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Heat And Thermodynamics Mcq
Quiz 5
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Q.1
An ideal gas is initially at P1, V1 is expanded to P2, V2 and then compressed adiabatically to the same volume V1 and pressure PIf W is the net work done by the gas in complete process which of the following is true.. [ IIT 2004]
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a)W > 0; P3 > P1
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b) W < 0; P3 > P1
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c)W > 0; P3 < P1
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d)W < 0; P3 < P1
Explanation
In the first process W is positive as ΔV is positive, in the second process W is negative as ΔV is negative and area under the curve of second process is more ∴ Net Work < 0 and also P3 > P1Answer: (b)
Q.2
An ideal gas is expanding such that PT2=constant. The coefficient of volume expansion of the gas is [ IIT 2008]
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a) 1/T
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b) 2/T
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c)3/T
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d)4/T
Explanation
Given PT2=constant also for ideal gas PV / T=constant From above two equation after eliminating P V=k T3 , where k is constant Thus dV/ V=3 ( dT/T) dV=(3/T) VdT -- eq(1) We know that change in volume due to thermal expansion is given by dV=VγdT --eq(2) where γ is coefficient of volume expansion From eq(1) and eq(2) we get γ=3/TAnswer: (c)
Q.3
Which statement is incorrect? [ AIEEE 2002]
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a) all reversible cycles have same efficiency
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b) reversible cycles has more efficiency than irreversible one
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c)carnot cycle is reversible one
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d)carnot cycle has the maximum efficiency in all cycles
Explanation
If temperature of source and sink are different , efficiency is different Answer:(a)
Q.4
Cooking gas container are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will .. [ IIT 2002]
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a) increase
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b) decrease
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c) remain same
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d) decrease for some, while increase for others
Explanation
Higher the velocity of gas molecules, higher is the average kinetic energy per mole and higher is the temperature of the gas Answer: (a)
Q.5
At what temperature is the r.m.s. velocity of a hydrogen molecule equals to that of an oxygen molecule at 47°C? [ AIEEE 2002]
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a) 80 K
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b) -73K
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c) 3K
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d) 20K
Explanation
Molecular weight of Oxygen Mo = 32 Molecular weight of Hydrogen MH= 2 For vrms to be equal Answer: (d)
Q.6
Even Carnot engine can not give 100% efficiency because we cannot .. [AIEEE 2002]
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a) prevent radiation
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b) find ideal sources
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c) reach absolute zero temperature
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d) eliminate friction
Explanation
Efficiency of Carnot cycle is 100% if sink temperature is absolute zero which can not be obtained Answer: (c)
Q.7
" Heat cannot by itself flow from body at lower temperature to a body at higher temperature" is a statement or consequence of ..[ AIEEE 2003]
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a) second law of thermodynamics
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b) conservation of momentum
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c) conservation of mass
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d) first law of thermodynamics
Explanation
Answer:(a)
Q.8
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio Cp / Cv for the gas is .. [ AIEEE 2003]
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a) 4/3
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b) 2
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c) 5/3
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d) 3/2
Explanation
P ∝ T3 Thus PT-3 = constant --eq(1) But for an adiabatic process, the pressure temperature relationship is given by P(1-γ) Tγ = constant from eq(1) and eq(2) γ = 3/2 Answer: (d)
Q.9
Which of the following parameters does not characterize the thermodynamic state of matter ..[AIEEE 2003]
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a)temperature
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b) pressure
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c)work
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d)volume
Explanation
Work is a path function. The remaining three parameters are state functionAnswer: (c)
Q.10
A Carnot engine takes 3×106 cal. from a reservoir at 627°C and gives it to sink at 27°C. The work done by the engine is ..[ AIEEE2003]
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a) 4.2×106J
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b) 8.4×106J
