Solved Problems on Thermodynamics:-

Problem 1:-A container holds a mixture of three nonreacting gases: n

_{1}moles of the first gas with molar specific heat at constant volume C_{1}, and so on. Find the molar specific heat at constant volume of the mixture, in terms of the molar specific heats and quantitites of the three separate gases.

Concept:-Heat capacity

Cof a body as the ratio of the amount of heat energyQtransferred to a body in any process to its corresponding temperature change ΔT.

C=Q/ΔTSo,

Q=CΔTEach species will experience the equal temperature change.

If the gas has

nmolecules, thenQwill be,

Q=nCΔT

Solution:-If the gas has

n_{1}moles, then the amount of heat energyQ_{1}transferred to a body having heat capacityC_{1}will be,

Q_{1}=n_{1}C_{1}ΔTSimilarly, if the gas has

n_{2}moles, then the amount of heat energyQ_{2}transferred to a body having heat capacityC_{2}will be,

Q_{2}=n_{2}C_{2}ΔTAnd

if the gas has

n_{3}moles, then the amount of heat energyQ_{3}transferred to a body having heat capacityC_{3}will be,

Q_{3}=n_{3}C_{3}ΔTAs, each species will experience the same temperature change, thus,

Q=Q_{1}+Q_{2}+Q_{3}=

n_{1}C_{1}ΔT+n_{2}C_{2}ΔT+n_{3}C_{3}ΔTDividing both the sides by

n=n_{1}+n_{2}+n_{3}and ΔT, then we will get,

Q/nΔT= (n_{1}C_{1}ΔT+n_{2}C_{2}ΔT+n_{3}C_{3}ΔT)/nΔTAs,

Q/nΔT=C, thus,

C=ΔT(n_{1}C_{1}+n_{2}C_{2}+n_{3}C_{3})/nΔT= (

n_{1}C_{1}+n_{2}C_{2}+n_{3}C_{3})/n=

n_{1}C_{1}+n_{2}C_{2}+n_{3}C_{3}/n_{1}+n_{2}+n_{3}From the above observation we conclude that, the molar specific heat at constant volume of the mixture would be

n_{1}C_{1}+n_{2}C_{2}+n_{3}C_{3}/n_{1}+n_{2}+n_{3.}

Problem 2:-A thermometer of mass 0.055 kg and heat capacity 46.1 J/K reads 15.0°C. It is then completely immersed in 0.300 kg of water and it comes to the same final temperature as the water. If the thermometer reads 44.4°C, what was the temperature of the water berfore insertion of the thermometer, neglecting other heat losses?

Concept:-In accordance to the law of conservation of energy, for a thermodynamic system, in which internal is the only type of energy the system may have, the law of conservation of energy may be expressed as,

Q+W= ΔE_{int}Here Q is the energy transferred between the system and its environment, W is the work done on or by the system and Δ

E_{int}is the change in the internal energy of the system.The heat capacity

Cof a body as the ratio of amount of heat energyQtransferred to a body in any process to its corresponding temperature change ΔT.

C=Q/ΔTSo,

Q=CΔTThe heat capacity per unit mass of a body, called specific heat capacity or usually just specific heat, is characteristic of the material of which the body is composed.

c=C/m=

Q/mΔTSo,

Q=c mΔT

Solution:-The heat transfers for the water

Q_{w}is,

Q_{w}=m_{w}c_{w}(T_{f}–T_{i})Here, mass of water is

m_{w}, specific heat capacity of water isc_{w}, final temperature isT_{f}and initial temperature isT_{i}.The heat transfers for the thermometer

Q_{t}is

Q_{t}=C_{t}ΔT_{t}Here, heat capacity of thermometer is

C_{t}and ΔT_{t}is the temperature difference.As the internal energy of the system is zero and there is no work is done, therefore substitute Δ

E_{int}= 0 andW= 0 in the equationQ+W= ΔE_{int},

Q+W= ΔE_{int}

Q+ 0= 0So,

Q= 0Or,

Q_{w}+Q_{t}= 0

m_{w}c_{w}(T_{f}–T_{i})+C_{t}ΔT_{t}= 0So,

T_{i}= (m_{w}c_{w}T_{f}+C_{t}ΔT_{t})/m_{w}c_{w}Here Δ

T_{t}= 44.4^{̊}C - 15.0^{̊}C= 29.4

^{̊}CTo obtain the temperature of the water before insertion

T_{i}of the thermometer, substitute 0.3 kg form_{w}, 4190 J/kg.m forc_{w}, 44.4^{̊}C forT_{f}, 46.1 J/K forC_{t}and 29.4^{̊}C for ΔT_{t}in the equationT_{i}= (m_{w}c_{w}T_{f}+C_{t}ΔT_{t})/m_{w}c_{w,}

T_{i}= (m_{w}c_{w}T_{f}+C_{t}ΔT_{t})/m_{w}c_{w}= [(0.3 kg) (4190 J/kg.m) (44.4

^{̊}C) + (46.1 J/K) (29.4^{̊}C)] /[(0.3 kg) (4190 J/kg.m)]=45.5

^{̊}CFrom the above observation we conclude that, the temperature of the water before insertion of the thermometer was 45.5

^{̊}C.

