In thermodynamics the Zeroth law is related to
Sol. Answer (2)
Zeroth law related to thermal equilibrium.
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In thermodynamics the Zeroth law is related to
Sol. Answer (2)
Zeroth law related to thermal equilibrium.
For a cyclic process
Sol. Answer (1)
Since initial and final points are at same, temperature so ΔU = 0
In following figures (a) to (d), variation of volume by change of pressure is shown in figure. The gas is taken along the path ABCDA. Change in internal energy of the gas will be (a) (b) P (c) (d) B B →V V
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Sol. Answer (4)
ΔU = 0 in all cases because cyclic process.
Which of the following laws of thermodynamics defines internal energy?
Sol. Answer (3)
Internal energy is defined in first law
: ΔQ = ΔU + ΔW
So, ΔU = ΔQ - ΔW
Select the correct statement for work, heat and change in internal energy.
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Sol. Answer (4)
All statements are correct.
Morning breakfast gives 5000 cal to a 60 kg person. The efficiency of person is 30%. The height upto which the person can climb up by using energy obtained from breakfast is
In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat when 8 J of work was done on the gas. If the initial internal energy of the gas was 30 J, then the final internal energy will be
Sol. Answer (2)
We know by 1st Law of Thermodynamics
ΔQ = ΔU + ΔW
- 20 J = ΔU - 8 J
: ΔU = Ufinal - Vinitial
ΔU = - 12 J
So, Ufinal = Uinitial + ΔU
= 30 + (-12) = 18 J
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A perfect gas goes from state A to state B by absorbing 8 × 105 joule and doing 6.5 × 105 joule of external work. If it is taken from same initial state A to final state B in another process in which it absorbs 105 J of heat, then in the second process work done
Sol. Answer (2)
ΔQ = ΔU + ΔW
8 × 105 = ΔU + 6.5 × 105
1.5 × 105 J = ΔU
Again using ΔQ = ΔU + W for the second case ΔU will stay the same.
Now, 105 = 1.5 × 105 + ΔW
- 0.5 × 105 = ΔW
negative sign indicates work is being done on the gas.
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A sample of an ideal gas undergoes an isothermal expansion. If dQ, dU and dW represent the amount of heat supplied, the change in internal energy and the work done respectively, then
Sol. Answer (2)
dQ = positive, dU = zero, dW = positive
* dQ = dU + dW
Sol. Answer (3)
ΔU= PiV, - P,Vi
is the correct relation.
V-1
Select the incorrect statement about the specific heats of a gaseous system.
A certain amount of an ideal monatomic gas needs 20 J of heat energy to raise its temperature by 10°C at constant pressure. The heat needed for the same temperature rise at constant volume will be
The specific heat of a gas in a polytropic process is given by R R R R R R R R
Sol. Answer (3)
R R R
C= Cv+
1-N r-1 N-1
If during an adiabatic process the pressure of mixture of gases is found to be proportional to square of its absolute temperature. The ratio of Cp/C, for mixture of gases is
Sol. Answer (3)
Cyclic process is anticlockwise then
Work done = -(Area of P-V graph)
W=-rR,R2
3ro -Po). ,3V0-v0)
=-П
2 2
-22
= -PoVo
7
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A gas undergoes a change at constant temperature. Which of the following quantities remain fixed?
Sol. Answer (4)
When temperature change = 0 then,
P,V, = P2Vz = constant
Rest may change.
For a certain process, pressure of diatomic gas varies according to the relation P = aV2, where a is constant. What is the molar heat capacity of the gas for this process? 17R 6R 13R 16R
Following figure shows P-T graph for four processes A, B, C and D. Select the correct alternative. (0, 0)
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Sol. Answer (3)
.C
(A) Temperature is constant - isothermal
(B) Pressure is constant - Isobaric -B
(C) Pressure Temperature - Isochoric process
A
(D) P1-yTv = constant - Adiabatic process
(0, o)
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An ideal gas with adiabatic exponent y is heated at constant pressure. It absorbs Q amount of heat. Fraction of heat absorbed in increasing the temperature is 1
Sol. Answer (2)
Heat absorbed in increasing temperature = ΔU = ΔQ - ΔW= nC,ΔT
Fraction of heat absorbed = Heat absorbed
Total heat
nCVΔT
=
пCpΔT
1
=.
