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Heat And Thermodynamics Mcq
Quiz 4
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Q.1
A carnot engine uses first an ideal monoatomic gas ( γ=5/3) and then ideal diatomic gas ( γ=7/5) as its working substance. The source and sink temperature are 411°C and 69°C respectively and the engine extract 1000J of heat from the source in each cycle. Then.. [ CPMT 1998]
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a) The efficiencies of the engine in the two cases are in ratio 21:25
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b) The area enclosed by the PV diagram in the first case only 500J
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c) The area enclosed by the PV diagram in both cases is 500J
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d)The heat energy rejected by the engine in the first case is 600J while that in the second case is 741.3J [ CPMT 1998]
Explanation
T1=411+273=684 KT2=69+273=684 KQ=1000J>br/>Efficiency η=In both the cases gas is ideal gas Hence option (c) is correct Answer:(c)
Q.2
The specific heat of 1 mole of an ideal gas at constant pressure is represented by Cp and that at constant volume by Cv. One of the following statement is correct [ MNR 1992]
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a) Cp of hydrogen gas is 5R/2
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b) Cv of hydrogen is 7R/2
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c) H2 has a very small value of Cp and Cv
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d) Cp -Cv=1.99 cal/ mole K for H2
Explanation
Answer: (d)
Q.3
A gas undergoes a process, changing from its initial state to some other state. Which of the following does not depend only on the initial and final states but also on the path followed
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a)heat exchanged with systems
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b) Internal energy
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c)entropy
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d)none of the above
Explanation
Answer: (a)
Q.4
A gas undergoes a change at constant temperature. Which of the following quantities remains fixed?
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a) Pressure
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b) Entropy
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c)Heat exchanged with the system
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d)All the above may change
Explanation
Answer: (d)
Q.5
A gas does 4.5 J of external work during adiabatic expansion. Its temperature falls by 2K. Its internal energy will
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a) Increase by 4.5J
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b) Decrease by 4.5J
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c)Decrease by 2.25J
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d)Decrease by 9J
Explanation
Answer:(b)
Q.6
An ideal gas is isothermally expanded. Its internal energy will [ CPMT 1990]
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a) increase
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b) decrease
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c) increase or decrease depending upon the nature of gas
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d) remains same
Explanation
Answer: (d)
Q.7
The difference between Cp and Cv is ...
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a)zero for solids and liquids
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b) same for solids, liquid and gases
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c)positive for solids, liquids and gases
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d)more for 1 gm molecules of oxygen than for 1 gm molecule of hydrogen
Explanation
Answer: (c)
Q.8
When 1g water freezes at at 0°C and 1×105N/m2 pressure, it is converted into 1.091 cc ice at 0°C. The external work done nearly
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a) +0.0091 J
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b) -0.0091J
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c)+0.0182J
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d)-0.0182J
Explanation
W=PΔV W=105× (0.091)×10×-6 W=0.0091JW is positive because the system is expandingAnswer: (a)
Q.9
When a gas is heated at constant pressure, the percentage of heat energy supplied, which goes as the internal energy of the gas is ..
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a) more for a diatomic gas for a monoatomic gas
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b) same for monoatomic, diatomic and triatomic gases
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c)100% for all gas
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d)less for a triatomic gas than for a diatomic gas
Explanation
Change in internal energy ∝ CpFor monoatomic gas Cp=5R/2 For Diatomic gas Cp=7R/2 Answer:(a)
Q.10
The molar specific heat of oxygen at constant pressure, Cp=7.03 Cal mole-1°C-1 and R=8.31 Joule mole-1°C-The amount of heat taken by 5 mole of oxygen when heated at constant volume from 10°C to 20°C will be approximately [ MPPMT 1987]
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a) 25 cal
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b) 50 cal
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c) 250 cal
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d) 500 cal
Explanation
dQ=µCvdT dQ=5 × (5R/2)×10 dQ=125R=(125×8.3)/ 4.2=250 cal Answer: (c)
Q.11
A bullet travelling with velocity 100 m/sec hits a concrete wall. All its kinetic energy is converted into heat energy in the bullet with the result that there is an increase of 50°C in the temperature of the bullet. What will be the increase in the temperature if the bullet were travelling with a velocity 200 m/sec?
