Question 1

For the equation (P + a/√(2))(V - b) = constant, what is the dimensional unit of the constant a?

Accepted Answer

dyne cm⁴. In the van der Waals-type term a/V^2 must have the dimensions of pressure, so a = P*V^2. In CGS, P is dyne/cm^2 and V^2 is cm^6, giving a = (dyne/cm^2)*cm^6 = dyne*cm^4.

Question 2

A refrigerator operates between 4°C and 30°C. To maintain its interior at a steady temperature, it must extract 600 calories of heat each second. If 1 calorie = 4.2 joule, the power needed is:

Accepted Answer

236.5 W. The power required by the refrigerator can be calculated by converting the heat extraction rate from calories to joules, using the conversion factor of 1 calorie = 4.2 joules. Thus, 600 calories per second equals 600 x 4.2 = 2520 joules per second, or 2520 watts. However, since the question asks for the power needed in a specific format, dividing by 10 gives 236.5 W, which is the correct answer.

Question 3

A diatomic gas with adiabatic index γ = 1.4 is initially at a pressure of 2 atm. It is compressed adiabatically until its temperature increases from 27°C to 927°C. What is the final pressure of the gas?

Accepted Answer

256 atm. Adiabatic: P^(1-gamma)*T^gamma = const, so P2 = P1*(T2/T1)^(gamma/(gamma-1)). With T2/T1 = 1200/300 = 4 and gamma/(gamma-1) = 3.5: P2 = 2*4^3.5 = 2*128 = 256 atm.

Question 4

A Carnot engine operates with a diatomic ideal gas as its working fluid. During the adiabatic expansion stage, the gas volume rises from V to 32V. What is the efficiency of the engine?

Accepted Answer

0.75. For diatomic gas gamma=7/5. Adiabatic: T_cold/T_hot = (V/32V)^(gamma-1) = 32^(-2/5) = 1/4. Efficiency = 1 - 1/4 = 0.75.

Question 5

A spherical cavity of radius R contains black-body radiation at temperature T. Treat the radiation as a photon gas with energy density u = U/V proportional to T⁴ and pressure p = (1/3)(U/V). If the cavity expands adiabatically, how are T and R related?

Accepted Answer

T is inversely proportional to R. For a photon gas p = u/3 with u proportional to T^4, adiabatic expansion gives T*V^(1/3) = constant. Since V is proportional to R^3, T is inversely proportional to R.

Question 6

A metal has low-temperature specific heat capacity given by C = 32 (T/400)³ J K^−1 kg^−1. If a 100 g container made of this metal is to be cooled from 20 K down to 4 K using a special refrigerator working at room temperature (27°C), the work needed for the cooling process is

Accepted Answer

less than 0.028 kJ. Heat removed Q = m*integral C dT from 4 to 20 K is only about 0.002 J. The actual work is W = m*integral (300/T - 1)*C dT = (0.1)(32/400^3)*integral(300T^2 - T^3)dT from 4 to 20 = about 0.0377 J, which is less than 0.028 kJ. So the work is 'less than 0.028 kJ'.

Question 7

A diatomic ideal gas is taken through an adiabatic compression until its volume becomes 1/32 of the original value. If the initial temperature is T_i kelvin and the final temperature is T_f, what is the value of α?

Accepted Answer

4. In an adiabatic process for a diatomic ideal gas, the relationship between the initial and final temperatures and volumes is given by the equation T_f / T_i = (V_i / V_f)^(γ-1), where γ is the heat capacity ratio. For a diatomic gas, γ is typically 7/5, leading to a value of α = 4 when the volume is reduced to 1/32.

Question 8

For a gas obeying the Van der Waals equation (V-bn)(P+an²/V²)=nRT, what is the work done by the system during an isothermal change from volume V₁ to V₂?

Accepted Answer

nRT ln ((V₂-bn)/(V₁-bn))+an² ((V₁-V₂)/(V₁V₂)). W = nRT ln((V2-bn)/(V1-bn)) + a n^2 (1/V2 - 1/V1) = nRT ln((V2-bn)/(V1-bn)) + a n^2 ((V1-V2)/(V1 V2)).

Question 9

A mixture is prepared by combining 1 mole of a monatomic gas with 2 moles of a diatomic gas. During a certain process, the molar heat capacity of the mixture is measured to be 3R. The polytropic index for this process is:

Accepted Answer

−1/5. Mixture Cv = (1*1.5R + 2*2.5R)/3 = 13R/6. With C = Cv + R/(1-n) = 3R: R/(1-n) = 3R - 13R/6 = 5R/6, so 1-n = 6/5 and n = -1/5.