Cp Y
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Two cylinders contain same amount of ideal monatomic gas. Same amount of heat is given to two cylinders. If temperature rise in cylinder A is To then temperature rise in cylinder B will be Free piston Fixed piston Tạạạnn Heat Heat To
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A mass of dry air at N.T.P. is compressed to 32 th of its original volume suddenly. If y = 1.4, the final pressure would be 1
The adiabatic elasticity of a diatomic gas at NTP is
For an isometric process
Sol. Answer (2)
For an isometric process, (i.e., isochoric) workdone = zero
So ΔQ = ΔU
A mixture of gases at NTP for which y = 1.5 is suddenly compressed to - th of its original volume. The final temperature of mixture is
In which process P-V diagram is a straight line parallel to the volume axis?
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Sol. Answer (1)
PVi = constant [equation of graphs]
^P
So for more y less the rate of change or slope of graph
and y is less for diatomic. .1 (diatomic)
*..... (monoatomic)
So graph 1 for 02
>V
Graph 2 for He.
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The pressure and volume of a gas are changed as shown in the P-V diagram in this figure. The temperature of the gas will B D → V
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The figure shows P-V diagram of a thermodynamic cycle. Which corresponding curve is correct? (0, 0) >V ALe B B
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Sol. Answer (1)
A → B V = constant PA
: PV = RT
P = RT V
Compare with y = mx
P-T graph is a straight line which must passes from origin
A → B volume constant, P-increasing, T-increasing.
B → C pressure constant, volume - increasing, temperature - increasing
B → C P = constant, origin P-T graph is a straightline parallel to v-axis
C → D V= constant then
P =
P-T graph is straight line must passes from origin
D → A P = constant
P-T graph is a straightline parallel to T-axis.
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During the thermodynamic process shown in figure for an ideal gas PA → V
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For P-V diagram of a thermodynamic cycle as shown in figure, process BC and DA are isothermal. Which of the corresponding graphs is correct? D • V
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Sol. Answer (2)
From A →B, volume increasing, pressure constant
A
1
BC, Pressure «Volume → Temperature constant
Same for D- → A
(0,0)
→D pressure decreasing, volume constant
So Po T
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Work done for the process shown in the figure is B(30 kPa, 25 cc) A(10 kPа, 10 cc) →
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During which of the following thermodynamic process represented by PV diagram the heat energy absorbed by system may be equal to area under PV graph? P4 P4
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In a thermodynamic process two moles of a monatomic ideal gas obeys p o V-2 . If temperature of the gas increases from 300 K to 400 K, then find work done by the gas (where R = universal gas constant).
Sol. Answer (2)
Px V2
PV- = constant Compare with PVN = constant then N = 2
W=M(FN)ΔT
W = MR 1- N(T2-T1)
2x R(400 - 300)
(1-2)
=-200 R
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If the efficiency of a carnot engine is n, then the coefficient of performance of a heat pump working between the same temperatures will be 1 1-n 1
Sol. Answer (3)
1 1
Coefficient of performance of heat pump =
efficiency of Carnot engine
In a Carnot engine, when heat is absorbed from the source, temperature of source
Sol. Answer (3)
Even when heat is taken out temperature stays the same. i.e., heat capacity of surface is infinite.
A Carnot engine working between 300 K and 600 K has a work output of 800 J per cycle. The amount of heat energy supplied to engine from the source in each cycle is
Sol. Answer (2)
W = 800 J
W -1-7,
800
@=1_300
600
1600 J = Q
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An ideal heat engine operates on Carnot cycle between 227°C and 127°C. It absorbs 6 × 10* cal at the higher temperature. The amount of heat converted into work equals to
Sol. Answer (4)
400
6x104=1-500
W = 1.2 × 104 cal
The maximum possible efficiency of a heat engine is
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Sol. Answer (4)
n=1_T2
T1
So it depends on source and sink temperature.