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a)100°C
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b) 50°C
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c)200°C
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d)400°C
Explanation
Kinetic energy converted in heat Thus Temperature ∝ v2Answer: (c)
Q.12
Carbon monoxide is carried around a closed cycle abc, in which bc is an isothermal process, as shown in the figure. The gas absorbs 7000J of heat, as its temperature increases from 300K to 1000 K in going from a to b. The quantity of heat rejected by the gas during the process ca is ..[ SCRA 1994]
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a) 4200 J
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b) 5000 J
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c)9800 J
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d)-9800 J
Explanation
dQab=7000J, dWab=0, as volume is constant.dUab=dQQ=7000 J7000=µ ×(5R/2)×700 µR=4Ta=300K Tb=Tc=1000K Ub=Uc dQab=dWac - 7000=-P1( V2 - V1) - 7000 dQab=- µ[ RT2 - RT1] - 7000dQ=-9800 JAnswer: (d)
Q.13
A bullet made of lead moving with velocity v hits a wall. Its 50% kinetic energy is converted into heat, then increase in temperature will be... [ Raj.PMT 1996]
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a) 2v2 / JS
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b) v2 / 4JS
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c)v2S / J
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d)v2S / 2J
Explanation
50% of kinetic energy converted in heat energy causing increase in temperature of the bulletS be the specific heat of the bullet Answer:(b)
Q.14
The molar specific heat of an ideal gas at constant pressure and volume are denoted by Cp and Cv respectively. Further Cp/Cv=γ and R is the gas constant for 1gm of gas. Then Cv is equal to ...
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a) R
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b) γR
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c) R / (γ-1)
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d) γR / (γ-1)
Explanation
Cp - Cv=R Dividing by Cp 1- γ=R / Cp Cp=R / ( 1 - γ) Answer: (c)
Q.15
If the ratio of specific heat of gas at constant pressure to that at constant volume is γ, the change in internal energy of a mass of gas when the volume change from V to 2V at constant pressure P is .. [ CBSE 1998]
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a)R / ( γ -1)
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b) PV
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c)PV / (γ-1)
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d)γPV / (γ-1)
Explanation
Answer: (c)
Q.16
The specific heat of hydrogen gas at constant pressure is Cp=3.4×103 calories /kg°C and at constant volume is Cv=2.4×103 calories/ kg°C. If one kilogram hydrogen is heated from 10°C to 20°C at constant pressure, the external work done on the gas to maintain it at constant pressure is [ MPPMT 1995]
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a) 103 calories
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b) 5 ×103 calories
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c)104 calories
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d)105 calories
Explanation
dW=dQ - dU dW=mCpdT - mCvdT dW=1×3.4×103 ×10 - 1×2.4×103 ×10=104 calAnswer: (c)
Q.17
Fig. below represents a cyclic process abca for one mole of an ideal gas. If ab is isothermal process then which of the following is the P-T diagram for the cyclic process
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a)
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b)
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c)
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d)
Explanation
For isothermal process temperature remain same Answer:(a)
Q.18
A constant volume gas thermometer works on [ IIT 1980]
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a) The principle of Archimedes
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b) Boyle's law
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c) Pascal's law
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d) Charles' law
Explanation
Volume is constant Answer: (d)
Q.19
An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P_V diagram. The work done during the cycle is .. [ IIT 1983]
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a)PV
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b) 2PV
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c)½ PV
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d)zero
Explanation
Work=area enclosed=PVAnswer: (a)
Q.20
One mole of a monoatomic gas ( γ=5/3) is mixed with one mole of a diatomic gas ( γ=7/5), the value of γ for mixture is .. [ IIT 1988]
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a) 1.4
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b) 1.5
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c)1.53
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d)3.07
Explanation
From the formula Answer: (b)
Q.21
From the following statements ideal gas at any given temperature T, select the correct one [ IIT 1995]
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a) The coefficient of volume expansion at constant pressure is the same for all ideal gases
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b) The average translational kinetic energy per molecule of oxygen gas is 3kT, k is being Boltzmann constant
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c)The mean free path of molecules increases with increase in the pressure