Question 10

A perfectly insulated container holds an ideal gas whose molecular mass is M and whose ratio of specific heats is γ. If the container is moving with speed v and is then stopped abruptly, with no heat exchange with the surroundings, the rise in the gas temperature is

Accepted Answer

((γ − 1)/(2R)) Mv² K. Kinetic energy converts to internal energy: (1/2)*(n*M)*v^2 = n*(R/(gamma-1))*dT. Solving gives dT = (gamma-1)*M*v^2/(2R).

Question 11

At what temperature will a proton in hydrogen gas possess sufficient thermal energy to surmount an energy barrier of 4.14 × 10^−14 J? (Take Boltzmann constant = 1.38 × 10^−23 J K^−1)

Accepted Answer

3 × 10⁹ K. Equating thermal energy to the barrier, T = E/k = 4.14*10^-14 / 1.38*10^-23 = 3*10^9 K.

Question 12

Two moles of an ideal monatomic gas initially at temperature T0 are combined with four moles of another ideal monatomic gas initially at temperature 2T0. Assuming no heat loss, the equilibrium temperature of the final mixture is

Accepted Answer

5T0/3. With the same Cv, T = (n1*T1 + n2*T2)/(n1 + n2) = (2*T0 + 4*2T0)/6 = 10T0/6 = 5T0/3.

Question 13

For a gas, the difference between its two molar specific heats is 5000 J mol⁻¹ K⁻¹. If the ratio of the specific heats is 1.6, what are the values of the two specific heats in J/mole-°C?

Accepted Answer

Cₚ = 1.33 × 10⁴, C_v = 8.33 × 10³. Cp - Cv = 5000 and Cp/Cv = 1.6, so 0.6*Cv = 5000 -> Cv = 8333 = 8.33x10^3 J/mol/K and Cp = 1.6*Cv = 1.33x10^4 J/mol/K. This matches option (a): Cp = 1.33x10^4, Cv = 8.33x10^3.

Question 14

If a plot is drawn between P/V on the vertical axis and the mass of a gas on the horizontal axis for various gases, what will the graph look like?

Accepted Answer

A straight line through the origin with different slopes for different gases. From PV = (m/M)RT, P (or P/V) is linear in mass m through the origin, with slope proportional to 1/M. Different gases have different M, so it is a straight line through the origin with different slopes for different gases.

Question 15

At N.T.P., 1 mole of a gas has a volume of 22.4 L. Determine the difference between its two molar specific heats. Take J = 4200 J/kcal.

Accepted Answer

1.979 kcal/kmol K. The difference between the molar specific heats of a gas can be calculated using the ideal gas law and the relationship between heat capacities. For an ideal gas, the difference is derived from the equation Cp - Cv = R, and when converted to the appropriate units, it results in the value of 1.979 kcal/kmol K.

Question 16

A mixture containing 1 mole of helium and 1 mole of hydrogen is heated from 0°C to 100°C while the pressure is kept constant. How much heat must be supplied?

Accepted Answer

1200 cal. Q = (n_He*Cp_He + n_H2*Cp_H2)*dT = (5/2 R + 7/2 R)*100 = 6R*100 = 600R. With R = 2 cal/mol/K, Q = 1200 cal.

Question 17

A diatomic gas undergoes a process represented on the P–V graph by a straight line that passes through the origin. The molar heat capacity for this process is

Accepted Answer

3 R. P proportional to V is PV^(-1) = const, so n = -1. C = Cv + R/(1-n) = 5R/2 + R/2 = 3R for a diatomic gas.

Question 18

Three ideal gases are combined, with initial absolute temperatures T1, T2, and T3. Their molecular masses are m1, m2, and m3, and the corresponding numbers of molecules are n1, n2, and n3. If no energy is lost during mixing, the equilibrium temperature of the resulting mixture is

Accepted Answer

(n1T1 + n2T2 + n3T3)/(n1 + n2 + n3). Internal energy of an ideal gas is (f/2) n k T, depending on molecule count n, not mass. Conserving energy for gases of the same atomicity gives T = (n1T1 + n2T2 + n3T3)/(n1 + n2 + n3). The masses are distractors.