Problem 3:-A mixture of 1.78 kg of water and 262 g of ice at 0°C is, in a reversible process, brought to a final equilibrium state where the water / ice ratio, by mass 1:1 at 0°C. (a) Calculate the entropy change of the system during this process. (b) The system is then returned to the first equilibrium state, but in an irreversible way (by using a Bunsen burner, for instance). Calculate the entropy change of the system during this process. (c) Show that your answer is consistent with the second law of thermodynamics.

Concept:-The entropy change Δ

Sfor a reversible isothermal process is defined as,Δ

S=Q/T= -

mL/THere

mis the mass,Lis the latent heat andTis the temperature.

Solution:-(a) Mass of water = 1.78 kg

Mass of ice = 262 g

So the total mass of ice and water mixture will be,

Mass of ice-water mixture = (Mass of water) + (Mass of ice)

= (1.78 kg) + (262 g)

= (1.78 kg) + (262 g×10

^{-3}kg/1 g)= 1.78 kg + 0.262 kg

= 2.04 kg

If eventually the ice and water have the same mass, then the final state will have 1.02 kg (2.04 kg/2) of each.

Thus the mass of the water that changed into ice

mwill be the difference of mass of waterm_{w }and mass of final statem_{s}.So,

m=m_{w}-m_{s}To obtain mass of water that changed into ice

m, substitute 1.78 kg for mass of waterm_{w}and 1.02 kg for mass of final statem_{s}in the equationm=m_{w}-m_{s},

m=m_{w}-m_{s}= 1.78 kg – 1.02 kg

= 0.76 kg

The change of water at 0

^{̊}C to ice at 0^{̊}C is isothermal.To obtain the change in entropy Δ

Sof the system during this process, substitute 0.76 kg for massm, 333×10^{3}J/kg for heat of fusion of waterLand 273 K forTin the equation ΔS= -mL/T,Δ

S= -mL/T= -(0.76 kg) (333×10

^{3}J/kg )/(273 K)= -927 J/K

From the above observation we conclude that, the change in entropy Δ

Sof the system during this process will be -927 J/K.(b) Now the system is returned to the first equilibrium state, but in an irreversible way. Thus the change in entropy Δ

Sof the system during this process is equal to the negative of previous case.So, Δ

S= -(- 927 J/K)= 927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system would be 927 J/K.

(c) In accordance to second law of thermodynamics, entropy change Δ

Sis always zero.The total change in entropy will be,

Δ

S= (-927 J/K) + (927 J/K)= 0

From the above observation we conclude that, our answer is consistent with the second law of thermodynamics.

Problem 4:-Apparatus that liquefies helium is in a laboratory at 296 K. The helium in the apparatus is at 4.0 K. If 150 mJ of heat is transferred from the helium, find the minimum amount of heat delivered to the laboratory.

Concept:-Coefficient of performance

Kof a Carnot refrigerator is defined as,

K=T_{L}/T_{H}-T_{L}…… (1)Here

T_{L}is the lower temperature of sink andT_{H}is the higher temperature of source.A refrigerator would like to extract as much heat

Q_{L}as possible from the low-temperature reservoir (“what you want”) for the least amount of workW(“what you pay for”). So the efficiency of a refrigerator is defined as,

K= (what you want)/(what you pay for)=

Q_{L}/Wand this is called coefficient of performance. The larger the value of

K, the more efficient is the refrigerator.Thus,

W=Q_{L}/K…… (2)Substitute the value of

Kfrom equation (1) in the equationW=Q_{L}/K,

W=Q_{L}/K=

Q_{L}/(T_{L}/T_{H}-T_{L})=

Q_{L}(T_{H}/T_{L}– 1) …… (3)The first law of thermodynamics, applied to the working substance of the refrigerator, gives,

W=Q_{H}–Q_{L}Here

Q_{H}is the exhausted heat.Thus exhausted heat will be,

Q_{H}=W+Q_{L}…… (4)Substitute the value of

Wfrom equation (3) in the equationQ_{H}=W+Q_{L},

Q_{H}=W+Q_{L}=

Q_{L}(T_{H}/T_{L}– 1) +Q_{L}=

Q_{L}(T_{H}/T_{L})

Solution:-To obtain the minimum amount of heat delivered to the laboratory, substitute 150 mJ for

Q_{L}, 296 K forT_{H}and 4.0 K forT_{L}in the equationQ_{H}=Q_{L}(T_{H}/T_{L}),

Q_{H}=Q_{L}(T_{H}/T_{L})= ((150 mJ) (10

^{-3}J/1 mJ)) (296 K/4.0 K)= 11 J

From the above observation we conclude that, the minimum amount of heat delivered to the laboratory would be 11 J.

Related Resources:You might like to refer some of the related resources listed below:Click here for the Detailed Syllabus of IIT JEE Physics.

Look into the Sample Papers of Previous Years to get a hint of the kinds of questions asked in the exam.