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Physics NEET MCQ
Quiz 7
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Q.1
PARAGRAPH In the figure a container is shown to have a movable (without friction)piston on top. The container and the piston are all made of perfectlyinsulating material allowing no heat transfer between outside and inside thecontainer. The container is divided into two compartments by a rigidpartition made of a thermally conducting material that allows slow transferof heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are Cv=3R/2 , Cp= 5R/2 and thosefor an ideal diatomic gas are CvM = 5R/2 , Cp=7R/2 . 187A) Consider the partition to be rigidly fixed so that it does not move. When equilibrium isachieved, the final temperature of the gases will be
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a) 550 K
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b) 525 K
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c) 513 K
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d) 490 K
Explanation
Now partition can move so during transfer of heat pressure will remain constant So we will use value of CP for both n1 Cp (700-T) = n2 Cp (T-400) 3500-5T=7T-2800 12T= 6300 T = 525K Work PΔV =nRΔT Work done by upper diatomic gas w1 = n1R ΔT1 = 2(R) 125 = 250 R Work done by lower monoatomic gas w2 = n2 R ΔT2 = 2 (R) (−175) = − 350 R Net work done = -350R +250R= -100R Answer:(d)
Q.2
than one correct answer Q188) A container of fixed volume has a mixture of one mole of hydrogen and one mole of heliumin equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s)is(are)
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a) The average energy per mole of the gas mixture is 2RT
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b) The ratio of speed of sound in the gas mixture to that in helium gas is √(6/5)
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c) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2
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d) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/√2
Explanation
Option a Thus average energy per mole = 2RT option “a” correct Option b ratio of speed of sound in the gas mixture option "d" Velocity of sound in gas For mixture of gases Cp = Cpm + Cpd Cv = Cvm + Cvd Helium Option b is correct Answer:(a, b, d)
Q.3
An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2, pressure P2 and volume VDuring this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are) ... [IIT Advance 2015]
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a) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is P1 V1/4
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b) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1
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c) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is 7P1V1/3
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d) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is 17 P1V1/6
Explanation
Work done by gas = Potential energy of spring Let surface area of piston be S, then change in volume of gas ΔV = Sx Force due to spring = kx Thus pressure on piston when piston displaced by x, P2= kx/S = k x2/ΔV ∴ P2ΔV=kx2 Potential energy of spring = (1/2) kx2 Work done by gas = P2 ΔV/2 Option a ΔV= V2 - V1 = 2V1 - V1 = V1 Work done by gas option A is wrong Option B Change in internal energy As f =3 for mono atomic gas ( degree of freedom) and by ideal gas equation As calculated P2=3P1/2 Option “b” is correct Option “c” Work done by gas = P.E. of spring . Thus option “c” wrong Option d Heat supplied = Work done by gas + change in internal energy option “d” wrong Answer:(b)
Q.4
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (ΔSsurr) in JK-1 is (1 L atm = 101.3 J) [IIT Advance 2016]
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a) 5.763
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b) 1.013
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c) -1.013
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d) -5.763
Explanation
ΔE = q + w 0 = q - Pext ΔV q = Pext ×V = 3 atm (2 - 1) L = 3 atm L q= (3 × 101.3) Joule Answer:(c)
Q.5
PARAGRAPH Answer 192A, 192B and 192C by appropriately matching the information given in the three columns of the following table. An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to stateW denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamics processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n. Q192A) Which of the following options is the only correct representation of a process in which ΔU = ΔQ – PΔV?
Column-1
Column-2
Column-3
(i) Isothermal
(ii) Isochoric
(iii) Isobaric
(iv) Adiabatic
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a) (II) (iv) (R)
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b) (II) (iii) (P)
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c) (II) (iii) (S)
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d) (III) (iii) (P)
Explanation
Now partition can move so during transfer of heat pressure will remain constant So we will use value of CP for both n1 Cp (700-T) = n2 Cp (T-400) 3500-5T=7T-2800 12T= 6300 T = 525K Work PΔV =nRΔT Work done by upper diatomic gas w1 = n1R ΔT1 = 2(R) 125 = 250 R Work done by lower monoatomic gas w2 = n2 R ΔT2 = 2 (R) (−175) = − 350 R Net work done = -150R Answer:(a)
Q.6
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.7
A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi = 105 Pa and volume Vi = 10-3 m3 changes to a final state at Pf = (1/32) × 105 Pa and Vf = 8 × 10⁻³ m3 in an adiabatic quasi-static process, such that P3V5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf. The amount of heat supplied to the system in the two-step process is approximately … [ IIT Advance 2016]
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a) 112 J
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b) 294 J
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c) 588 J
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d) 813 J
Explanation
For the Adiabatic process A to C P2V5 = constant thus γ = 5/2, monoatomic gas For process A to B pressure is constant Heat Q1 Q1 = nCPΔT Q1 = 1750 J For process B to C , volume is constant Heat Q2 Q2 = nCvΔT Q2 = -1162.5 J net Q= 1750 J -1162.5 J = 588 J Answer:(c)
Q.8
The maximum energy in the thermal radiation from a heat source occurs at a wave length of 11×10⁻⁵cm. according to wein's displacement Law the temperature of this source will be 'n' times. The temperature of another source for which the wavelength at maximum energy is 5.5×10⁻⁵cm. Then the value of n is [ AFMC 2000]
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a) 1/2
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b) 1
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c) 2
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d) 4
Explanation
According to Wein's displacement law λ1T1=λ2T2 11×10⁻⁵nT=5.5×10⁻⁵T n=1/2 Answer: (a)
Q.9
The density of steam is 0.6 kg/m3, 1gm water at 100°C and 1×105 N/m2 pressure is converted into steam. The external work done nearly
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a) 0.6×105 J
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b) 1700 J
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c)17 J
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d)170 J
Explanation
density of steam 0.6 kg/m3=0.6×10⁻³ g/cc Volume of steam=1 / (0.6×10⁻³ )=1666.66 cc=1666.66×10⁻⁶ m3 Volume of 1 cc water=10-6 m3 W=pΔP W=105 [ 1667 - 1]×10-6 W=166.7 J=170 JAnswer: (d)
Q.10
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.11
Figure below shows two paths that may be taken by a gas to go from a state A to a state C In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be :
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a) 300 J
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b) 380 J
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c) 500 J
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d) 460 J
Explanation
We know that Thus value of Q – W depends only on initial and final state of the system Qabc - Wabc = Qac – Wab…(i) Work down for A to B = 0 as process is isochoric and Work down for B to C = PΔV= 6×104×2×10⁻³ =120 J Thus Wabc= 0+120 = 120 J and Qabc=400+100 = 500 J Work for A to C = Wac = Area under AC Which is trapezium Substituting values in (i) 500- 120=Qac – 80 ⇒ Qac = 460 J Answer:(d)
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