First Law of Thermodynamics

103. The first law of thermodynamics was developed by-
(a) Joule (b) Kelvin

RPSC AE 2018

UKPSC AE-2013, Paper-II

Ans. (a) : The first law of thermodynamics was developed by Joule.

First law of thermodynamics—When a closed system executes a complete cycle the sum of heat interactions is equal to the sum of work interaction. ∑Q = ∑W

The summations being over the entire cycle.

104. Joules law states the specific internal energy of a gas depends only on

(a) the pressure of the gas

(b) the volume of the gas

(d) pressure and volume of the gas

TNPSC AE 2018

Ans. (c) : Joules law states the specific internal energy of a gas depends only on the temperature of the gas. u = f (T) only

105. Internal energy of an isolated system :

(a) Increases with addition on heat

(b) Increases with increase in mass of the system

(d) Increases with addition of work

OPSC Civil Services Pre. 2011

Ans. (c) : Internal energy of an isolated system remains same.

According to first law of thermodynamics

δQ = dU + δW

For isolated system

δW = 0

δQ = 0

dU = 0

Internal energy of perfect gas is only function of temperature only.

106. The identity δQ = dU + PdV is valid for :

(a) Any process occurring in an open system

(b) Any process occurring in a closed system

(d) Any process in any system

OPSC Civil Services Pre. 2011

Ans. (c) : δQ = dU + PdV

This equation is valid for closed system, quasi-static process without dissipation in a closed cycle and reversible process.

107. Heat transferred to a closed stationary system constant volume is equal to

(a) work transfer

(b) increased in internal energy

(d) increase in gibbs function

RPSC INSP. OF FACTORIES AND BOILER 2016

Ans : (b) δQ = dU + δW

for constant volume close system

W = 0

Hence, δQ = dU

108. The short coming of first law of thermodynamics is

(a) Direction of process

(b) Possibility of process

(d) Quantity of energy

JPSC AE PRE 2019

Ans. (d) : The short coming of first law of thermodynamics is quantity of energy.

• The short coming of second law of thermodynamics is quality of energy.

109. In a steady flow process, across the control volume mass and energy flow

(a) Varies continuously

(b) Remain constant

(d) Depends on type of process

JPSC AE PRE 2019

Ans. (b) : Assumptions made in steady flow processes are

(i) Control volume moves relative to the coordinate frame.

(ii) The state of the mass at each discrete area of flow on the control surface does not vary with time. The rate at which heat and work cross the control surface remain constant.

110.During throttling process :

(a) internal energy does not change

(b) pressure does not change

(d) enthalpy does not change

UKPSC AE-2013, 2007 Paper-II

RPSC INSP. OF FACTORIES AND BOILER 2016

OPSC AEE 2019 PAPER - II

Gujarat PSC AE 2019

(KPSC AE. 2015)

UPRVUNL AE 2014

Ans.(d) During throttling process enthalpy does not change.

Throttling process:- Steam is said to be throttled when it passes through a restricted opening such as a narrow aperture or a slightly opened valve. The leakage of a fluid through a crack in the vessel is an example of throttling.

It may be noted that throttling process:-

(i) No heat is supplied or rejected (1Q2 = 0)

(ii) No Work is done by the expending fluid (1W2=0)

(iii) No Change in the internal energy (du=0)

(iv) The enthalpy by or total heat of the fluid remains constant (h1=h2)

111. During an isothermal expansion process of a gas :

(a) pressure remains constant

(b) temperature remains constant

(d) none of the above

(HPPSC LECT. 2016)

Ans : (b) During an isothermal expansion process of a gas temperature remains constant.

112. An ideal gas is filled in a balloon kept in an evacuated and insulated room When the balloon ruptures, the gas fills up the entire room. Now internal energy of gas.......... and the enthalpy of gas.......... at the end of this process.

(a) increases, increases

(b) constant, decreases

(d) decreases, increases

(HPPSC LECT. 2016)

Ans : (c) An ideal gas is filled in a balloon kept in an evacuated and insulated room. when the balloon ruptures, the gas fills up the entire room. Now internal energy of gas constant and the enthalpy of gas constant at the end of this process.

