NCERT Solutions: Thermodynamics
Q11.1: A geyser heats water flowing at the rate of 3.0 liters per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g?
Ans: Volume of water heated = 3.0 litre per minute
Mass of water heated, m = 3000g per minute
Increase in temperature, ΔT - 77°C - 27°C = 50°C
Specific heat of water, c = 4.2 Jg-1°C-1
amount of heat used, Q = mc ΔT
or
Heat required per minute, Q = m c ΔT = 3000 × 4.2 × 50 = 630000 J per minute.
Heat of combustion of fuel = 4.0 × 104 J g-1.
Rate of consumption of fuel = Q / (heat per gram) = 630000 / (4.0 × 104) = 15.75 g per minute ≈ 15.8 g per minute.
Rate of combustion of fuel
Therefore, the fuel is consumed at about 15.8 g per minute to supply the required heat.
Q11.2: What amount of heat must be supplied to 2.0 × 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.)
Ans: Here, mass of gas,
rise in temperature,
Heat required.
Molecular mass, M = 28
Number of moles,
As nitrogen is a diatomic gas, molar specific heat at constant pressure is
As
∴
Enhanced answer (steps and numerical evaluation):
Mass of nitrogen = 2.0 × 10-2 kg = 0.02 kg = 20 g.
Molar mass M = 28 g mol-1 = 0.028 kg mol-1.
Number of moles, n = mass / M = 0.02 / 0.028 = 0.7143 mol.
For a diatomic gas at ordinary temperatures, molar heat capacity at constant pressure, Cp = (7/2) R = 3.5 × 8.3 = 29.05 J mol-1 K-1.
Temperature rise ΔT = 45 K.
Heat required, Q = n Cp ΔT = 0.7143 × 29.05 × 45 = 933.8 J ≈ 9.34 × 102 J.
Thus, about 9.34 × 102 J of heat must be supplied.
Q11.3: Explain why
(a) Two bodies at different temperatures T1 and T2, if brought in thermal contact, do not necessarily settle to the mean temperature (T1 + T2)/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Ans: (a) When two bodies are put in thermal contact, heat flows from the hotter to the colder body until they reach the same temperature. The final temperature will equal the arithmetic mean (T1 + T2)/2 only if the two bodies have equal thermal capacities (that is, equal products of mass and specific heat). If their masses or specific heats differ, the body with the larger thermal capacity shifts the equilibrium temperature closer to its initial temperature.
(b) A coolant should have a high specific heat because such a liquid can absorb a large amount of heat for a small rise in temperature. This property allows the coolant to remove and store heat effectively, keeping machinery or reactors at safe operating temperatures without large temperature swings.
(c) While driving, tyres undergo deformation and friction against the road; moreover, air inside the tyre is compressed repeatedly in places. The mechanical work done against friction and deformation is converted into heat, which raises the temperature of the tyre and the air inside it. For a given amount of gas at higher temperature, pressure increases (for nearly constant volume), so tyre pressure rises during driving.
(d) Large bodies of water, like the sea at a harbour, have a high heat capacity and so heat up and cool down slowly. They absorb heat in daytime or summer and release it at night or in winter, moderating the air temperature near the coast. In contrast, desert surfaces (sand and rock) have low heat capacity and change temperature quickly, producing much larger daily and seasonal temperature variations. Hence a harbour town has a more temperate climate than an inland desert town at the same latitude.
Q11.4: A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Ans: Here the process is adiabatic compression and
atm and for hydrogen (a diatomic gas) γ = 1.4.
⇒ P2 = (2)1.4 atm = 2.639 atm,
Enhanced explanation and steps:
For an adiabatic process PVγ = constant, so P2/P1 = (V1/V2)γ.
Given V2 = V1/2, therefore V1/V2 = 2.
For hydrogen γ ≈ 1.4, hence P2/P1 = 21.4 ≈ 2.639.
Thus the pressure increases by a factor of about 2.64.
Q11.5: In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Ans: Here, when the change is adiabatic,
If
is change in internal energy of the system, then
In the second case.
ΔW = ?
As
+ ΔW = ΔQΔW = ΔQ -
= 39.3 -22.3 - 17.0 J.
Enhanced answer (clear signs and arithmetic):
When the process A → B is adiabatic, ΔQ = 0. Using the first law, ΔQ = ΔU + Wby, where Wby is work done by the system.
For the adiabatic path 0 = ΔU + Wby,ad. We are told an amount of 22.3 J is done on the system, so Wby,ad = -22.3 J (negative because work is done on the system). Thus ΔU = -Wby,ad = 22.3 J. The internal energy of the final state B is higher by 22.3 J than state A.
