datavorous/entrance-exam-dataset huggingface dataset:
One gas bleaches the colour of flowers by reduction while the other by oxidation

Options:
A H2S and Br2
B CO and Cl2
C NH2 and SO3
D SO2 and Cl2

The correct answer is:
SO2 and Cl2

Solution:
SO2 bleaches flowers by reduction SO2+2H2O→H2SO4+2[H]
Cl2 bleaches flowers by oxidation Cl2+2H2O→2HCl+[O]

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
One gene- one enzyme relationship was established for the first time in:

Options:
A Salmonella typhimurium
B Escherichia coli
C Diplococcus pneumoniae
D Neurospora crassa.

The correct answer is:
Neurospora crassa.

Solution:
Beadle and Tatum gave one gene one hypothesis which states that each gene in an organism controls the production of a specific enzyme. It is these enzymes that catalyse the reactions that lead to the phenotype of the organism.

Topic: Biology, Biomolecules (B), NEET, NEET 2007
One gene-one enzyme hypothesis was postulated by:

Options:
A Beadle and Tatum
B R. Franklin
C Hershey and Chase
D A. Garrod

The correct answer is:
Beadle and Tatum

Solution:
Beadle and Tatum gave one gene one hypothesis which states that each gene in an organism controls the production of a specific enzyme. It is these enzymes that catalyse the reactions that lead to the phenotype of the organism. It was established first time in Neurospora. Rosalind Franklin discover the structure of DNA, by the use of X-ray diffraction. Hershey and Chase confirmed that DNA is the genetic material using bacteriophage. Archibald Garrod was the first to connect a human disorder, alkaptonuria with Mendel's laws of inheritance.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2006
One glucose on complete oxidation yields 36 ATP. During this oxidation of glucose, what is the contribution of the Krebs cycle via oxidative phosphorylation?

Options:
A 11 ATP
B 12 ATP
C 22 ATP
D 24 ATP

The correct answer is:
22 ATP

Solution:
During oxidation of glucose through Kreb's cycle 6 molecules of NADH+H+ are produced and 2 molecule of FADH2 is produced. One molecule of NADH +H+ produces 3 ATP and one molecule of FADH2 produced 2 ATP in oxidative phosphorylation. Thus total 22 ATP are produced.

Topic: Biology, Respiration in Plants, NEET
One goes from the centre of the earth to a distance two third the radius of the earth. The acceleration due to gravity is highest at

Options:
A the centre of the earth
B a distance half the radius of the earth
C a distance one-third of the radius of the earth
D a distance two-third of the radius of the earth

The correct answer is:
a distance two-third of the radius of the earth

Solution:
The acceleration due to gravity at a depth d inside the earth is
g′=g1-dR=gR-dR=grR
where, R-d=r= distance of a place from the centre of earth, therefore, g′∝r

Topic: Physics, Gravitation, NEET
One gram of hydrogen and 112 g of nitrogen are enclosed in two separate containers each of volume 5 L and at 27°C . If the pressure of the hydrogen is 1 atm, then the pressure of nitrogen would be

Options:
A 16 atm
B 12 atm
C 8 atm
D 4 atm

The correct answer is:
8 atm

Solution:
Mol of H2=12 mol
Mol of N2=112 g28 gmol-1=4 mol
p1V=n1RT (Constant T and V)
p2V=n2RT
So p1p2=n1n2
i.e., 1 atmp2=12 mol4 mol
Hence, p2=8 atm.

Topic: Chemistry, States of Matter, NEET
One hormone hastens maturity period in juvenile conifers, a second hormone controls xylem differentiation, while the third hormone increases the tolerance of plants to various stresses. They are respectively:

Options:
A Gibberellin, Auxin, Ethylene
B Auxin, Gibberellin, Cytokinin
C Gibberellin, Auxin, ABA
D Auxin, Gibberellin, ABA

The correct answer is:
Gibberellin, Auxin, ABA

Solution:
Spraying juvenile conifers with gibberellins hastens their maturity period, thus leading to early seed production. Auxin controls xylem differentiation. Abscisic acid (ABA) increases the tolerance of plants to various kinds of stresses, therefore, is also known as the stress hormone.