A frictionless heat engine can be 100% efficient only if its exhaust temperature is
Sol. Answer (3)
n=1-12
If exhaust temperature zero kelvin then n = 100%.
A reversible engine and an irreversible engine are working between the same temperatures. The efficiency of
Sol. Answer (2)
Efficiency of reversible engine is greater, because there is no loss of heat.
Which of the following can be coefficient of performance of refrigerator?
Sol. Answer (4)
B=1-n
B=l-1
n is less than 1 so →1
n
1
→ --1>0
→ ß >0
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The temperature inside and outside a refrigerator are 273 K and 300 K respectively. Assuming that the refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surrounding will be nearly
Sol. Answer (1)
n=1-. 12 : n=1- 273 •= 9
T1 300 100
B=1-n =100 0-1=91 -~ 11J
9
Q
B=W
For W = 1 J
Q = B
Q = 11 J
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By opening the door of a refrigerator placed inside a room you
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Sol. Answer (3)
Ultimately warm the room because work is being done by the refrigerator.
A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased to as to increase its efficiency by 50% of original efficiency?
Sol. Answer (2)
40 300
100 = 1-. 50% increase in efficiency
150
T1 = 500K 100 X0.4 = 0.6
new efficiency = 0.6 = 60
100 =1 - 300
T, =750 K
Difference between 2 Temperatures = 250 K
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Entropy of a system decreases
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Sol. Answer (2)
Entropy of a system decreases when heat is taken out of the system at constant temperature.
Internal energy of a non-ideal gas depends on
Sol. Answer (4)
Depends on Temperature, Pressure, Volume.
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A container is filled with 20 moles of an ideal diatomic gas at absolute temperature T. When heat is supplied to gas temperature remains constant but 8 moles dissociate into atoms. Heat energy given to gas is
Sol. Answer (2)
It is a cyclic system → ΔU = 0
and work done is (+)ive, so heat is supplied to system.
A triatomic, diatomic and monatomic gas is supplied same amount of heat at constant pressure, then
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Sol. Answer (3)
ΔU nCvΔT = Cy -= 1
ΔQ nCpΔT Cp Y
1 3
,ΔQ/mono = Ymono = 5
(ΔU) =- 1 5
ΔQ/dia Ydia 7
(ΔU) = 1 3
ΔQ/tria Ytria 4
Fractional energy used to change internal energy is maximum in Triatomic gas.
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If a gas is taken from A to C through B then heat absorbed by the gas is 8 J. Heat absorbed by the gas in taking it from A to C directly is P (kPа) 20 10 200 400 → V(cc)
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Sol. Answer (2)
When taken through ABC [ΔU + work = heat absorbed]
Heat absorbed = area under graph + ΔU = 8
ΔU = 8_10x200 -= 6
1000
when taken directly to C
W + ΔU = Q
Г10x200 -+1 2000|
-X- + 6 = Q → Q = 9 J
1000 1000
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Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which one of the following graphs represents the variation of temperature with time? Temperature Temperature Temperature
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Sol. Answer (3)
Liquid oxygen when heated will observe
a rise in temperature as well as change
Temperature
in state one time, which can be
represented as
105 calories of heat is required to raise the temperature of 3 moles of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required in calories to raise the temperature of the gas through the range (60°C to 65°C) at constant volume is Y= G,=14)
Sol. Answer (2)
At constant pressure heat absorbed = ΔQ = nCpΔT...(1)
At constant volume heat absorbed = ΔU = nC,ΔT ...(2)
Dividing (1) by (2),
105
ΔQ_Ce =r=1.4→ -= 1.4
ΔU
: ΔUv = 75 cal
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A gas may expand either adiabatically or isothermally. A number of P-V curves are drawn for the two processes over different range of pressure and volume. It will be found that
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Sol. Answer (4)
Slope for isothermal and adiabatic are not same so they will intersect.
