I85PREP
⌘ K
IIT JAM MA study materials
IIT JAM MA study resources

IIT JAM MA practice questions

One question at a time. Build your IIT JAM MA understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Biotechnology (BT)

In the lac operon of E. coli, which protein binds the operator region to block transcription when lactose is absent?

  1. RNA polymerase sigma factor
  2. LacI repressor Correct answer.
  3. LacZ beta-galactosidase
  4. CAP (catabolite activator protein)

Explanation

The LacI gene product is a repressor that binds the operator and physically blocks RNA polymerase from initiating transcription of the lac operon when inducer is absent. CAP instead acts as an activator at the promoter when cAMP levels are high.

Question 2Biotechnology (BT)

Which enzyme catalyzes the formation of a phosphodiester bond between adjacent nucleotides during DNA replication?

  1. Topoisomerase
  2. Helicase
  3. Primase
  4. DNA ligase Correct answer.

Explanation

DNA ligase seals nicks by forming a phosphodiester bond between a 3'-OH and a 5'-phosphate, joining Okazaki fragments and completing replication. Helicase unwinds duplex DNA, topoisomerase relieves supercoiling, and primase synthesizes RNA primers.

Question 3Biotechnology (BT)

A protein has a molecular mass of 60 kDa and contains 500 amino acids. What is the approximate average mass per amino acid residue?

  1. 180 Da
  2. 110 Da
  3. 120 Da Correct answer.
  4. 150 Da

Explanation

Average residue mass in proteins is approximately 110 Da, so 500 residues give about 55 kDa, close to the stated 60 kDa. Dividing 60,000 Da by 500 residues gives 120 Da, which is the closest match among the options.

Question 4Biotechnology (BT)

Which molecule serves as the final electron acceptor in the mitochondrial electron transport chain under aerobic conditions?

  1. Molecular oxygen Correct answer.
  2. Cytochrome c
  3. Ubiquinone
  4. NAD+

Explanation

Complex IV transfers electrons to molecular oxygen, which is reduced to water, making oxygen the terminal electron acceptor. NAD+, cytochrome c, and ubiquinone are intermediate carriers within the chain, not the final acceptor.

Question 5Biotechnology (BT)

In a polymerase chain reaction, what is the primary role of the annealing step?

  1. Allowing primers to hybridize to complementary template sequences Correct answer.
  2. Inactivating the polymerase before the next cycle
  3. Denaturing double-stranded DNA into single strands
  4. Extending the primer using thermostable DNA polymerase

Explanation

During annealing, the temperature is lowered so that sequence-specific primers base-pair with complementary regions on the single-stranded template. Extension by Taq polymerase occurs in the following step at a higher temperature.

Question 6Biotechnology (BT)

Which type of RNA carries amino acids to the ribosome during translation?

  1. rRNA
  2. tRNA Correct answer.
  3. snRNA
  4. mRNA

Explanation

Transfer RNA (tRNA) is charged with a specific amino acid by aminoacyl-tRNA synthetases and delivers it to the ribosome by matching its anticodon to the mRNA codon. mRNA carries the coding sequence and rRNA forms the ribosome core.

Question 7Biotechnology (BT)

Which technique separates proteins primarily on the basis of their size under denaturing conditions?

  1. Affinity chromatography
  2. SDS-PAGE Correct answer.
  3. Ion-exchange chromatography
  4. Isoelectric focusing

Explanation

SDS-PAGE denatures proteins and coats them with uniform negative charge, so migration depends on molecular size through the gel mesh. Ion exchange and affinity methods separate on charge or binding specificity, and isoelectric focusing separates by pI.

Question 8Biotechnology (BT)

Which bacterial structure is the primary target of lysozyme?