Topic: Biology, Plant - Growth and Development, NEET
One hormone helps in petiole elongation in deepwater rice plants, while the other stimulates closure of stomata. These are respectively:

Options:
A ABA and auxin
B Ethylene and ABA
C Ethylene and gibberellic acid
D Gibberellic acid and ABA

The correct answer is:
Ethylene and ABA

Solution:
The submerged rice water plants are unable to grow and elongate at times due to various reasons. Ethylene is a gaseous hormone which helps in the internode elongation or petiole elongation in plants. It is a multifunctional phytohormone which promotes or inhibits the growth and senescence of leaves depending on the plant species. ABA is abscisic acid which is known as an inhibitor and plays an important role in seed development and maturation by inducing the closure of stomata and also involved in seed dormancy.

Topic: Biology, Plant - Growth and Development, NEET
One key function of nuclear pores is to

Options:
A aid in the production of new nuclei
B allow cells to communicate with each other
C form connections between different organelles
D allow molecules like proteins to move in and out of the nucleus

The correct answer is:
allow molecules like proteins to move in and out of the nucleus

Solution:
A nuclear pore is a protein channel found in the nuclear membrane. These pores are formed when there is a fusion of the two membranes in the nuclear envelope and functions to allow the transport of RNA and protein molecules from the nucleus to the cytoplasm and the DNA proteins, carbohydrates, and lipids from the cytoplasm to the nucleus. In short, they act as passageways for the proteins and other molecules to move in and out of the nucleus.

Topic: Biology, Cell - The Unit of Life, NEET
One likes to sit under sunshine in winter season, because

Options:
A we get heat from the sun by conduction
B we get heat from the sun by convection
C we get heat from the sun by radiation
D none of the above

The correct answer is:
we get heat from the sun by radiation

Solution:
Sun produces radiation which reaches earth without any propagating medium. When we sit under sunshine, radiation falls on our body and we feel comfortable.

Topic: Physics, Thermal Properties of Matter, NEET
One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm, are introduced into a vessel of volume 1 L. If there is no change in temperature, the final pressure of the mixture of gas (in atm ) is

Options:
A 1.5
B 1
C 2
D 4

The correct answer is:
2

Solution:
Ideal gas equation is given by
PV=nRT ...(i)
For oxygen, P=1 atm, V=1 L, n=nO2
Therefore, Eq. (i) becomes
∴  1×1= n O 2 RT
nO2=1RT
For nitrogen P=0.5 atm, V=2 L, n=nN2
∴  0.5×2=nN2RT
⇒ n N 2 = 1 RT
For mixture of gas
PmixVmix=nmixRT
Here, n mix = n O 2 + n N 2
∴  PmixVmixRT=1RT+1RT
PmixVmix=2Vmix=1

Topic: Physics, Thermodynamics, NEET
One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm is introduced into a vessel of volume 1 L. If there is no change in temperature, the final pressure of the mixture of gasses (in atm) is

Options:
A 1.5
B 2.5
C 2
D 4

The correct answer is:
2

Solution:
Ideal gas equation is given by,
pV = nRT ...(i)
For oxygen, p = 1 atm, V = 1 L, n = n O 2
Therefore Eq. (i) becomes
1 × 1 = n O 2 RT
⇒ n O 2 = 1 RT
For nitrogen p = 0.5 atm, V = 2 L, n = n N 2
∴ 0.5 × 2 = n N 2 RT
⇒ n N 2 = 1 RT
For mixture of gasses,
PmixVmix=nmixRT⁡
Here n mix = n O 2 + n N 2
∴ P mix V mix RT⁡ = 1 RT⁡ + 1 RT
⇒ P mix V mix = 2

Topic: Physics, Thermodynamics, NEET
One microspore mother cell undergoes microsporogenesis to produce ______ functional microspores