A hydrogen cylinder is designed to withstand an internal pressure of 100 atm. At 27°C, hydrogen is pumped into the cylinder which exerts a pressure of 20 atm. At what temperature does the danger of explosion first sets in?
The variation of pressure P with volume V for an ideal diatomic gas is parabolic as shown in the figure. The molar specific heat of the gas during this process is →V
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When 1 kg of ice at 0°C melts to water at 0° C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/°C is
For an isobaric process, the ratio of ΔQ (amount of heat supplied) to the ΔW (work done by the gas) is (r=av) Y Y
3 moles of an ideal gas are contained within a cylinder by a frictionless piston and are initially at temperature T. The pressure of the gas remains constant while it is heated and its volume doubles. If R is molar gas constant, the work done by the gas in increasing its volume is 3 1) RTIn2
Sol. Answer (4)
W = PAV
= PV
= nRT
= 3RT
Two moles of a gas at temperature T and volume V are heated to twice its volume at constant pressure. If Cp=y then increase in internal energy of the gas is RT 2RT 2RT 2T
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To an ideal triatomic gas 800 cal heat energy is given at constant pressure. If vibrational mode is neglected, then energy used by gas in work done against surroundings is
Sol. Answer (1)
Heat at constant pressure
ΔQ = nCpΔT
Heat for doing work
ΔW = nRΔT
Then nCpΔT
ΔW (4)
800
ΔW -=1_1
800
B00 -1-3
ΔW = 200 cal
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A closed cylindrical vessel contains N moles of an ideal diatomic gas at a temperature T. On supplying heat, temperature remains same, but n moles get dissociated into atoms. The heat supplied is E(N-n)RT
An ideal monatomic gas at 300 K expands adiabatically to 8 times its volume. What is the final temperature?
Slope of isotherm for a gas (having v =3) is 3 × 105 N/m?. If the same gas is undergoing adiabatic change then adiabatic elasticity at that instant is
Sol. Answer (2)
Adiabatic elasticity = yP
5
-x3x105 = 5x105 N/m?
3
Figure shows, the adiabatic curve on a log T and log V scale performed on ideal gas. The gas is 2- 1 23 45> 109V
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Sol. Answer (1)
TVY-1 = K
logT + (Y -1)logV = 0
logT = -(v-1)logV
y = - (V-1) x
У = -(r-1) = slope = 2-4
4-1
2
→ - (Y - 1) =- 3
5
Y = 3
Monoatomic.
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A diatomic gas undergoes a process represented by PV1.3 = constant. Choose the incorrect statement
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Sol. Answer (4)
PV1.3 = K
W=PzV2 -PV1
: N> 1, so W is negative.
1- N
Heat supplied by surrounding heat goes to do work.
Down when expands.
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The process CD is shown in the diagram. As system is taken from C to D, what happens to the temperature of the system? P 3p..c Po. Vo 3Vo
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A P-T graph is shown for a cyclic process. Select correct statement regarding this B/ D →T
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Sol. Answer (3)
In process BC (isochoric process) where ΔT is (+)ve.
So ΔU = nG,ΔT
* ΔT is positive → U increases
An ideal gas of volume V and pressure P expands isothermally to volume 16 V and then compressed adiabatically to volume V. The final pressure of gas is [y = 1.5]
The pressure P of an ideal diatomic gas varies with its absolute temperature T as shown in figure. The molar heat capacity of gas during this process is [R is gas constant] P. / →T
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Sol. Answer (3)
5
Cv of diatomic = R
2
An ideal gas expands according to the law PZV = constant. The internal energy of the gas
Neon gas of a given mass expands isothermally to double volume. What should be the further fractional decrease in pressure, so that the gas when adiabatically compressed from that state, reaches the original state?