  1. The 30S ribosomal subunit
  2. The cytoplasmic membrane bilayer
  3. Peptidoglycan in the cell wall Correct answer.
  4. Lipopolysaccharide in the outer membrane

Explanation

Lysozyme hydrolyzes the beta-1,4 glycosidic bonds between N-acetylglucosamine and N-acetylmuramic acid in peptidoglycan, weakening the cell wall. This is why Gram-positive cells are especially sensitive to it.

Question 9Biotechnology (BT)

A DNA sample has an absorbance ratio A260/A280 of 1.9. What does this value most strongly indicate?

  1. The sample contains degraded RNA
  2. The sample is heavily contaminated with protein
  3. The sample is contaminated with phenol
  4. The sample is pure double-stranded DNA Correct answer.

Explanation

Pure double-stranded DNA typically shows an A260/A280 ratio near 1.8 to 1.9, reflecting low protein contamination. Lower ratios suggest protein or phenol carryover, while RNA usually gives a higher ratio around 2.0.

Question 10Biotechnology (BT)

In a dihybrid cross of two heterozygotes (AaBb x AaBb) with independent assortment, what proportion of offspring is expected to show both recessive phenotypes?

  1. 9/16
  2. 1/16 Correct answer.
  3. 3/16
  4. 1/4

Explanation

Each recessive phenotype appears with probability 1/4 in a monohybrid cross, and independent assortment allows multiplication, giving 1/4 x 1/4 = 1/16 for the double recessive class. The 9:3:3:1 ratio confirms this smallest class.

Question 11Biotechnology (BT)

Which feature distinguishes a competitive inhibitor from a non-competitive inhibitor in enzyme kinetics?

  1. Non-competitive inhibitors lower Km
  2. Non-competitive inhibitors raise Vmax
  3. Competitive inhibitors increase apparent Km without changing Vmax Correct answer.
  4. Competitive inhibitors increase Vmax

Explanation

A competitive inhibitor competes with substrate at the active site, so higher substrate can overcome it: apparent Km rises while Vmax is unchanged. Non-competitive inhibition instead lowers Vmax without affecting Km.

Question 12Biotechnology (BT)

During which phase of the eukaryotic cell cycle does DNA replication occur?

  1. S phase Correct answer.
  2. M phase
  3. G2 phase
  4. G1 phase

Explanation

DNA synthesis occurs during S phase, between G1 and G2, when each chromosome is duplicated into sister chromatids. G1 and G2 are gap phases for growth and checkpoint control, and M phase is where segregation occurs.

Question 13Biotechnology (BT)

A plasmid vector carries an ampicillin resistance gene and a lacZ gene with a multiple cloning site inside lacZ. After transformation into lacZ-negative E. coli on X-gal plus ampicillin, which colonies are recombinant?

  1. Colonies that grow but turn green
  2. White colonies Correct answer.
  3. Only colonies that fail to grow
  4. Blue colonies

Explanation

Insertion of foreign DNA into lacZ disrupts beta-galactosidase, preventing cleavage of X-gal, so recombinant clones remain white. Non-recombinant transformants retain functional lacZ and appear blue on X-gal.

Question 14Biotechnology (BT)

Which property of the genetic code explains why a single base substitution may not change the encoded amino acid?

  1. The code is universal
  2. The code is degenerate Correct answer.
  3. The code is ambiguous
  4. The code is overlapping

Explanation

Degeneracy means several codons specify the same amino acid, so a third-position change often leaves the protein sequence unchanged. This redundancy buffers some mutations, though not all substitutions are silent.

Question 15Biotechnology (BT)

Which step in glycolysis produces ATP by substrate-level phosphorylation?

  1. Conversion of glucose to glucose-6-phosphate
  2. Conversion of pyruvate to acetyl-CoA
  3. Conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate Correct answer.
  4. Conversion of fructose-6-phosphate to fructose-1,6-bisphosphate

Explanation

Phosphoglycerate kinase transfers a phosphate from 1,3-bisphosphoglycerate to ADP, generating ATP directly from a high-energy substrate. This is substrate-level phosphorylation, distinct from oxidative phosphorylation.