For the second (non-adiabatic) path, heat absorbed ΔQ = 9.35 cal = 9.35 × 4.19 = 39.2 J (approximately).
Use ΔQ = ΔU + Wby. We know ΔU = 22.3 J, so Wby = ΔQ - ΔU = 39.2 - 22.3 = 16.9 J ≈ 16.9 J.
Therefore, the net work done by the system in the latter case is about 16.9 J (approximately 17.0 J).
Q 11.6: Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?
Ans: (a) Since the external temperature and initial temperature remain the same,
P2V2 = P1V1
But P1 = 1 atm, V1 = V, V2 = 2V and P2 - ?
(b) Since the temperature of the system remains unchanged, change in internal energy is zero.
(c) The system being thermally insulated, there is no change in temperature (because of free expansion)
(d) The expansion is a free expansion. Therefore, the intermediate states are non equilibrium states and the gas equation is not satisfied in these states. As a result, the gas can not return to an equilibrium state which lie on the P-V-T surface.
Enhanced and clarified answers:
(a) The final pressure is half the initial pressure. When the two equal volumes are connected without heat exchange and no external work, the gas expands freely into the evacuated cylinder; final volume is 2V, so Pf = Pi × (V/Vf) = Pi/2 = 0.5 atm.
(b) For an ideal gas, internal energy depends only on temperature. Since the free expansion is adiabatic and no work is done on or by the gas externally, the temperature does not change; therefore ΔU = 0.
(c) There is no change in temperature for an ideal gas undergoing free expansion (ΔT = 0).
(d) The intermediate states during the sudden opening of the stopcock are not equilibrium states; they are irreversible and cannot be described by the equilibrium P-V-T relation. Only the initial and final equilibrium states lie on the equilibrium P-V-T surface.
Q11.7: An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Ans: According to law of conservation of energy
Total energy = work done + internal energy
ΔQ = ΔW + ΔU
Here,
Rate at which heat is supplied ΔQ = 100 W
Rate at which work is done ΔW = 75 Js-1
Rate of change of internal energy is ΔU
ΔU = ΔQ - ΔW
ΔU = 100 - 75
We get,
ΔU = 25 J s-1
Hence,
The internal energy of the system is increasing at a rate of 25 W.
Enhanced concise explanation:
Using the first law in rate form, dU/dt = dQ/dt - dW/dt = 100 - 75 = 25 J s-1.
Thus the internal energy increases at 25 W.
Q11.8: A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Ans: As is clear from Fig.
Change in pressure, ΔP = EF = 5.0 - 2-0 " 3-0 atm = 3.0 * 105 Nm-2
Change in volume, ΔV=DF = 600 - 300 = 300 cc
= 300 * 10-6 m3
Work done by the gas from D to E to F = area of ΔDEF
Enhanced calculation:
Convert the pressure difference to SI units: ΔP = 3.0 atm = 3.0 × 105 N m-2.
Change in volume ΔV = 300 cm3 = 300 × 10-6 m3 = 3.0 × 10-4 m3.
The shaded triangular area ΔDEF on the P-V diagram equals (1/2) ΔP × ΔV.
Work = 0.5 × (3.0 × 105) × (3.0 × 10-4) = 0.5 × 90 = 45 J.
Therefore, the total work done by the gas in going from D → E → F is 45 J.
Old NCERT Solutions
Q1. A steam engine delivers 5.4×108 J of work per minute and services 3.6 × 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?Ans. Work done by the steam engine per minute, W = 5.4 × 108 J
Heat supplied from the boiler, H = 3.6 × 109 J
Efficiency η = W / H = (5.4 × 108) / (3.6 × 109) = 0.15 = 15%
Heat energy wasted per minute = Heat supplied - Useful work = H - W = 3.6 × 109 - 5.4 × 108 = 3.06 × 109 J per minute.
Since ΔQ = + ΔW
∴ Change in internal energy,
= ΔQ - ΔW= 100 - 75 = 25 J/s.
Enhanced concise presentation:
Efficiency = 15%; heat wasted per minute = 3.06 × 109 J.
Q2. A refrigerator is to maintain eatables kept inside at 9 °C, if room temperature is 36 °C. Calculate the coefficient of performance.
Ans. Here, T1 = 36°C = (36 + 273) K = 309 K
T2 = 9°C = (9 + 273) K = 282 K
Coefficient of performance,
Enhanced calculation:
For a reversible refrigerator, COP = Tcold / (Thot - Tcold).
Here Thot = 309 K and Tcold = 282 K.
COP = 282 / (309 - 282) = 282 / 27 = 10.444... ≈ 10.44.
Therefore, the coefficient of performance is approximately 10.44.
FAQs on NCERT Solutions: Thermodynamics
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