Sol. Answer (1)
AB is isothermal expansion. BC is adiabatic expansion
CD is isothermal compression
ΔA = adiabatic compression.
Efficiency of a heat engine working between a given source and sink is 0.5. Coefficient of performance of the refrigerator working between the same source and the sink will be
Sol. Answer (1)
1
n =
1+B
1
0.5 = • → ß = 1
1+B
A heat engine rejects 600 cal to the sink at 27°C. Amount of work done by the engine will be (Temperature of source is 227°C & J = 4.2 J/cal)
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A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 105 Nm-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is [NEET-2018]
The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is [NEET-2018]
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Sol. Answer (1)
Given process is isobaric, dQ = nC,dT
dQ=nER)aT
dW = PdV = nRdT
dW nRdT 2
Required ratio = = =
dQ 5
1ER)aT
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The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is [NEET-2018]
Sol. Answer (1)
Efficiency of ideal heat engine, n = (-77)
T2 : Sink temperature
T, : Source temperature
273
%n = 1[2 x100 =1- × 100
T, 373
100
= × 100 = 26.8%
373
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Muaseigme naung an eniciency of 10 as heat engine, is used as a refrigrator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is [NEET-2017]
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Sol. Answer (2)
11
1-n
B =- =- 10 =10
1
10 10
B = 9
Q2
ß=-
W
Q2 = 9 × 10 = 90J
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Thermodynamic processes are indicated in the following diagram. P. •500 K 1700 K *300 Kv Match the following Column-l Column-Il P. Process I a. Adiabatic Q. Process II b. Isobaric R. Process IIII c. Isochoric S. Process IV d. Isothermal [NEET-2017]
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Sol. Answer (2)
Process, |= Isochoric
|| = Adiabatic
Ill = Isothermal
IV = Isobaric
One mole of an ideal monatomic gas undergoes a process described by the equation PV3 = constant. The heat capacity of the gas during this process is [NEET (Phase-2)-2016]
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Sol. Answer (4)
PV3 = constant polytropic process with n = 3
R
C = Cv +
1- n
R R R R
= =5 -+ =R
r-1*1-n 1- 3
- - 1
3
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The temperature inside a refrigerator is tz°C and the room temperature is t,°C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be INEET (Phase-2) - 2016] 4, +273
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A refrigerator works between 4°C and 30°C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is [Take 1 cal = 4.2 J) [NEET-2016]
A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then [NEET-2016]
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Sol. Answer (3)
Adiabatic
Isothermal
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4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0 JK-' mol-'. If the speed of sound in this gas at NTP is 952 ms-', then the heat capacity at constant pressure is (Take gas constant R = 8.3 JK-' mol-1) [Re-AIPMT-2015]
Sol. Answer (2)
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is -20°C, the temperature of the surroundings to which it rejects heat is [Re-AIPMT-2015]
Sol. Answer (2)
An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas? [Re-AIPMT-2015]
Sol. Answer (2)
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure P (in kPa) 4 5 2---- 4 6 V (in m') The change in internal energy of the gas during the transition is [AIPMT-2015]
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A Carnot engine, having an efficiency of n = To as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is [AIPMT-2015]
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A a voarom loas the Tinal pres sure of nge gas is: (Vexpa S/sothermally to a volume 2V and the ApiaT-2014) to a volume 16V. the final pressure of the gas is: (take y = 5/3)
The molar specific heats of an ideal gas at constant pressure and volume are denoted by C, and C, respectively. If Y = and R is the universal gas constant, then C, is equal to [NEET-2013] R (Y -1) i+Y
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During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of - for the gas is: [NEET-2013] 5 3 4
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Sol. Answer (3)
Po TB
PT-3 = constant
Compare with PT = constant
Y 3
Then, = -3→ y=
1-Y 2
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One mole of an ideal gas goes from an initial state A to final state B via two processes: It first undergoes isothermal expansion from volume V to 3V and then its volume is reduced from 3V to V at constant pressure. The correct P-V diagram representing the two processes is [AIPMT (Prelims)-2012] A 1 1
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Sol. Answer (2)
1
3V
→V
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A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is [AIPMT (Prelims)-2012] -2PD Pressure -P 4 V 3V Volume
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Sol. Answer (3)
Heat rejected = workdone by surrounding = area of PV graph = P × 2V = 2PV
Sol. Answer (1)
Q, > Q2 > Q3 and ΔU, = ΔU2 = ΔU3
During an isothermal expansion, a confined ideal gas does -150 J of work against its surroundings. This implies that [AIPMT (Prelims)-2011]
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Sol. Answer (2)
It implies 150 J heat has been removed from the gas.
