁T₂ + p₂T₁)

(b) Let final pressure in LHS = P₁’ and in RHS = P₂’

Number of mole remains constant.

So, P₁’V/2RT = P₁V/2RT₁ → P₁’ = p₁T/T₁

Or, P₁’ = p₁T₂ (p₁ + p₂)/λ on the left

Similarly, P₂’ = p₂T₁ (p₁ + p₂)/λ on the right.

(c) The final equilibrium temperature, T = (p₁ + p₂) T₁T₂/λ.

(d) For RHS Q = ΔU (As W = 0)

Or, Q = 1.5 n₂RΔT

Or, Q = 1.5 * (P₂V/2RT₂) * R * (T – T₂)

Or, Q = 3p₁p₂ (T₂ – T₁) V/4λ.

22. An adiabatic vessel of total volume V is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas (U = 1.5 nRT) and the part on the right contains two moles of the same gas. Initially, the pressure on each side is p. The system is left for sufficient time so that a steady state is reached. Find (a) the work done by the gas in the left part during the process, (b) the temperature on the two sides in the beginning, (c) the final common temperature reached by the gases, (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part.

Sol:

Given: internal energy, U = 1.5 nRT; ΔU = 1.5 nRΔT; Q = 0 (adiabatic).

(a) Since the wall cannot be moved thus change of volume is zero.

Hence, work done by the gas on the left part, W = 0.

(b) For left side, n = 1, pressure = p, volume = V/2,

Let initial temperature be T₁.

For ideal gas, pV = nRT

Or, T₁ = pV/2(mol) R.

For right side, n = 2, pressure = p, volume = V/2,

Let initial temperature be T₂.

For ideal gas, pV = nRT

Or, T₂ = pV/4(mol) R.

(c) Let the final Temperature = T

Final Pressure = p

No. of mole = 1 mole + 2 moles = 3 moles

∴ T = pV/3(mol) R.

(d) For RHS Q = ΔU (As W = 0)

Or, Q = 1.5 n₂RΔT

Or, Q = 1.5 * 2 * R * (T – T₂)

Or, Q = 3R [pV/3(mol) R – pV/4(mol) R]

Or, Q = pV/4.

(e) As heat loss by left side, so Q = – pV/4 and W = 0.

1^st Law: Q = ΔU + W → ΔU = – pV/4.

Download HC Verma's Concepts Of Physics Chapter 26 (Laws of Thermodynamics) Solutions in PDF: Click Here

Discussion - If you have any Query or Feedback comment below.

Tagged With: Previous year Chapter-wise JEE Questions Solutions, Chapter-wise previous year solutions of IIT JEE, AIEEE, Other JEE Exams, HC Verma Solutions, Concepts ofPhysics Volume 1, Concepts of Physics Volume 2, HC Verma Solutions Part 1 andPart 2, Chapter wise solutions of hc verma’s Concepts of Physics, HC Verma, HCVerma Part 1 PDF Solution Download, HCVERMA SOLUTION (CHAPTERWISE), HC Verma Part 2 PDF Solution Download, hcverma part 1 solutions, hc verma part 2 solutions, hc verma objectivesolutions, Concepts of Physics solutions download, Solutions of HC VermaConcepts of Physics Volume 1 & 2.