Question 16Biotechnology (BT)

Which statement correctly describes the role of a selectable marker in a cloning vector?

  1. It encodes the restriction enzyme used in cloning
  2. It provides the origin of replication
  3. It allows identification of cells that have taken up the plasmid Correct answer.
  4. It determines the size of the insert

Explanation

A selectable marker, such as an antibiotic resistance gene, lets transformed cells survive on selective medium while non-transformants die. This makes recovery of plasmid-bearing cells possible after transformation.

Question 17Biotechnology (BT)

A mutation changes a codon from UAC to UAA. What is the most likely consequence for the protein?

  1. A frameshift affecting all downstream codons
  2. No change because the code is degenerate
  3. A conservative amino acid substitution
  4. Premature termination of translation Correct answer.

Explanation

UAA is a stop codon, so the ribosome terminates translation at that point, producing a truncated protein. This is a nonsense mutation, distinct from missense or frameshift changes.

Question 18Biotechnology (BT)

Which method is best suited to quantify the expression level of a specific mRNA across many samples simultaneously?

  1. Western blotting
  2. Quantitative real-time PCR with reverse transcription Correct answer.
  3. SDS-PAGE followed by Coomassie staining
  4. Southern blotting

Explanation

Reverse transcription followed by quantitative PCR converts mRNA to cDNA and measures amplification in real time, giving sensitive, quantitative mRNA estimates. Western blot and SDS-PAGE measure protein, while Southern blot detects DNA.

Question 19Biotechnology (BT)

In a population at Hardy-Weinberg equilibrium, the frequency of a recessive allele is 0.2. What is the expected frequency of heterozygotes?

  1. 0.04
  2. 0.32 Correct answer.
  3. 0.16
  4. 0.64

Explanation

If q = 0.2, then p = 0.8 and the heterozygote frequency is 2pq = 2 x 0.8 x 0.2 = 0.32. The homozygous recessive frequency would be q squared = 0.04.

Question 20Biotechnology (BT)

Which feature of the CRISPR-Cas9 system determines the specificity of DNA targeting?

  1. Complementarity between the guide RNA and the target sequence Correct answer.
  2. The Cas9 nuclease domain alone
  3. The PAM sequence length only
  4. The size of the donor template

Explanation

The single guide RNA base-pairs with the complementary target DNA adjacent to a PAM, directing Cas9 cleavage to that site. Cas9 itself is a nonspecific nuclease whose specificity comes from the guide.

Question 21Biotechnology (BT)

A protein with a pI of 5.0 is placed in a buffer at pH 7.0 during ion-exchange chromatography. What is the expected behavior on a cation-exchange resin?

  1. It binds because the resin is positively charged
  2. It does not bind because it carries a net negative charge Correct answer.
  3. It binds only after the pH is raised further
  4. It binds strongly because it is positively charged

Explanation

At pH above its pI, the protein carries a net negative charge because carboxyl groups are deprotonated. A cation-exchange resin is negatively charged and repels the protein, so it flows through without binding.

Question 22Biotechnology (BT)

An enzyme follows Michaelis-Menten kinetics with Km = 2 mM. At what substrate concentration is the initial velocity equal to 80% of Vmax?

  1. 4 mM
  2. 2 mM
  3. 8 mM Correct answer.
  4. 0.5 mM

Explanation

Using v = Vmax[S]/(Km + [S]), setting v/Vmax = 0.8 gives [S] = 4Km = 8 mM. At [S] equal to Km the velocity is only 50% of Vmax.

Question 23Biotechnology (BT)

During the light reactions of photosynthesis, what is the immediate consequence of cyclic electron flow around photosystem I?

  1. Production of ATP without NADPH Correct answer.
  2. Splitting of water to release oxygen
  3. Reduction of carbon dioxide to sugar
  4. Production of NADPH without ATP

Explanation

Cyclic electron flow returns electrons to the cytochrome b6f complex, generating a proton gradient and ATP while bypassing NADP+ reduction. This supplies the extra ATP needed when the Calvin cycle demand exceeds the ATP/NADPH ratio from linear flow.