A mass of diatomic gas (y = 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27°C to 927°C. The pressure of the gas in the final state is [AIPMT (Mains)-2011]
If ΔU and ΔW represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true? [AIPMT (Prelims)-2010]
Sol. Answer (2)
In adiabatic process Q = 0
So ΔU = - ΔW [: ΔQ = ΔW + ΔU]
Sol. Answer (3)
Cp-c=R
Because C, & C, are given per unit mass
And Gp - C, = R is for 1 mole
So here we use RIM where M is molecular mass.
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Sol. Answer (4)
: PVi = constant
v,)
= P(8)% = 32P,
P2 = P1
V2)
In thermodynamic processes which of the following statements is not true? [AIPMT (Prelims)-2009]
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Sol. Answer (1)
In isochoric processes volume remains constant.
The internal energy change in a system that has absorbed 2 Kcals of heat and done 500 J of work is [AIPMT (Prelims)-2009]
Sol. Answer (3)
2 × 4.2 × 1000 = dU + 500 → dU = 7900 J
If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then [AIPMT (Prelims)-2008]
Sol. Answer (4)
E = change in U and in cyclic process ΔU = 0
→ E = 0
At 10°C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110°C this ratio is [AIPMT (Prelims)-2008] 283 383 10
Sol. Answer (1)
P=x at 10°C
PV=*
Molecular mass × number of moles - = X
RXT
1
7**
383
283 = =-
x =283 383*
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An engine has an efficiency of 1/6. When the temperature of sink is reduced by 62°C, its efficiency is doubled. Temperature of the source is [AIPMT (Prelims)-2007]
A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency? [AIPMT (Prelims)-2006]
Sol. Answer (3)
Where T2 = Sink Temperature
n =1-. 11 T, = Source Temperature
Temperature of sink is given to be 300 K.
7 = 0.4
so 0.4=1_300
→ T, = 500 K
Now, n is increased by 50%.
→ n' = 100 150 -xп= 10 15 +2x0.4 = 0.6
To maintain same sink temperature new source temperature is
0.6 = 1- 300
T, =750 K
. Increase in temperature = 750 - 500 = 250 K
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The molar specific heat at constant pressure of an ideal gas is -R. The ratio of specific heat at constant pressure to that at constant volume is : [AIPMT (Prelims)-2006]
Question image retained because the diagram/text is not fully readable.
Which of the following processes is reversible ? [AIPMT (Prelims)-2005]
Sol. Answer (4)
Isothermal compression takes place slowly at constant pressure, also ΔU is zero so it is a reversible process.
An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs 6 × 10* cal of heat at higher temperature. Amount of heat converted to work is [AIPMT (Prelims)-2005]
Sol. Answer (3)
A system is taken from state a to state c by two paths adc and abc as shown in the figure. The internal energy at a is Ua = 10 J. Along the path adc the amount of heat absorbed SQ, = 50 J and the work obtained oW, = 20 J whereas along the path abc the heat absorbed 8Q2 = 36 J. The amount of work along the path abc is d 1 a b V →
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Consider two insulated chambers (A, B) of same volume connected by a closed knob, S. 1 mole of perfect gas is confined in chamber A. What is the change in entropy of gas when knob S is opened? R = 8.31 J mol-'K-1. B
Question image retained because the diagram/text is not fully readable.