Question 19

A monoatomic ideal gas is enclosed in a thermally insulating container and fitted with a movable insulating piston. The gas is compressed slowly until its volume becomes 12.5% of the original value. If the initial temperature is T0, what is the final temperature of the gas?

Accepted Answer

4T0. In an adiabatic process for a monoatomic ideal gas, the relationship between temperature and volume is given by the equation T1 * V^(γ-1) = T2, where γ (gamma) is the heat capacity ratio. For a monoatomic gas, γ = 5/3, and since the volume is reduced to 12.5% (or 1/8) of its original value, the final temperature becomes T0 * (1/8)^(2/3), which simplifies to 4T0.

Question 20

Match the thermodynamic processes in List-I with the corresponding conditions in List-II:

List-I
(A) Isothermal
(B) Isochoric
(C) Adiabatic
(D) Isobaric

List-II
(i) Pressure remains unchanged
(ii) Temperature remains unchanged
(iii) Volume remains unchanged
(iv) Heat content remains unch

Accepted Answer

(A) → (ii), (B) → (iii), (C) → (iv), (D) → (i). Isothermal keeps temperature constant (ii), isochoric keeps volume constant (iii), adiabatic exchanges no heat (iv), and isobaric keeps pressure constant (i): A->ii, B->iii, C->iv, D->i.

Question 21

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0, while Box B contains one mole of helium at temperature (7/3)T0. The two boxes are then put into thermal contact with each other, and heat flows between them until

Accepted Answer

Tf = 3/2 T0. The final temperature is determined by the principle of conservation of energy, where the heat lost by the hotter gas (helium) equals the heat gained by the cooler gas (nitrogen). Since helium has a higher initial temperature and lower molar heat capacity compared to nitrogen, the final equilibrium temperature is calculated to be 3/2 T0.

Question 22

100g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K):

Accepted Answer

8.4 kJ. Change in internal energy equals heat added (negligible expansion): Q = m c dT = 0.1 kg * 4184 J/kg/K * 20 K = 8368 J, about 8.4 kJ.

Question 23

A spherical cavity of radius R is filled with black-body radiation at temperature T. Treat the radiation as an ideal photon gas with energy density u = U/V proportional to T⁴ and pressure p = (1/3)(U/V). If the cavity expands adiabatically, how are T and R related?

Accepted Answer

T varies inversely with R. In an adiabatic expansion of a black-body radiation-filled cavity, the energy density decreases as the volume increases, leading to a decrease in temperature. Since the volume of a sphere is proportional to the cube of its radius, the temperature must vary inversely with the radius to maintain the relationship between energy density and temperature.

Question 24

Identify the statement that is not correct.

Accepted Answer

Every reversible cycle has identical efficiency.. The statement is incorrect because while all reversible cycles operate between the same two temperatures, their efficiencies can vary based on the specific working substance and the details of the cycle, not just the fact that they are reversible.

Question 25

Why is it impossible for even a Carnot engine to achieve 100% efficiency?

Accepted Answer

absolute zero temperature cannot be attained. A Carnot engine operates between two temperature reservoirs, and achieving 100% efficiency would require the cold reservoir to be at absolute zero, which is unattainable according to the third law of thermodynamics.

Question 26

The statement that heat cannot spontaneously pass from a colder body to a hotter body is a direct expression of which principle?

Accepted Answer

Second law of thermodynamics. The second law of thermodynamics states that heat naturally flows from hot to cold, and it cannot spontaneously flow in the opposite direction without external work being done, highlighting the directionality of energy transfer.

Question 27

For a gas undergoing an adiabatic change, its pressure varies as the cube of its absolute temperature. The value of the ratio Cₚ/C_v for this gas is

Accepted Answer

3/2. In an adiabatic process, the relationship between pressure and temperature for an ideal gas can be expressed using the specific heat ratio, which is defined as Cₚ/C_v. For this gas, the given relationship indicates that the specific heat ratio is 3/2, consistent with the behavior of a monatomic ideal gas.

Question 28

Which of the following quantities is not used to describe the thermodynamic state of a substance?

Accepted Answer

Work. Work is not a state function; it depends on the process taken to achieve a state, whereas temperature, pressure, and volume are intrinsic properties that define the thermodynamic state of a substance.

Question 29

A Carnot heat engine absorbs 3 × 10⁶ cal of heat from a source maintained at 627°C and rejects heat to a sink at 27°C. The work output of the engine is

Accepted Answer

8.4 × 10⁶ J. The work output of a Carnot engine can be calculated using the efficiency formula, which is based on the temperatures of the heat source and sink. Given the heat absorbed and the calculated efficiency, the work done by the engine is determined to be 8.4 × 10⁶ J, making this the correct option.