Options:
A One
B Two
C Three
D Four

The correct answer is:
Four

Solution:
One microspore mother cell undergoes meiosis to form four microspores / pollen grains. All these pollen grains are functional and they further undergo mitosis. Pollen grain is generally released in 2 celled stage in angiosperms. The two cells are generative and vegetative cell. The generative cell later undergoes division and forms two male gametes. The mature fully developed, fertilization ready pollen grain has 3 nuclei in it after the division of generative cell.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
One milligram of matter converted into energy will give

Options:
A 9Ã107Â J
B 9Ã103Â J
C 9Ã1010 J
D 9Ã105Â J

The correct answer is:
9Ã1010 J

Solution:
Given, the mass of the matter =1 mg=1Ã10-6 kg
As we know, c=3Ã108 m s-1
The matter will be converted into energy by the formula, E=mc2
=1Ã10-6Ã3Ã1082
=1Ã10-6Ã9Ã1016
=9Ã1010 J

Topic: Physics, Nuclear Physics, NEET
One mole of a diatomic gas undergoes a thermodynamic process, whose process equation is P∝V2. The molar specific heat of the gas is

Options:
A 17R3
B 17R6
C 15R4
D 15R8

The correct answer is:
17R6

Solution:
Given PV-2= constant
Comparing with PVN= constant, we get
∴N=-2
C=Cv+R1-N=5R2+R1+2=5R2+R3
=15R+2R6=17R6

Topic: Physics, Thermodynamics, NEET
One mole of a monoatomic gas behaving as per PV=nRT at 27 Â°C is subjected to reversible adiabatic compression until the final temperature reaches 327 Â°C. If the initial pressure was 1.0 atm then the value of lnPfinal is
(given ln 2=0.7)

Options:
A 1.75
B 0.176
C 1.0395
D 2.0

The correct answer is:
1.75

Solution:
Isoentropic process is an adiabatic process. So,
T1T2γ=P1P2γ-1⇒30060053=PiPf23
⇒1253=1Pf23
⇒53ln 2=23ln Pf⇒lnPf=52ln 2=1.75

Topic: Physics, Thermodynamics, NEET
One mole of a monoatomic ideal gas is mixed with one mole of a diatomic ideal gas. The molar specific heat of the mixture at constant volume is

Options:
A 3/2R
B 5/2R
C 2R
D 4R

The correct answer is:
2R

Solution:
By the conservation of energy at constant volume
CV=n1CV1+n2CV2n1+n2
=1Ã32R+1Ã52R1+1=2R

Topic: Physics, Kinetic Theory of Gases, NEET
One mole of a monoatomic ideal gas undergoes the process A â¶ B, as in the given P-V diagram. The specific heat for this process is

Options:
A 3R2
B 15R7
C 30R7
D 20R7

The correct answer is:
15R7

Solution:
ÎQ=ÎU+W (From 1st law)
Here, ÎU=nCVÎT
Temperature at A,
TA=2P0V0R
At B, TB=16P0V0R
So, ÎU=32R14P0V0R=21P0V0
W=2P03V0+123V02P0 [Area under P-V diagram]
â9P0V0
So, ÎQ=21P0V0+9P0V0
â30P0V0
ÎQ=nCÎT
C=ÎQnÎT=30P0V014P0V0R=157R

Topic: Physics, Thermal Properties of Matter, NEET
One mole of a non-ideal gas undergoes a change of state (2.0atom,3.0L,95K) →(4atm,5L,245K) with a change in internal energy, ΔE=30.0Latm. The change in enthalpy, ΔH , of the process in L atm is

Options:
A 40.0
B 42.3
C 44.0
D not defined, because pressure is not constant

The correct answer is:
44.0

Solution:
ΔH=ΔE+Δ(PV)
ΔH=30+4×5-2×3=44Latm

Topic: Chemistry, Thermodynamics (C), NEET
One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) → (4.0 atm, 5.0 L, 245 K) with a change in internal energy, ΔU = 30.0 L atm. The change in enthalpy (ΔH) of the process in L atm is

Options:
A 40.0
B 22.0
C 44.0
D 48.0

The correct answer is:
44.0

Solution:
ÎH=ÎU+Î(PV)
=ÎU+(P2V2-P1V1)
=30+(4Ã5-2Ã3)
=30+14=44 Latm

Topic: Chemistry, Thermodynamics (C), NEET
One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vaporisation?