Sol. Answer (3)
AS = 2.303 nRloge
If initially volume is taken as V, then final volume = ZV, as volume of both chambers is given to be same.
2V
ΔS = 2.303x1x8.31xloge7
V
AS = 5.46 J/K
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A Carnot engine has efficiency 25%. It operates between reservoirs of constant temperatures with temperature difference of 80°C. What is the temperature of the low-temperature reservoir?
Sol. Answer (3)
n=1_IL
TH
1
TH
4
TH=3T
Also, TH- TL= 80
→ T, = 240 K = -33°C
Solution image retained because the diagram/text is not fully readable.
In an adiabatic change, the pressure and temperature of a monatomic gas are related as Poc T°, where c equals 2 5
Sol. Answer (4)
PxTO → PT-C = K
And compare with pT lF) [condition from adiabatic process]
= constant
Y
Then, -C =
1-Y
5/3 5/3 5
C =-- =- =2
1-5/3 -2/3
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An ideal Carnot engine, whose efficiency is 40%, receives heat at 500 K. If its efficiency is 50%, then the intake temperature for the same exhaust temperature is
Sol. Answer (3)
n=1-.
T1
T2
100 =1- 500
T2 = 300 K
If n = 50%
50 -=1_300
100 T1
→ T, = 600 K
Solution image retained because the diagram/text is not fully readable.
A monatomic gas initially at 18°C is compressed adiabatically to one eighth of its original volume. The temperature after compression will be
Sol. Answer (2)
PV' = constant
PV=RT
p([)' = constant
pl-v. TY = constant
Solution image retained because the diagram/text is not fully readable.
A sample of gas expands from volume V, to V. The amount of work done by the gas is greatest, when the expansion is
Sol. Answer (4)
Work done is maximum in isobaric proces, W = P.AV = P(V, - V,) = nR(T, - T,).
The efficiency of a Carnot engine operating with reservoir temperature of 100°C and - 23°C will be 373 + 250 373 - 250 100 + 23 100 - 23
We consider a thermodynamic system. If ΔU represents the increase in its internal energy and W the work done by the system, which of the following statements is true?
Sol. Answer (3)
As Q = zero for adiabatic process
So ΔU = -W for adiabatic process
If the ratio of specific heat of a gas at constant pressure to that at constant volume is y, the change in internal energy of a mass of gas, when the volume changes from V to 2V at constant pressure P, is PV R YPV
An ideal gas at 27°C is compressed adiabatically to 8/27 of its original volume. The rise in temperature is (Take v = 5/3)
Two Carnot engines A and B are operated in series. The engine A receives heat from the source at temperature T, and rejects the heat to the sink at temperature T. The second engine B receives the heat at temperature T and rejects to its sink at temperature T2. For what value of T the efficiencies of the two engines are equal? T, + T2
Question image retained because the diagram/text is not fully readable.
The (W/Q) of a Carnot engine is 1/6. Now the temperature of sink is reduced by 62°C, then this ratio becomes twice, therefore the initial temperature of the sink and source are respectively
A scientist says that the efficiency of his heat engine which works at source temperature 127°C and sink temperature 27°C is 26%, then
The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be
An ideal gas heat engine operates in a Carnot cycle between 227°C and 127°C. It absorbs 6 kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal to
Sol. Answer (4)
W
TH
W 400
-=1- 500 -= 1.2 J
One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be
Question image retained because the diagram/text is not fully readable.
The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from T, K to T2 K is 3 3 N,KB (Tz - T4)
Which of the following relations does not give the equation of an adiabatic process, where terms have their usual meaning?
Sol. Answer (1)
It is Pl-yTv = K
According to C.E. van der Waal, the interatomic potential varies with the average interatomic distance (R) as
Sol. Answer (4)
According to van der Waal's formulae, interatomic potential is inversely proportion to Rô.