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c) 16.8×106J
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d)zero
Explanation
But η=W/QW=η×QW=(2/3)×3×106W=2×106 cal W=2×106×4.2J=8.4×106 JAnswer: (b)
Q.11
Which of the following statement is correct for any thermodynamic system? [ AIEEE 2004]
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a) the change in entropy can naver be zero
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b) Internal energy and entropy and state functions
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c)the internal energy changes in all processes
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d)the work done in an adiabatic process is always zero
Explanation
Internal energy and entropy are state function, they do not depend upon path taken Answer:(b)
Q.12
Two thermally insulated vessel 1 and 2 are filled with air at temperature ( T1 , T2 ), volume (V1, V2 ) and pressure (P1 , P2 ) respectively. If the valve joining the two vessel is opened, the temperature inside the vessel at equilibrium will be [ AIEEE 2004]
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a)
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b)
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c) T1 + T2
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d)
Explanation
Here Q = 0 and W = 0. Therefore from first law of thermodynamics ΔU = Q + W = 0 ∴ Internal energy of the system with partition = Internal energy without partition Answer: (a)
Q.13
Which of the following is incorrect regarding the first law of thermodynamics? [ AIEEE 2005]
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a) It is a restatement of the principle of conservation of energy
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b) It is applicable to any cyclic process
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c) It introduces the concept of entropy
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d) It introduces the concept of internal energy
Explanation
Concept of entropy is introduced by the second law Answer: (c)
Q.14
A system goes from A to B via two process I and II as shown in figure. If ΔU1 and ΔU2 are the changes in internal energies in the process I and II respectively, then [ AIEEE 2005]
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a) relation between ΔU1 and ΔU2 can not be determined
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b) ΔU1 = ΔU2
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c) ΔU2 < ΔU1
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d) ΔU2 > ΔU1
Explanation
Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states i.e. ΔU1 = ΔU2 Answer: (b)
Q.15
The temperature -entropy diagram of a reversible engine cycle is given in figure. Its efficiency is [ AIEEE 2005]
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a) 1/4
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b) 1/2
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c) 2/3
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d) 1/3
Explanation
Answer:(d)
Q.16
Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature (7To)/The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature ( ignore the heat capacity of boxes). Then, the final temperature of the gases, T in terms of To is ..[ AIEEE 2006]
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a) Tf = (3To) / 7
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b) Tf = (7To) / 3
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c) Tf = (3To) / 2
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d) Tf = (5To) / 2
Explanation
Heat lost by He = Heat gained by N2 n1Cv1ΔT1 = n2Cv2ΔT2 Answer: (c)
Q.17
The work of 146kJ is performed in order to compresses one kilo mole of gas adiabatic ally and in this process the temperature of the gas increased by 7°C. The gas is ( R=8.3 J mol-1K-1) [ AIEEE 2006]
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a)diatomic
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b) triatomic
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c)a mixture of monoatomic and diatomic
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d)monoatomic
Explanation
Hence the gas is diatomicAnswer: (a)
Q.18
When a system is taken from state i to state f along the path iaf, it is found that Q=50 cal and W=20 cal. Along the path ibf Q=36 cal. W along the path ibf is ...[ AIEEE 2007]
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a) 14 cal
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b) 6 cal
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c)16 cal
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d)66 cal
Explanation
For path iaf, Q=5 cal, W=20 calBy first law of thermodynamicsΔU=Q - WΔU=50 - 20=30 calFor path ibfQ=36 calW=Q - ΔU W=36 - 30=6 cal.( since internal energy does not depend on the path, therefore ΔU=30 cal)Answer: (b)
Q.19
A carnot engine, having an efficiency of η=1/10 s heat engine, is used as refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at low temperature is [ AIEEE 2007]
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a) 100 J
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b) 99 J
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c)90 J
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d)1 J
Explanation
The efficiency (η) of Carnot engine and the coefficient of performance (β) of a refrigerator are related as Also Coefficient of performance β=Q2 /Wwhere Q2 is the energy absorbed from the reservoirQ2=β×WQ2=9×10=90J Answer:(c)
