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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}+Q After 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}+W In 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}+W It may be noted that ‘U_{1}’, ‘U_{2}’,‘Q’ and ‘W’ all are being taken in same units. So, Q = (U_{2} – U_{1}) + W When 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 Q is the energy transferred (as heat) between the system and environment, W is 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 Q to be positive when heat is transferred into the system and W to 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 for Q and +214 J for W in the equation ΔE_{int} = Q + W, ΔE_{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 ΔE_{int} is the change in the internal energy of the system. The heat capacity C of a body as the ratio of amount of heat energy Q transferred to a body in any process to its corresponding temperature change ΔT. C = Q/ΔT So, Q = C ΔT The 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/mΔT So, 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 is c_{w}, final temperature is T_{f} and initial temperature is T_{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 ΔE_{int} = 0 and W = 0 in the equation Q + W = ΔE_{int}, Q + W = ΔE_{int} Q + 0= 0 So, Q = 0 Or, Q_{w} + Q_{t} = 0 m_{w}c_{w} (T_{f} –T_{i})+ C_{t}ΔT_{t} = 0 So, T_{i} = (m_{w}c_{w} T_{f} + C_{t}ΔT_{t} )/ m_{w}c_{w} Here ΔT_{t} = 44.4 ^{°} C - 15.0 ^{°} C = 29.4 ^{°} C To obtain the temperature of the water before insertion T_{i} of the thermometer, substitute 0.3 kg for m_{w}, 4190 J/kg.m for c_{w}, 44.4 ^{°} C for T_{f}, 46.1 J/K for C_{t} and 29.4 ^{°} C for ΔT_{t} in the equation T_{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 ^{°} C From 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} = -nRT ln V_{2}/V_{1} As from ideal gas equation, p_{1}V_{1} = p_{2}V_{2} = nRT, thus W_{1} = -nRT ln V_{2}/V_{1} = - p_{1}V_{1 }ln p_{1}/p_{2} Here n is the number of moles, R is the gas constant, T is the temperature, V_{1} is the initial volume, V_{2} is the final volume, p_{1} is the initial pressure and p_{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 W on the gas will be equal to the sum of work done W_{1} by isothermal process and work done W_{2} at constant pressure during cooling. So, W = W_{1} + W_{2} = (- p_{1}V_{1 }ln p_{1}/p_{2}) + (V_{1} (p_{1}-p_{2})) To find out the work done W on the gas, substitute 204×10^{3} Pa for p_{1}, 0.142 m^{3} for V_{1}, 101×10^{3} Pa for p_{2} in the equation W = (- p_{1}V_{1 }ln p_{1}/p_{2}) + (V_{1} (p_{1}-p_{2})), W = (- p_{1}V_{1 }ln p_{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} J From the above observation we conclude that, work done W on 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 = -nRT ln V_{f} / V_{i} = -nRT ln V_{2}/ V_{1} …… (1) In accordance to ideal gas equation, PV = nRT …… (2) Here P is the pressure of the gas, V is the volume of the gas, n is the number of moles, r is the gas constant and T is the temperature of the gas. At constant temperature if the gas compress from its initial pressure P_{1}, intial volume V_{1} to its final pressure P_{2}, final volume V_{2}, then P_{1} V_{1} = P_{2} V_{2} = nRT …… (3) To find work W in terms of P_{1} and V_{1}, substitute P_{1} V_{1} for nRT in the equation W = -nRT ln V_{2} / V_{1} , W = -nRT ln V_{2}/ V_{1} = -P_{1} V_{1} ln V_{2}/ V_{1} Solution:- To obtain work done, substitute 1.32 atm for pressure P_{1}, 22.4 L for initial volume V_{1} and 15.3 L for initial volume V_{2} in the equation W = -P_{1} V_{1} ln V_{2}/ V_{1}, W = -P_{1} V_{1} ln V_{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} J From 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 the pV diagram in below figure. Concept:- Work is a path function. So work done on the gas depends upon the path. The area under pv diagram gives the work done on the gas between initial state i and final state f. Work done W is defined as, W = -pdV Here p is the pressure and dV is 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 J The 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 W on 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 A under 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 J The 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 W on 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:

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}+Q

After 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}+W

In 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}+W

It may be noted that ‘U_{1}’, ‘U_{2}’,‘Q’ and ‘W’ all are being taken in same units.