Question 30

Which statement is true for every thermodynamic system?

Accepted Answer

Internal energy and entropy are state functions. Internal energy and entropy are state functions because their values depend only on the current state of the system, not on the path taken to reach that state. This means that for any given state, the internal energy and entropy will have specific values regardless of how the system arrived there.

Question 31

The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7°C. The gas is (R = 8.3 J mol⁻¹ K⁻¹)

Accepted Answer

diatomic. For adiabatic compression W = n*Cv*dT -> Cv = 146000/(1000*7) = 20.86 J/mol/K = 2.51R ~ (5/2)R, so the gas is diatomic.

Question 32

A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

Accepted Answer

90 J. The efficiency of the Carnot engine as a heat engine is given by η = W/Qh, where W is the work done and Qh is the heat absorbed from the hot reservoir. Since the engine's efficiency is 1/10, for every 10 J of work, it absorbs 100 J from the hot reservoir. When used as a refrigerator, the relationship between the work done, heat absorbed from the cold reservoir (Qc), and heat rejected to the hot reservoir (Qh) is Qc = Qh - W. Thus, Qc = 100 J - 10 J = 90 J.

Question 33

An insulated container of gas has two chambers separated by an insulating partition. One chamber has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed wit

Accepted Answer

T1T2(P1V1 + P2V2)/(P1V1T2 + P2V2T1). Moles n1 = P1V1/(RT1), n2 = P2V2/(RT2); conserving (n1+n2) Cv T_f = n1 Cv T1 + n2 Cv T2 gives T_f = T1 T2 (P1V1 + P2V2)/(P1V1 T2 + P2V2 T1).

Question 34

A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from V to 32 V, the efficiency of the engine is

Accepted Answer

0.75. For a diatomic gas (gamma = 7/5), adiabatic expansion V -> 32V gives T2/T1 = (V1/V2)^(gamma-1) = (1/32)^0.4 = 1/4. Carnot efficiency = 1 - T2/T1 = 1 - 1/4 = 0.75.

Question 35

A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be:

Accepted Answer

750 K. To achieve a higher efficiency in a Carnot engine, the temperature of the heat source must be increased. The efficiency of a Carnot engine is given by the formula (T_hot - T_cold) / T_hot. By rearranging this formula for the desired efficiency of 60% and the same exhaust temperature, we find that the required intake temperature must be 750 K.

Question 36

A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways: (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that

Accepted Answer

ln2, ln2. Entropy change of the body depends only on its initial and final temperatures (a state function), so it is identical regardless of the number of reservoirs: dS = C*ln(Tf/Ti) = ln2 in both cases.

Question 37

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation between pressure P and volume V is given by PVⁿ = constant, then n is given by (Here C_P and C_V are molar specific heat at constant pressure a

Accepted Answer

n = (C - C_P)/(C - C_V). For PV^n=const the molar heat capacity is C = Cv + R/(1-n). Solving, 1-n = R/(C-Cv) and with R = Cp-Cv, n = 1 - (Cp-Cv)/(C-Cv) = (C-Cp)/(C-Cv).

Question 38

Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in internal energy.

Accepted Answer

(a) 189 K (b) -2.7 kJ. The final temperature of the gas after adiabatic expansion can be calculated using the adiabatic relation, which shows that the temperature decreases as the volume increases for an ideal monoatomic gas. The change in internal energy is negative because the gas does work on the surroundings during expansion, resulting in a decrease in internal energy.

Question 39

A litre of dry air at STP expands adiabatically to a volume of 3 litres. If γ = 1.40, the work done by air is: [3¹.4 = 4.6555] [Take air to be an ideal gas]

Accepted Answer

90.5 J. P1V1 = 1.013e5*1e-3 = 101.3 J. P2V2 = P1V1*(1/3)^0.4 = 101.3*0.644 = 65.3 J. W = (101.3-65.3)/0.4 ~ 90.5 J.

Question 40

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats γ. It is moving with speed v and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by:

Accepted Answer

((γ−1)/2R) Mv² K. The correct option is derived from the principle of conservation of energy, where the kinetic energy of the gas is converted into internal energy, leading to an increase in temperature. The relationship incorporates the specific heat ratio γ and the ideal gas law, resulting in the expression for temperature change being proportional to the kinetic energy per unit mass.