Internal energy of gas (dU) = mcvdt

Enthalpy of gas (dH) = mcpdt

Internal energy and Enthalpy of an ideal gas is a function of temperature

dU = f (T)

dH = f (T)

113. The change in enthalpy of a closed system is equal to the heat transferred, if the reversible process takes place at constant

(a) Temperature

(b) Pressure

(d) Entropy

RPSC Vice Principal ITI 2018

Ans. (b) : From Tds equation,

Tds = dh – VdP Tds = dQ (for reversible process)

dQ = dh when, dP = 0

Hence for constant pressure process, dQ = dh

114. Gas expands for a definite volume in a closed vessel. The maximum work will be done when the process is at constant

(a) Volume

(b) Temperature

(d) Enthalpy

UPPSC AE 12.04.2016 Paper-II

Ans : (c) Gas expands for a definite volume in a closed vessel. The maximum work will be done when the process is at constant pressure.

Work done during a Non-flow process:-

Work done for non-flow process from state 1 to state 2 From above, we see that the work done is given by the area under the P - V diagram.

115. During mixing of steams in case of air conditioning, the process is associated with:

(a) Throttling

(b) Adiabatic

(d) Isochoric

UPPSC AE 12.04.2016 Paper-II

Ans : (b) During mixing of steams in case of air conditioning, the process is associated with adiabatic.

116. The heat absorbed or rejected during a polytropic process is equal to :

MPPSC AE 2016

Ans : (d)

Heat supplied or heat transferred

work done during polytropic expension

117. What is the rise in temperature of 80 litres of water in 40 min by a heater of 2 kW. If whole of the heater energy used to raise the water temperature?

(a) 14.3ºC (b) 1.43ºC

OPSC AEE 2019 PAPER - II

Ans : (a) :Heat taken by water = Heat given by heater

Heat (Q) = mCp ∆T = Power × time

80 × 4.2 ×∆T = 2× 40 × 60

= 14.28ºC

≈ 14.3ºC

118. In a reversible isothermal expansion process,the fluid expands from 10 bar and 2 m^3 to 2 bar and 10 m^3. During this expansion process, 100 kW of heat is supplied. Then the work done during the process is

(a) 33.3 kW (b) 80 kW

BPSC AE Mains 2017 Paper - V

Ans : (c) : Assuming ideal gas and reversible isothermal process,

δQ = dU + δW

δQ = δW

for isothermal and ideal gas

U = f(T)

As heat transfer is 100 kW is given, work will also be equal to 100 kW.

119.In an adiabatic process 5000 J of work is performed on a system. The system returns to its original state while 1000 J of heat is added. The work done during non adiabatic process.

(a) +4000J (b) –4000J

RPSC INSP. OF FACTORIES AND BOILER 2016

Ans : (c) Adiabatic Process 1–2

1Q2 = 0

1Q2 – 1W2 = 1U2

0 – (–5000) = 1U2

U2 – U1 = 5000 J

Non Adiabatic Process 2–1

2Q1 – 2W1 = 2U1 = U1 – U2

= – (U2 – U1)

– 1000 – 2W1 = – 5000

2W1 = 6000 J

120. In a heat exchanger, 50 kg of water is heatedper minute from 50°C to 110°C by hot gases which enter the heat exchanger at 250°C. The value of Cp for water is 4.186 kJ/kg.K and for air is 1 kJ/kg.K. If the flow rate of gases is 100 kg/min, the net change of enthalpy of air will be nearly

(a) 17.6 MJ/min (b) 15.0 MJ/min

ESE 2020

121. A system executes a cycle during which there are four heat transfers Q12 = 220 kJ,

Q23 = -25 kJ, Q34 = -180 kJ, Q41 = 50 kJ. The work during tube of the processes is

W12 = 15 kJ, W23 = -10 kJ, W34 = 60 kJ. The work done during the process 4 – 1 is

(a) –230 kJ (b) 0 kJ

TNPSC 2019

Ans. (b) : According to first law of thermodynamics

ΣQ = ΣW [for cyclic process]

Q1-2 + Q2-3 + Q3-4 + Q4-1 = W1-2 + W2-3 + W3-4 + W4-1

220 + (-25) + (-180) + 50 = 15 + (-10) + 60 + W4-1

W4-1 = 0kJ

122. Consider the following devices :

1. Internal combustion engine working on Otto cycle

2. Internal combustion engine working on Diesel cycle

3. Gas turbines

4. Steam turbines

In which of the above devices, the equation dQ = dU + Pdv is not applicable ?