So, Q = (U_{2} – U_{1}) + W

When 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.

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 Q is the energy transferred (as heat) between the system and environment, W is 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 Q to be positive when heat is transferred into the system and W to be positive when work is done on the system.

(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.

Thus, Q = -293 J.

ΔE_{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.

Calculate (i) q (ii) w and (iii) ΔU

(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}

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?

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,

Here Q is the energy transferred between the system and its environment, W is the work done on or by the system and ΔE_{int} is the change in the internal energy of the system.

The heat capacity C of a body as the ratio of amount of heat energy Q transferred to a body in any process to its corresponding temperature change ΔT.

C = Q/ΔT

So, Q = C ΔT

The 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/mΔT

So, Q = c mΔT

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 is c_{w}, final temperature is T_{f} and initial temperature is T_{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 ΔE_{int} = 0 and W = 0 in the equation Q + W = ΔE_{int},

Q + 0= 0

So, Q = 0

Or, Q_{w} + Q_{t} = 0

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

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

Here ΔT_{t} = 44.4 ^{°} C - 15.0 ^{°} C

= 29.4 ^{°} C

To obtain the temperature of the water before insertion T_{i} of the thermometer, substitute 0.3 kg for m_{w}, 4190 J/kg.m for c_{w}, 44.4 ^{°} C for T_{f}, 46.1 J/K for C_{t} and 29.4 ^{°} C for ΔT_{t} in the equation T_{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 ^{°} C

From the above observation we conclude that, the temperature of the water before insertion of the thermometer was 45.5 ^{°} C.

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.

In an isothermal process the work done W_{1} is defined as,

W_{1} = -nRT ln V_{2}/V_{1}

As from ideal gas equation, p_{1}V_{1} = p_{2}V_{2} = nRT, thus

= - p_{1}V_{1 }ln p_{1}/p_{2}

Here n is the number of moles, R is the gas constant, T is the temperature, V_{1} is the initial volume, V_{2} is the final volume, p_{1} is the initial pressure and p_{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})

The total work done W on the gas will be equal to the sum of work done W_{1} by isothermal process and work done W_{2} at constant pressure during cooling.

So, W = W_{1} + W_{2}

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

To find out the work done W on the gas, substitute 204×10^{3} Pa for p_{1}, 0.142 m^{3} for V_{1}, 101×10^{3} Pa for p_{2} in the equation W = (- p_{1}V_{1 }ln p_{1}/p_{2}) + (V_{1} (p_{1}-p_{2})),

W = (- p_{1}V_{1 }ln p_{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} J

From the above observation we conclude that, work done W on the gas would be -5.74×10^{3} J.

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.

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 = -nRT ln V_{f} / V_{i}

= -nRT ln V_{2}/ V_{1} …… (1)

In accordance to ideal gas equation,

PV = nRT …… (2)

Here P is the pressure of the gas, V is the volume of the gas, n is the number of moles, r is the gas constant and T is the temperature of the gas.

At constant temperature if the gas compress from its initial pressure P_{1}, intial volume V_{1} to its final pressure P_{2}, final volume V_{2}, then

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

To find work W in terms of P_{1} and V_{1}, substitute P_{1} V_{1} for nRT in the equation W = -nRT ln V_{2} / V_{1} ,

W = -nRT ln V_{2}/ V_{1}

= -P_{1} V_{1} ln V_{2}/ V_{1}

To obtain work done, substitute 1.32 atm for pressure P_{1}, 22.4 L for initial volume V_{1} and 15.3 L for initial volume V_{2} in the equation W = -P_{1} V_{1} ln V_{2}/ V_{1},

W = -P_{1} V_{1} ln V_{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} J

From 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.

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 the pV diagram in below figure.

Work is a path function. So work done on the gas depends upon the path. The area under pv diagram gives the work done on the gas between initial state i and final state f.

Work done W is defined as,

W = -pdV

Here p is the pressure and dV is the change in volume.

(a) In the process 1, the work done W_{1} will be zero for vertical path.

So W_{1} = 0 J

The 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 W on the gas will be,

W = W_{1} + W_{2}

_{ }= 0 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 A under 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 J

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

W = W_{3} + W_{4}

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

From the above observation we conclude that, the net work done on the gas would be -20 J.

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