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d)In gaseous mixture, the average translational kinetic energy of the molecules of each component is different
Explanation
Answer:(a)
Q.22
A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity. Then the pressure in the compartment is ..[ IIT 1999]
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a) same everywhere
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b) lower in the front side
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c) lower in te rear side
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d) lower in the upperside
Explanation
When a enclosed gas is accelerated in the positive x-direction then the pressure of the gas decreases along the positive x-axis and follows the equation ΔP=-ρsdx where ρ is the density and a the acceleration of the container Result will be more pressure on the rear side and less pressure on the front side Answer: (b)
Q.23
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is [ IIT 1999s]
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a)4 RT
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b) 15 RT
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c)9 RT
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d)11 RT
Explanation
The internal energy of the n moles of gas is U=½ nfRTwhere f=number of degree of freedom The internal energy of 2 moles of oxygen at temperature T is U1=½ ×2×5RT=5RT [ f=5 for oxygen]U2=½ 4×3RT=6RT Total internal energy=11RTAnswer: (d)
Q.24
A monoatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the length of the gas column before and after expansion respectively, then T1 / T2 is given by .. [ IIT 2000]
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a)
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b)
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c)
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d)
Explanation
here TVγ- 1=constant as γ=5/3, hence TV2/3=constant we know that V ∝ L Answer: (d)
Q.25
An ideal gas is initially at temperature T and volume V. Its volume is increased by ΔV due to an increase in temperature ΔT, pressure remains constant. The quantity δ=ΔV / ( VΔT) varies with temperature as .. [ IIT 2000]
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a)
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b)
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c)
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d)
Explanation
We know that V/T=constant Answer:(c)
Q.26
In a given process on an ideal gas, dW=0 and dQ
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a) the temperature will decrease
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b) the volume will increase
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c) the pressure will remain constant
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d) the temperature will increase
Explanation
form first law if dW=0 the dQ=dU given dQ is negative thus dU is negative temperature will decrease Answer: (a)
Q.27
P-V plots for two gases during adiabatic process are shown in the figure. Plot 1 and 2 should correspond respectively to .. [ IIT 2001]
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a)He and O2
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b) O2 and He
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c)He and Ar
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d)O2 and N2
Explanation
For adiabatic process PVγ=constantAlso for mono atomic gas γ=1.67 Also for diatomic gas γ=1.4Since γ diatomic < γ monatomic∴ Pdiatomic > Pmonoatomic∴ Graph 1 is for diatomic gas O2 and graph 2 for monatomic gasAnswer: (b)
Q.28
Which of the following graph correctly represents the variation of with P for an ideal gas at constant temperature? [ IIT 2002]
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a)
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b)
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c)
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d)
Explanation
PV=constantDifferentiating ∴ Graph between β and P will be rectangular hyperbolaAnswer: (a)
Q.29
The PT diagram for an ideal gas is shown in figure, where AC is an adiabatic process, find the corresponding PV diagram ..[ IIT 20003]
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a)
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b)
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c)
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d)
Explanation
If we study the P-T graph we find AB to be a isothermal process, AC is adiabatic process given. Also for an expansion process, the slop of adiabatic curve is more ( or we can say that the area under P-V graph for isothermal process is more than adiabatic process for same increase in volume ) only graph (b) fits the above criteria) Answer:(b)
Q.30
If liquefied oxygen at 1 atmospheric pressure is heated from 50K tp 300K by supplying heat at constant rate. The graph of temperature vs time will be ..[ IIT 2004]
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a)
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b)
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c)
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d)
Explanation
Q=mcΔT Q=mc(T - to) ... eq(1)∴ From 50k to boiling temperature, T increases inearly. During boiling, equation is Q=mL Temperature remains constant till boiling is completed. After that, again eq(1) is followed and temperature increases linearlyAnswer: (c)
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