Q.20
If Cp and Cv denotes the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively then .. [ AIEEE 2007]
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a) Cp - Cv=28R
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b) Cp - Cv=R/28
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c) Cp - Cv=R/14
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d) Cp - Cv=R
Explanation
Cp - Cv=R Thus Answer: (b)
Q.21
One kg of diatomic gas is at a pressure of 8×104N/mThe density of the gas is 4kg/mWhat is the energy of the gas due to thermal motion? [ AIEEE 2009]
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a)5×104J
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b) 6×104J
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c)7×104J
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d)3×104J
Explanation
Volume=mass/ density=(1/4) m3degree of freedom of diatomic gas=5K.E.=(5/2)PVK.E.=(5/2) ×8×104 ×(1/4)=5×104JAnswer: (a)
Q.22
At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to that of oxygen molecules at 47°C? [ CPMT 1985]
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a) 20 K
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b) 80 K
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c)-73 K
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d)3 K
Explanation
r.m.s velocity is given by If r.ms. velocity of both the gases is same then Answer: (a)
Q.23
A gas at a temperature 250K is contained in a closed vessel. If the gas is heated through 1°C the percentage increase in its pressure is
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a) 0.4%
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b) 0.6%
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c)0.8%
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d)1.0%
Explanation
Since volume is constant Answer:(a)
Q.24
The density of a gas is 6×10-2 kg/m3 and the root mean square velocity of the gas molecules is 500 m/s. The pressure exerted by the gas on the walls of the vessel is
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a) 5×103N/m2
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b) 1.2×10-4N/m2
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c) 0.83×10-N/m2
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d) 30 N/m2
Explanation
From formula Answer: (a)
Q.25
N molecules, each of mass m of gas A and 2N molecules, each of mass 2m of gas B are contained in the same vessel at temperature T. The mean square of the velocity of molecules of type B is denoted by v2 and the mean square of the X component of the velocity of A type is denoted by w2, then w2/v2 is :
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a)2
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b) 1
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c)1/3
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d)2/3
Explanation
For gas A mean square of the X component of the velocity of A type is denoted by w2 thus mean square velocity C2=3 ×mean square of the X component of the velocityC2=3 ×w2Kinetic energy for each freedom=½ mw2=½ kTThus w2=kT /m3 ×w2=3KT/mw2=KT/mAnswer: (d)
Q.26
A sample of Oxygen gas and a sample of Hydrogen gas both have the same mass, the same volume and the same pressure. The ratio of their absolute temperature is
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a) 1:4
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b) 4:1
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c)1:16
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d)16:1
Explanation
PV=nRT PV=(m/M)RTPV=(m/2)RT1 and PV=(m/32)RT2 Thus T1 / 2=T2 /30T2 / T1=16:1Answer: (d)
Q.27
By what percentage should the pressure of a given mass of gas be increased so as to decrease its volume by 10% at constant temperature?
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a) 8.1%
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b) 9.1%
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c)10.1%
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d)11.1%
Explanation
Using Boyle's law PV=P'×(0.9V)P'=(10/9)P=1.111PThus Increase in P=11.1% Answer:(d)
Q.28
At room temperature the r.m.s speed of the molecules of certain diatomic gas is found to be 1930 m/s. The gas is
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a) H2
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b) F2
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c) O2
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d) Cl2
Explanation
From the formula for r.m.s speed gas must be Hydrogen Answer: (a)
Q.29
Four molecules of gas are having speeds of 1,4,8 and 16 m/s. the root mean square velocity of the gas molecules is
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a)7.25 m/s
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b) 52.56 m/s
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c)84.25 m/s
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d)9.2 m/s
Explanation
Answer: (d)
Q.30
Two vessels of equal volume contain equal masses of oxygen in one at 400 K. Then the ratio of the r.m.s speed of the molecules of hydrogen to that of the molecules of oxygen is
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a) 3/4
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b) 4/3
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c)2√3
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d)3√2
Explanation
r.m.s velocity Answer: (c)
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