Options:
A Acetone
B Water
C Both has the same enthalpy of vaporisation
D The enthalpy of vaporization for both acetone and water is zero

The correct answer is:
Water

Solution:
Correct Option is : (B)
Water

Topic: Chemistry, Thermodynamics (C), NEET
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,

The change in internal energy of the gas during the transition is:

Options:
A 20 kJ
B -20 kJ

C 20 J

D -12 kJ

The correct answer is:
-20 kJ

Solution:
ÎU=nCvÎT
T=PVnR
ÎT=P2V2-P1V2nR
ÎU=nRÎ³-1 P2V2-P1V1nR
=P2V2-P1V1Î³-1
=-8Ã10325=-20 kJ

Topic: Physics, Thermodynamics, NEET, NEET 2015 (Phase 1)
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,

The change in internal energy of the gas during the transition is:

Options:
A 20 kJ
B -20 kJ

C 20 J

D -12 kJ

The correct answer is:
-20 kJ

Solution:
ÎU=nCvÎT
T=PVnR
ÎT=P2V2-P1V2nR
ÎU=nRÎ³-1 P2V2-P1V1nR
=P2V2-P1V1Î³-1
=-8Ã10325=-20 kJ

Topic: Physics, Thermodynamics, NEET
One mole of an ideal gas at an initial temperature of T K does 6R joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 53, the final temperature of the gas will be

Options:
A (T+2.4) K
B (T-2.4) K

C (T+4) K

D (T-4) K

The correct answer is:
(T-4) K

Solution:
Work done in the adiabatic process is given as
W = - 1 γ - 1 P f⁡ V f⁡ - P i⁡ V i⁡
6R=-15/3-1RTf⁡-Ti⁡ [using PV = RT]
⇒ T f⁡ - T i⁡ = - 4
⇒ Tf⁡=T-4 K.

Topic: Physics, Thermodynamics, NEET
One mole of an ideal gas at an initial temperature of $T~ K$ does $6~ R$ 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:

Options:
A $(T+2.4) K$
B $(T-2.4) K$
C $(T+4) K$
D $(T-4) K$

The correct answer is:
$(T-4) K$

Solution:
We know that work done in adiabatic process is:
$$
\begin{aligned}
& W=\frac{-1}{V-1}\left(P_f V_f-P_i V_i\right) \\
& \Rightarrow \quad 6 R=\frac{-1}{\frac{5}{3}-1}\left(T_f-T_i\right) \\
& \text { Since } P V=R T\} \\
& \Rightarrow \quad T_f-T_i=-4 \\
& \Rightarrow \quad T_f=(T-4) \mathrm{K} \\
&
\end{aligned}
$$

Topic: Physics, Thermodynamics, NEET, NEET 2004
One mole of an ideal gas at temperature T1 expands according to the law (PV) = constant. Find the work done when the final temperature becomes T2.

Options:
A R(T2- T1)

B R2(T2- T1)
C R4(T2- T1)
D PV (T2-T1)

The correct answer is:
R2(T2- T1)

Solution:
W=∫V1V2PdV=∫V1V2KV dV ∵pV=K=constant
∴ W=12kV22-V12
PV = RT
But P= RT
∴ KV2=RT
or KV22-V12=RT2-T1
∴ W=R2T2-T1

Topic: Physics, Thermodynamics, NEET
One mole of an ideal gas CV=20JK-1mol-1 initially at STP is heated at constant volume to twice the initial temperature. For the process, W and q will be-

Options:
A W=0;q=5.46kJ
B W=0;q=0
C W=-5.46kJ;q=5.46kJ
D W=5.46kJ;q=5.46kJ

The correct answer is:
W=0;q=5.46kJ

Solution:
T1=273K
T2=2Ã273K
⇒âT=273
Volume is constant
So, âV=0
W=PÎV=PÃ0=0
q=nCVÎT=20Ã273=5460J=5.46kJ