(a) 2 and 3 (b) 1 and 4

UPSC JWM 2017

Ans. (c) : δQ=dU+ δW ...(1)

δQ=dU+PdV ...(2)

The equation (2) is applicable for the closed system.

123. The steady flow energy equation of a perfect gas flowing through a nozzle with the initial and final velocities of V1 and V2 are given by

TNPSC AE 2013

Ans. (b) : We know that SFEE,

For nozzle

We assume that

Z1 = Z2, Q = 0, W = 0

then we get

124. Heat transfer in a cyclic process are +20 kJ, -5 kJ, -10 kJ and +15 kJ. Net work done for this cycle will be given by:

(a) +0 kJ (b) -20 kJ

UPRVUNL AE 2016

Ans. (c) : From first law of thermodynamics.

QNet = Wnet

Wnet = 20 - 5 - 10 + 15

Wnet = + 20 kJ

125. 1 kg liquid (specific heat = 3.0 kJ/kg-K) is stirred in closed chamber and its temperature is raised by 10ºC. Heat loss to the surrounding is 3.0 kJ. The work done on the water during the process will be:

(a) 36 kJ (b) 30 kJ

UPRVUNL AE 2016

Ans. (d) : From energy balance equation.

Work done on the closed system (W) = Change in internal energy of the system + heat loss to the surrounding.

W = mCv∆T + QLoss

W = 1 × 3 × 10 + 3

W = 33 kJ

126. A system is composed of a gas contained in a cylinder fitted with a piston. The gas expands from the state 1 for which E1 = 75 kJ to a state 2 for which E2 = – 25 kJ. During the expansion, the gas does 60 kJ of work surroundings. The heat transferred to or from the systems during process is:

(a) – 30 kJ (b) – 40 kJ

TRB Polytechnic Lecturer 2017

Ans. (b) : Given as,

E1 = 75 kJ,

E2 = –25 kJ

W = 60 kJ

Q = ?

We know that

Q = ∆E + W

Q = (–25 – 75) + 60

Q = –40 kJ

127. For a heat engine cycle, which of the following relation is always true [Q = heat transfer, W= work transfer]

SJVN ET 2019

Ans. (b) :

128. Flow work is analogous to

(a) Stirring work

(b) Electrical work

(d) Shaft work

JPSC AE PRE 2019

Ans. (c) : Work in steady flow process = ∫VdP

Work in non-flow process = ∫PdV

129. If H be the heat supplied to the system to do work W and change in internal energy as

∆U, then,

(a) H = ∆U + W

(b) ∆U = H + W

(d) H = W/∆U

SJVN ET 2013

Ans. (a) : First Law of thermodynamics

H= ∆U + W

where, H = Heat supplied

∆U = Internal energy

W = work

130. If δQ is the heat transferred to the system and δW is the work done by the system, then

which of the following is an exact differential

(a) δQ (b) δW

Nagaland PSC CTSE 2017 Paper-2

Ans. (d) :δQ–δW, will shows the exact differential form

131. An elastic generator coupled to a windmill produces an average electric power of 6 kW. The power is used to charge a storage battery. Heat transfer from the battery to the surroundings is 0.3 kW. What will be the amount of energy stored in the battery in 2 hours?

(a) 41040 kW (b) 114 kW

SJVN ET 2019

Ans. (d) : Given,

δQ = – 0.3 kW

δW = – 6 kW

dU = ?

dU = δQ – δW

dU = – 0.3 + 6

= 5.7 kW

dU = 5.7 × 2 × 3600

= 41040 kJ

132. Statement I: The energy of an isolated system is constant.

Statement II: The entropy of an isolated system can increase but cannot decrease.

ESE 2018

Ans. (b) : Energy of the isolated system is constant as there is no energy interaction.

Entropy of an isolated system always increases and never decreases.