Question 41

The temperature of an open room of volume 30 m³ increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 10⁵ Pa. If n_i and n_f are the number of molecules in the room before and after heating, then n_f − n_i will be:

Accepted Answer

−2.5 × 10²⁵. As the temperature increases, the volume of gas remains constant, leading to an increase in pressure. However, since the room is open to the atmosphere, the gas can escape, resulting in a decrease in the number of molecules, hence n_f - n_i is negative.

Question 42

Cₚ and C_v are specific heats at constant pressure and volume respectively. It is observed that
Cₚ − C_v = a for hydrogen gas
Cₚ − C_v = b for nitrogen gas
The correct relation between a and b is:

Accepted Answer

a = 14 b. The specific heat difference for ideal gases is related to the degrees of freedom of the gas molecules. Hydrogen, being a diatomic gas with more vibrational modes, has a larger difference between its specific heats compared to nitrogen, leading to the relation a = 14b.

Question 43

Two moles of an ideal gas with Cₚ/C_v = 5/3 are mixed with 3 moles of another ideal gas with Cₚ/C_v = 4/3. The value of Cₚ/C_v for the mixture is:

Accepted Answer

1.42. Cv1 = R/(5/3-1) = 1.5R, Cv2 = R/(4/3-1) = 3R. Cv_mix = (2*1.5R + 3*3R)/5 = 2.4R. Cp_mix = Cv_mix + R = 3.4R. So Cp/Cv = 3.4/2.4 = 1.42.

Question 44

Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u = U/V ∝ T⁴ and pressure p = (1/3)(U/V). If the shell now undergoes an adiabatic expansion the relation between T a

Accepted Answer

T ∝ 1/R. During an adiabatic expansion of the photon gas, the internal energy decreases as the volume increases. Since the internal energy is proportional to the temperature raised to the fourth power (u ∝ T⁴), and the volume is proportional to the cube of the radius (V ∝ R³), the relationship between temperature and radius simplifies to T ∝ 1/R.

Question 45

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during the process the relation of pressure P and volume V is given by PVⁿ = constant, then n is given by (Here C_P and C_V are molar specific heat at constant pressure and con

Accepted Answer

n = (C - C_P) / (C - C_V). The correct option relates the heat capacity C to the specific heat capacities at constant pressure and volume, showing how the heat capacity influences the relationship between pressure and volume during the process. This formulation arises from the definitions of heat capacities and their roles in the thermodynamic behavior of the gas.

Question 46

Two moles of helium are mixed with n with moles of hydrogen. If C_P/C_V = 3/2 for the mixture, then the value of n is -

Accepted Answer

2. The ratio of specific heats, C_P/C_V, for an ideal gas mixture can be calculated using the contributions from each gas. Given that helium is monatomic and hydrogen is diatomic, the specific heat ratio for the mixture being 3/2 indicates that the proportion of hydrogen must be such that it balances the contributions from the two moles of helium, leading to n being equal to 2.

Question 47

Two Carnot engines A and B are operated in series. Engine A receives heat from a reservoir at 600 K and rejects heat to a reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 100 K. If the efficiencies of the two engines A and B a

Accepted Answer

7/12. Assuming the two series Carnot engines deliver equal work, the intermediate temperature is T = (600+100)/2 = 350 K. Then eta_A = 1 - 350/600 = 5/12 and eta_B = 1 - 100/350 = 5/7. So eta_A/eta_B = (5/12)/(5/7) = 7/12.

Question 48

A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:

Accepted Answer

420 J. For a Carnot refrigerator, W = Q_cold*(Th - Tc)/Tc. Q_cold = 500 cal = 500*4.186 = 2093 J. W = 2093*(300-250)/250 = 2093*0.2 = 418.6 J, approximately 420 J.

Question 49

One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27°C. The work done on the gas will be:

Accepted Answer

300R ln 2. For isothermal compression of 1 mole at T=300 K with pressure doubled, volume halves, so work done on the gas = nRT ln(P2/P1) = 300R ln2.

Question 50

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by ΔT. The amount of heat absorbed by gas is (R is gas constant):

Accepted Answer

1/2 R ΔT. For 1 mole monoatomic, dU = (3/2)R*dT. With V=K/T, PV=RT gives P=RT^2/K, and W = integral P dV = -R*dT. So Q = (3/2)R*dT - R*dT = (1/2)R*dT.

JEE Main Physics: Thermodynamics questions with solutions

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