Topic: Chemistry, Thermodynamics (C), NEET
One mole of an ideal gas C V =20âJ K â1 mol â1 initially at STP is heated at constant volume to twice the initial temperature. For the process, W and q will be

Options:
A W=0;q=5.46 kJ
B W=0;q=0
C W=-5.46Â kJ;q=5.46Â kJ
D W=5.46Â kJ;q=5.46Â kJ

The correct answer is:
W=0;q=5.46 kJ

Solution:
One mole of an ideal gas initially at STP is heated at constant volume to twice the initial temperature, Therefore
W=-PâV=-PÃ0=0
Q=CvâT=20Ã273=5460J=5.46 kJ

Topic: Chemistry, Thermodynamics (C), NEET
One mole of an ideal gas expands adiabatically such that its temperature Tâ1VThe value of Î³ = CPCV of the gas is

Options:
A 1.30
B 1.50
C 1.67
D 2.00

The correct answer is:
1.50

Solution:
In an adiabatic expansion,
PVγ=constant
Put P=RTV∴RTVVγ=constant
TVγ-1=constantorT∝V1-γ
As T∝1V∴1-γ=-12
γ=1+12=32=1.5

Topic: Physics, Thermodynamics, NEET
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 $3 V$ and then its volume is reduced from $3 V$ to $V$ at constant pressure. The correct $p-V$ diagram representing the two processes is

Options:
A

B

C

D

The correct answer is:

Solution:
According to question first gas goes from volume $V$ to $3 V$ and after this volume is reduced from $3 V$ to $V$ at constant pressure. In isothermal expansion $p-V$ curve is rectangular hyperbola.
Option (4) is correct

Topic: Physics, Thermodynamics, NEET, NEET 2012 (Screening)
One mole of an ideal gas has internal energy given by U=U0+2PV, where P is the pressure and V is the volume of the gas. U0 is a constant. This gas undergoes the quasi-static cyclic process ABCD as shown in the U-V diagram

The molar heat capacity of the gas at constant pressure is

Options:
A 2R
B 3R
C 52R
D 4R

The correct answer is:
3R

Solution:
Given relation for internal energy is U=U0+2PV.

For this cyclic process change in internal energy is given by

∆U=2∆PV=2R∆T
⇒CV=2R
⇒Cp=3R

Topic: Physics, Thermodynamics, NEET
One mole of an ideal monatomic gas at temperature T0 expands slowly according to the law P=kV (k is constant). If the final temperature is 4T0 then heat supplied to gas is

Options:
A 2 RT0
B 32 RT0
C 6 RT0
D RT02

The correct answer is:
6 RT0

Solution:
P=kV
∴ PV-1=k
∴ x=-1
C=Cv+R1-x=3R2+R1+1=2R
Q=nC∆T=2R×3T0=6RT0

Topic: Physics, Thermodynamics, NEET
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

Options:
A 32R
B 52R
C 2R
D R

The correct answer is:
R

Solution:
PV3= constant polytropic process with n = 3
C=Cv+R1-n
For monoatomic gas : Cv=3R2
C=3R2+R1-3
=3R2-R2
=2R2
C=R

Topic: Physics, Thermodynamics, NEET, NEET 2016 (Phase 2)
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

Options:
A 32R
B 52R
C 2R
D R

The correct answer is:
R

Solution:
PV3= constant polytropic process with n = 3
C=Cv+R1-n
For monoatomic gas : Cv=3R2
C=3R2+R1-3
=3R2-R2
=2R2
C=R

Topic: Physics, Thermodynamics, NEET
One mole of an ideal monoatomic gas γ=53 is mixed with one mole of diatomic gas γ=75. The value of γ for the mixture is (where, γ represents the ratio of specific heat capacities at constant pressure and constant volume)