So, U o R-6
In a vessel, the gas is at a pressure P. If the mass of all the molecules is halved and their speed is doubled, then the resultant pressure will be
Sol. Answer (2)
P=3MAVz
P: =3XMn(2V)3=2x-MNV2=2P
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The mean free path of collision of gas molecules varies with its diameter (d) of the molecules as
Sol. Answer (2)
1
2 0c
d?
At O K, which of the following properties of a gas will be zero?
Sol. Answer (3)
at 0 K Vrms = 0 so K.E. = 0
The value of critical temperature in terms of van der Waals' constants a and b is given by 8a 27a
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Sol. Answer (1)
8a
Tc =
27Rb
The degrees of freedom of a triatomic gas is (Consider moderate temperature)
Sol. Answer (1)
Degree of freedom = 3 rotational + 3 translational + 0 vibrational [T is moderate] = 6
Sol. Answer (1)
fR
:Cv=:
2
20v = 2Cv 2 2
Then, f = = =-
R Cp-Cv Cp-1 Y-1
Cv
Solution image retained because the diagram/text is not fully readable.
Sol. Answer (1)
: PV=nRT = m RT
M)
5
PV =- RT
32
A : Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion adiabatically. R : Temperature remains constant in isothermal expansion and not in adiabatic expansion.
Sol. Answer (2)
A : is true
R : is true, but not correct explanation
correct explanation is, in isothermal expansion.
• ΔT = 0 so ΔU = 0
→ ΔQ = ΔW
all the heat goes in doing work.
Whereas in adiabatic process
Heat goes to work as well as in increasing internal energy.
:. Wisothermal > Wadiabatic
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A : Efficiency of heat engine can never be 100%. R : Second law of thermodynamics puts a limitation on the efficiency of a heat engine.
Sol. Answer (1)
A : is true
R : is true, and correct explanation
A : Heat absorbed in a cyclic process is zero. R : Work done in a cyclic process is zero.
Sol. Answer (4)
A : is false, in cyclic process only ΔU = 0, ΔQ = ΔW.
R : is false, work done is not zero only change in internal energy is zero.
A : Coefficient of performance of a refrigerator is always greater than 1. R : Efficiency of heat engine is greater than 1.
Sol. Answer (4)
A : is false.
R : is false
Because efficiency of heat engine can never be equal to greater to 1.
n$1
all the heat cannot be converted to work.
and coefficient of performane of refrigerator
ß = 1-n = --1
: n<1 so ß may be less than 1.
Solution image retained because the diagram/text is not fully readable.
A : Adiabatic expansion causes cooling. R : In adiabatic expansion, internal energy is used up in doing work.
Sol. Answer (1)
A : is true
R : is true, and correct explanation
A : The specific heat of an ideal gas is zero in an adiabatic process. R : Specific heat of a qas is process independent.
Sol. Answer (3)
A : is true
R : is false
Because specific heat depends on the process.
A : The change in internal energy does not depend on the path of process. R : The internal energy of an ideal qas is independent of the configuration of its molecules.
Sol. Answer (2)
A : is true
R : is true, but not the correct explanation, because internal energy depends on the temperature of the gas.
A : Heat supplied to a gaseous system in an isothermal process is used to do work against surroundings. R : During isothermal process there is no change in internal energy of the system.
Sol. Answer (1)
A : true
R : true and correct explanation
A : In nature all thermodynamic processes are irreversible. R : Durina a thermodynamic process it is not possible to eliminate dissipative effects.
Sol. Answer (1)
A : is true
R : is true and correct explanation
A : During a cyclic process work done by the system is zero. R : Heat supplied to a svstem in the cyclic process converts into internal eneray of the svstem.
Sol. Answer (4)
A : is false, in cyclic process, work done is not zero, internal energy change is zero.
R : is false, heat supplied converts to work as initial state is equal to final state.
.. No change in internal energy.
Solution image retained because the diagram/text is not fully readable.