The First Law of Thermodynamics:-The first law of thermodynamics states that, ”

If the quantity of heat supplied to a system is capable of doing work, then the quantity of heat absorbed by the system is equal to the sum of the increase in the internal energy of the system, and the external work done by it.”

Mathematical Expression of First Law:-Consider some gas enclosed in a barrel having insulating walls and conducting bottom. Let an amount of heat ‘Q’ be added to the system through the bottom. If ‘U

_{1}’ is the initial energy of the system, then,Total energy of the system in the beginning = U

_{1}+QAfter gaining heat the gas tends to expand, pushing the piston from A to B as shown in below figure. As a result of this, some work ‘W’ is done by the gas. The work is external work, since the system undergoes a displacement. If ‘U

_{2}’ is final internal energy of the system, then,Total energy of the system at end = U

_{2}+WIn accordance to the law of conservation of energy, total energy of the system in the beginning will be equal to total energy of the system at end.

So, U

_{1}+Q = U_{2}+WIt may be noted that ‘U

_{1}’, ‘U_{2}’,‘Q’ and ‘W’ all are being taken in same units.So, Q = (U

_{2}– U_{1}) + WWhen infinitesimal amount of heat ‘dQ’ is added to the system, corresponding changes in internal energy ‘dQ’ and external work done ‘dW’ are so small.

Then, dQ = dU + dW

Or, dQ = dU+ pdV

This is mathematical statement of the first law of thermodynamics.

Therefore, first law of thermodynamics signifies that, “energy can neither be created nor destroyed, but it can only be transformed from one form to another”.

Thus, dQ= dU+dW

If work is done by the surroundings on the system (as during the compression of a gas), W is taken as positive so that dQ = dU+W. if however work is done by the system on the surroundings (as during the expansion of a gas), W is taken as negative so that dQ = dU – dW.

Refer This Video:-

Problem 1:-Consider that 214 J of work done on a system, and 293 J of heat are extracted from the system. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of (a) W, (b) Q, and (c) ∆E

_{n}?

Concept:-For a thermodynamic system, in which internal energy 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

Qis the energy transferred (as heat) between the system and environment,Wis the work done on (or by) the system and ΔE_{int}is the change in the internal energy of the system.By convention we have chosen

Qto be positive when heat is transferred into the system andWto be positive when work is done on the system.

Solution:-(a) Since work is done on the system, therefore algebraic sign of the work done will be positive and the magnitude of work done is 214 J.

Thus,

W= +214 J.(b) Since heat is extracted from the system, therefore algebraic sign of the heat will be negative and the magnitude of heat is 293 J.Thus,

Q= -293 J.(c) To obtain internal energy of the system, substitute -293 J forQand +214 J forWin the equation ΔE_{int}=Q+W,Δ

E_{int}=Q+W= (-293 J) + (+214 J)

= -79.0 J

From the above observation we conclude that, the internal energy of the system would be -79.0 J.

Problem 2:1 mole of ideal monoatomic gas at 27°C expands adiabatically against a constant external pressure of 1.5 atm from a volume of 4dm^{3}to 16 dm^{3}.Calculate (i) q (ii) w and (iii) ΔU

Solution:-(i) Since process is adiabatic q = 0(ii) As the gas expands against the constant external pressure.

W = –PΔV = –1.6(V

_{2}–V_{1})= –1.5 (16–4) = –18 atm dm

^{3}(iii) ΔU = q + w = 0 + (–18) = –18 atm dm

^{3}

Problem 3: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 before 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 4:-Air that occupies 0.142 m

^{3}at 103 kPa gauge pressure is expanded isothermally to zero gauge pressure and then cooled at constant pressure until it reaches its initial volume. Compute the work done on the gas.

Concept:-In an isothermal process the work done

W_{1}is defined as,

W_{1}= -nRTlnV_{2}/V_{1}As from ideal gas equation,

p_{1}V_{1}=p_{2}V_{2}=nRT, thus

W_{1}= -nRTlnV_{2}/V_{1}= -

p_{1}V_{1 }lnp_{1}/p_{2}Here

nis the number of moles,Ris the gas constant,Tis the temperature,V_{1}is the initial volume,V_{2}is the final volume,p_{1}is the initial pressure andp_{2}is the final pressure.The work done

W_{2}at constant pressure,

W_{2}= -p_{2}ΔV= -

p_{2}(V_{1}-V_{2}) (Since, ΔV=V_{1}-V_{2})= -

p_{2}V_{1}(1-p_{1}/p_{2})=

V_{1}(p_{1}-p_{2})

Solution:-The total work done

Won the gas will be equal to the sum of work doneW_{1}by isothermal process and work doneW_{2}at constant pressure during cooling.So,

W=W_{1}+W_{2}= (-

p_{1}V_{1 }lnp_{1}/p_{2}) + (V_{1}(p_{1}-p_{2}))To find out the work done

Won the gas, substitute 204×10^{3}Pa forp_{1}, 0.142 m^{3}forV_{1}, 101×10^{3}Pa forp_{2}in the equationW= (-p_{1}V_{1 }lnp_{1}/p_{2}) + (V_{1}(p_{1}-p_{2})),

W= (-p_{1}V_{1 }lnp_{1}/p_{2}) + (V_{1}(p_{1}-p_{2}))= [-(204×10

^{3}Pa) (0.142 m^{3}) ln (204×10^{3}Pa)/( 101×10^{3}Pa) ] + [(0.142 m^{3}) ((204×10^{3}Pa) - (101×10^{3}Pa))]= -5.74×10

^{3}Pa. m^{3}= (-5.74×10

^{3}Pa. m^{3}) (1 J/1 Pa.m^{3})= -5.74×10

^{3}JFrom the above observation we conclude that, work done

Won the gas would be -5.74×10^{3}J.