Options:
A 1.5
B 1.75
C 1.33
D 1.85

The correct answer is:
1.5

Solution:
Given, n1=1,ââÎ³1=53
n2=1,ââÎ³2=75
Since, we know,
Î³mix=n1Î³1Î³1-1+n2Î³2Î³2-1n1Î³1-1+n2Î³2-1 ... (i)
Putting the given values in Equation (i), we get
Î³mix= 1Ã 5 3 5 3 â1 +1. 7 5 7 5 â1 1 5 3 â1 + 1 7 5 â1
=5/32/3+7/52/512/3+12/5=52+7232+52
â =12/28/2
â´âÎ³mix=128=32=1.5

Topic: Physics, Thermodynamics, NEET
One mole of an ideal monoatomic gas at temperature T0 expands slowly according to the law P V = constant. If the final temperature is 2T0, heat supplied to the gas isâ

Options:
A 2RT0

B RT0

C 32RT0
D 12RT0

The correct answer is:
2RT0

Solution:
In a process PVx = constant, molar heat capacity is given by
C = R γ - 1 + R 1 - x
As the process is P V = constant.
i.e PV-1 = constant, therefor, x = -1
For ideal monoatomic gas, γ = 5/3
∴ C = R 5 3 - 1 + R 1 - -1 = 3 2 R + R 2 = 2R
Δ Q = n C Δ T = 1 2 R 2 T 0 - T 0 = 2 RT 0

Topic: Physics, Thermodynamics, NEET
One mole of an ideal monoatomic gas at temperature T0 expands slowly according to the law PV=constant. If the final temperature is 2 T0, heat supplied to the gas isâ

Options:
A 2RT0

B RT0

C 32RT0
D 12RT0

The correct answer is:
2RT0

Solution:
In a process PVx = constant, molar heat capacity is given by
C=Rγ-1+R1-x
As the process is PV=constant.
i.e PV-1 = constant, therefor, x = -1
For ideal monoatomic gas, γ = 5/3
∴C=R53-1+R1--1=3R2+R2=2R
ΔQ=nCΔT=12R2T0-T0=2RT0

Topic: Physics, Thermodynamics, NEET
One mole of an ideal monoatomic gas undergoes a cyclic process, as shown in the figure. If the temperature of the gas at state 1 is 300 K and at state 4 is 500 K, then heat exchanged during process 2â3, is

Options:
A 1000R
B 600R
C 750R
D 800R

The correct answer is:
1000R

Solution:
Q23=nCPT3âT2

The processes 1â2 and 3â4 are isochoric in nature. Hence, for these processes TâP. As a result, we obtain,
T3=1000 K
T2=600 K
Q23=1Ã5R2Ã400=1000R

Topic: Physics, Thermodynamics, NEET
One mole of calcium nitride on the reaction with an excess of water gives

Options:
A two moles of ammonia
B two moles of nitric acid
C one mole of ammonia
D one mole of nitric acid

The correct answer is:
two moles of ammonia

Solution:
Ca3N2s1mol+6H2Ol→3CaOH2+2NH3g2mol
One mole of calcium nitride on the reaction with an excess of water gives two moles of ammonia and three moles of calcium hydroxide.

Topic: Chemistry, s Block Elements, NEET
One mole of carbon atom weighs 12g, the number of atoms in it is equal to, (Mass of carbon -12 is 1.9926×10-23g

Options:
A 1.2×1023
B 6.022×1022
C 12×1022
D 6.022×1023

The correct answer is:
6.022×1023

Solution:
1 mole of carbon atom \(=12 \mathrm{~g}\)
1 mole of a carbon atoms \(=6.022 \times 10^{23}\) atoms
Hence, mass of \(6.022 \times 10^{23}\) atoms of carbon \(=12 \mathrm{~g}\)
Mass of 1 atom of carbon \(=\frac{12}{6.022 \times 10^{23}}=1.9 \times 10^{-23} \mathrm{~g}\)

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2020 (Phase 2)
One mole of Ethylamine when reacts with nitrous acid will produce dinitrogen gas (at 0oC and 1 atmospheric pressure) equal to

Options:
A 22.4 L
B 1 L
C 11.2 L
D 24.8 L

The correct answer is:
22.4 L

Solution:

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation 2H2O(g) + CO(g) ⇌ H2(g) + CO2(g) Calculate the equilibrium constant for the reaction.