133. A system absorbs 100 kJ as heat and does 60 kJ work along the path 1-2-3. The same

system does 20 kJ work along the path 1-4-3. The heat absorbed during the path 1-4- 3 is

(a) –140 kJ (b) –80 kJ

ESE 2018

Ans. (d) : Given,

Q123 = 100 kJ

W123 = 60 kJ

W143 = 20 kJ

For path 1−2−3

Q123 = U3− U1 + W123

U3− U1 = 100 − 60 = 40 kJ

For path 1−4−3

Q143 = U3− U1 + W143

= 40 + 20 = 60 kJ

134. During a constant pressure expansion of a gas, 33.3% heat is converted into work while the temperature rises by 20K. The specific heat of the gas at constant pressure as a proportion of work, W is

(a) 8% (b) 10%

ESE 2017

Ans. (d) : Work (W) = 33.3% Q

W = 0.33 Q

Temperature rise (∆T) = 20 K

We know that Q = CP∆T

CP = 0.15 W

i.e. specific heat of the gas at constant pressure is 15% of the work.

135. A cylinder contains 10m3 of an ideal gas at a pressure of 2 bar. This gas is compressed

in a reversible isothermal process till its pressure increases to 15 bar. What quantum of

work will be required for this process?

(You can use the table given herewith.)

(a) 4500 kJ (b) 4030 kJ

ESE 2017

Ans. (b) : Given,

V1 = 10 m3

P1 = 2 bar

P = 15 bar

For reversible isothermal process work done is given by

136. During a certain compression process, 1 kJ of mechanical work input is supplied to 2 kg of a gas enclosed in a cylinder piston assembly and 400 J of heat is rejected to the cooling water being circulated in the jacket encasing the cylinder. This brings about a change in the specific internal energy by:

(a) –700 J/kg (b) 600 J/kg

OPSC AEE 2015 PAPER - II

Ans : (c)

Work input = 1000J

Rejected heat = 400J

mass = 2 kg.

first law of thermodynamics δQ = dU + δW.

dU =δQ - δW.

dU = -400+1000

dU = 600J

Specific internal energy = Internal energy/mass

u =600/2

u = 300 J/Kg

(δ→ show that heat and work both are inexact differential)

137. The internal energy of a certain system is a function of temperature alone and is given by the formula E = 25+0.25t kJ. If this system executes a process for which the work done done by it per degree temperature increases is 0.75 KN-m, dE/dt=Q−W,

the heat interaction per degree temperature increase, in kJ. is

(a) –1.00 (b) 1.00

MPPSC AE 2016

Ans : (b) E = 25 + 0.25t KJ

and

from the first law of thermodynamics

δQ = δW + dE

δQ = 0.25 + 0.75

δQ = 1.00 kJ

138. A system undergoes process a process in which the heat transfer to the system is

30 KJ and the work done by the system is 35000 Nm. The change in internal energy of the

system is

(a) +5 KJ (b) –5 KJ

TSPSC AEE 2015

Ans : (b) According to first law of thermodynamic

δQ = dU + δW

dU = δQ – δW

dU = –5 KJ.

139. Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The work done in the process is given by

(a) −∫ pdv (b) +∫pdv

UPPSC AE 12.04.2016 Paper-II

Ans : (c) Open system work:-

For reversible adiabatic compression work done = ∫dw = -∫vdp

140. For a non-flow constant pressure process the heat exchange is equal to:

(a) Zero

(b) The work done

(d) The change in enthalpy

OPSC AEE 2019 PAPER - II

Ans : (d) : Since enthalpy change of a system undergoing a process

dH = Tds + VdP

For reversible process

∵Tds = Qδ

∴dH = Qδ + VdP

If pressure is constant

dP = 0

dH = Qδ

141. A piston-cylinder device with air at an initial temperature of 30°C undergoes an expansion process for which pressure and volume are related as given below:

(a) 4.8 kJ (b) 6.8 kJ

ESE 2020

Ans. (d) :Work done,

W = 10.6 kJ

142. A carnot heat pump absorbs heat from atmosphere at 10 °C and supplies it to a room maintained at 25 °C. A temperature difference of 5 °C exists between working fluid and atmosphere on one hand, and the required room temperature on the other hand. If the heat pump consumes 1 kW power, the heat delivered to the room will be

(a) 12.1 kW (b) 14.9 kW

UKPSC AE 2012 Paper–II

Ans. (a) : 12.1 kW

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