Problem 5:-Calculate the work done by an external agent in compressing 1.12 mol of oxygen from a volume of 22.4 L and 1.32 atm pressure to 15.3 L at the same temperature.

Concept:-In an isothermal process the work done by an external agent in compressing gas from its initial volume (

V_{i }=V_{1}) to its final volume (V_{f}=V_{2}) is,

W= -nRTlnV_{f}/V_{i}= -

nRTlnV_{2}/V_{1}…… (1)In accordance to ideal gas equation,

PV=nRT…… (2)Here

Pis the pressure of the gas,Vis the volume of the gas,nis the number of moles,ris the gas constant andTis the temperature of the gas.At constant temperature if the gas compress from its initial pressure

P_{1}, intial volumeV_{1}to its final pressureP_{2}, final volumeV_{2}, then

P_{1}V_{1}=P_{2}V_{2}=nRT…… (3)To find work

Win terms ofP_{1}andV_{1}, substituteP_{1}V_{1}fornRTin the equationW= -nRTlnV_{2}/V_{1},

W= -nRTlnV_{2}/V_{1}= -

P_{1}V_{1}lnV_{2}/V_{1}

Solution:-To obtain work done, substitute 1.32 atm for pressure

P_{1}, 22.4 L for initial volumeV_{1}and 15.3 L for initial volumeV_{2}in the equationW= -P_{1}V_{1}lnV_{2}/V_{1},

W= -P_{1}V_{1}lnV_{2}/V_{1}= -(1.32 atm) (1.01×10

^{5}Pa / 1 atm) (22.4 L) (10^{-3}m^{3}/ 1 L) ln ((15.3 L) (10^{-3}m^{3}/1 L) /(22.4 L) (10^{-3}m^{3}/ 1 L))= 1.14×10

^{3}Pa. m^{3}= (1.14×10

^{3}Pa. m^{3}) ( 1 J/1 Pa. m^{3})= 1.14×10

^{3}JFrom the above observation we conclude that, the work done by an external agent in compressing the oxygen molecule would be 1.14×10

^{3}J.

Problem 6:-A sample of gas expands from 1.0 to 5.0 m

^{3}while its pressure decreases from 15 to 5.0 Pa. How much work is done on the gas if its pressure changes with volume according to each of the three processes shown in thepVdiagram in below figure.

Concept:-Work is a path function. So work done on the gas depends upon the path. The area under

pvdiagram gives the work done on the gas between initial stateiand final statef.Work done

Wis defined as,

W= -pdVHere

pis the pressure anddVis the change in volume.

Solution:-(a) In the process 1, the work done

W_{1}will be zero for vertical path.So

W_{1}= 0 JThe work done is present only in horizontal path.

Thus the work done

W_{2}for horizontal path will be,

W_{2}= -pdV(negative sign is due to work is done on the gas= - (15 Pa) (4m

^{3})= -60 Pa.m

^{3}

^{ }= (-60 Pa.m^{3}) (1 J/1 Pa.m^{3})= -60 J

Therefore the net work done

Won the gas will be,

W=W_{1}+W_{2}

_{ }= 0 J+(-60 J)= -60 J

From the above observation we conclude that, the net work done on the gas would be -60 J.

(b) For the process 2, work done is negative of the area under the curve.

So the area

Aunder the curve for process 2 will be,

A= ½ (15 Pa+5 Pa) (4m^{3})= 40 Pa.m

^{3}

^{ }= (40 Pa.m^{3}) (1 J/1 Pa.m^{3})= 40 J

As for the process 2, work done is negative of the area under the curve; therefore the net work done on the gas will be -40 J.

(c) In the process 3, the work done

W_{3}will be zero for vertical path.So

W_{3}= 0 JThe work done is present only in horizontal path.

Thus the work done

W_{4}for horizontal path will be,

W_{4}= -pdV(negative sign is due to work is done on the gas)= - (5 Pa) (4m

^{3})= -20 Pa.m

^{3}

^{ }= (-20 Pa.m^{3}) (1 J/1 Pa.m^{3})= -20 J

Therefore the net work done

Won the gas will be,

W=W_{3}+W_{4}

_{ }= 0 J+(-20 J)= -20 J

From the above observation we conclude that, the net work done on the gas would be -20 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.
- You can get the knowledge of Useful Books of Physics.

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