Options:
A 2.08
B 1.43
C 0.44
D 0.56D

The correct answer is:
0.44

Solution:
Correct Option is : (C)
0.44

Topic: Chemistry, Ionic Equilibrium, NEET
One mole of ideal gas goes through process P=2V21+V2 then change in temperature of the gas when volume changes from V=1 m3 to 2 m3 is:

Options:
A -45R K
B 115R K
C -52R K
D 2 K

The correct answer is:
115R K

Solution:
P=2V21+V2
P1=1
P2=85
ΔT=T2-T1=P2V2nR-P1V1nR=115R

Topic: Physics, Thermodynamics, NEET
One mole of ideal gas monoatomic gas ( Î³=53 ) is mixed with one mole of diatomic gas ( Î³=75 ). What is Î³ for the mixture? Î³ denotes the ratio of specific heat at constant pressure, to that at constant volume.

Options:
A 32
B 2315
C 3523
D 43

The correct answer is:
32

Solution:
For mixture of gases,
n 1 + n 2 Î³ m - 1 = n 1 Î³ 1 - 1 + n 2 Î³ 2 - 1
or 1 + 1 Î³ m - 1 = 1 5 3 - 1 + 1 7 5 - 1
or 2 Î³ m - 1 = 3 2 + 5 2 = 4 â Î³ m - 1 = 0.5
â´ Î³ m = 1.5 = 3/2.

Topic: Physics, Thermodynamics, NEET
One mole of methanol, when burnt in oxygen gives out 723 kJ mol-1 heat. If one mole of oxygen is used, what will be the amount of heat evolved?

Options:
A 723 kJ
B 482 kJ
C 964 kJ
D 289.2 kJ

The correct answer is:
482 kJ

Solution:
CH3OH+32O2âCO2+2H2O

ΔH= 2 3 ×723=2×241=482 kJ

Topic: Chemistry, Thermodynamics (C), NEET
One mole of methanol, when burnt in oxygen, gives out 723 kJ mol-1 heat. If one mole of oxygen is used, what will be the amount of heat evolved?

Options:
A 723 kJ
B 964 kJ
C 482 kJ
D 241 kJ

The correct answer is:
482 kJ

Solution:
2CH3OH2mol+3O23mol→2CO2+4H2O
ΔH=-723kJ/mol
1 mol O2≡23molCH3OH
1 mol CH3OH on combustion gives out 723 kJ heat
So 23molCH3OH combustion gives out =723×23 kJ heat
=482 kJ heat.

Topic: Chemistry, Thermodynamics (C), NEET
One mole of monoatomic gas is taken round the cyclic process. Heat absorbed by gas in process AB is :

Options:
A 2.5P0V0
B 3P0V0
C 3.5P0V0
D ​4.5P0V0

The correct answer is:
3P0V0

Solution:
Heat absorbed by the gas on the path AB during process
Δ Q AB = μ C v Δ T = 1 × 3 R 2 T B - T A
= 3 R 2 3 P 0 V 0 1 × R - P 0 V 0 1 × R = 3 P 0 V 0

Topic: Physics, Thermodynamics, NEET
One mole of water is converted to vapour at its boiling point 100 °C and at '1' atmospheric pressure. For this process, which one of following statement is correct?

Options:
A ΔS=0
B ΔG=0
C ΔH=0
D ΔE=0

The correct answer is:
ΔG=0

Solution:
One mole of water is converted to vapour at its boiling point 100 °C and at '1' atmospheric pressure. For this process ΔG=0 .
As phase transformation of 1 atm is an equilibrium process and at equilibrium, free energy change is always zero.

Topic: Chemistry, Thermodynamics (C), NEET
One molecule of glucose and one molecule of galactose form:

Options:
A Lactose
B Maltose
C Sucrose
D Inulin

The correct answer is:
Lactose

Solution:
One molecule of glucose and one molecule of galactose combine to form lactose.
Maltose is formed by the combination of two molecules of glucose.
Sucrose is formed by the combination of one molecule of glucose and one molecule of fructose.

Topic: Chemistry, Biomolecules, NEET