Question 24Biotechnology (BT)

A bacterial culture grows exponentially with a doubling time of 30 minutes. Starting from 10^4 cells, what is the approximate population after 3 hours?

  1. 6.4 x 10^5 Correct answer.
  2. 6.4 x 10^4
  3. 1.28 x 10^6
  4. 3.2 x 10^5

Explanation

Three hours contains six 30-minute doubling intervals, so the population increases 2^6 = 64-fold. Multiplying 10^4 by 64 gives 6.4 x 10^5 cells.

Question 25Biotechnology (BT)

A restriction digest of a 5 kb plasmid with a single-site enzyme yields one band on an agarose gel, whereas the same digest with a second enzyme yields two bands of 3 kb and 2 kb. What can be concluded about the second enzyme?

  1. It fails to cut the plasmid
  2. It removes a 2 kb fragment from the plasmid
  3. It has two recognition sites in the plasmid Correct answer.
  4. It linearizes the plasmid at one site

Explanation

Two fragments summing to the full 5 kb indicate that the enzyme cuts the circular plasmid at two sites. A single cut would linearize the plasmid and give one band of 5 kb.

Question 26Biotechnology (BT)

In the context of the genetic code, what does the term 'degeneracy' specifically refer to?

  1. Each codon specifies only one amino acid.
  2. The code is nearly universal across all organisms.
  3. The code is read in triplets without overlapping.
  4. Multiple codons can specify the same amino acid. Correct answer.

Explanation

Degeneracy means that the genetic code has redundancy: most amino acids are encoded by more than one codon (e.g., leucine has six codons). This is possible because there are 64 codons but only 20 standard amino acids. The other properties listed are distinct features of the code.

Question 27Biotechnology (BT)

A culture of Escherichia coli is grown in a medium containing both glucose and lactose. Initially, glucose is present, and the lac operon is repressed. After glucose is exhausted, the culture enters a lag phase before growth resumes on lactose. Which molecular event best explains the relief of repression and the subsequent induction of the lac operon?

  1. Phosphorylation of the lac repressor by a glucose-sensing kinase, leading to its degradation
  2. Binding of lactose to the lac repressor, causing it to dissociate from the operator Correct answer.
  3. Direct activation of RNA polymerase by glucose, overriding repressor binding
  4. Increased binding of cAMP to the lac repressor, reducing its affinity for the operator

Explanation

The lac operon is under negative control by the LacI repressor. When glucose is depleted, cAMP levels rise and CRP binds near the promoter, but induction specifically requires lactose (or allolactose) to bind the repressor, causing a conformational change that reduces its affinity for the operator. This allows RNA polymerase to transcribe the operon.

Question 28Biotechnology (BT)

A novel protein is found to have a molecular mass of 50 kDa and an isoelectric point (pI) of 6.5. When subjected to 2D-PAGE, it appears as a single spot. If the protein is treated with a reagent that adds a phosphate group to a specific serine residue, what is the most likely effect on its position in 2D-PAGE?

  1. The spot will shift toward the basic end in the first dimension and remain at the same molecular weight in the second dimension.
  2. The spot will shift toward the basic end in the first dimension and toward a higher molecular weight in the second dimension.
  3. The spot will shift toward the acidic end in the first dimension and remain at the same molecular weight in the second dimension. Correct answer.
  4. The spot will shift toward the acidic end in the first dimension and toward a lower molecular weight in the second dimension.

Explanation

Phosphorylation adds negative charges (phosphate groups) to the protein, lowering its isoelectric point (pI). In the first dimension (isoelectric focusing), the protein will migrate to a more acidic pH to become neutral, shifting toward the acidic end. In the second dimension (SDS-PAGE), the molecular weight is largely unchanged because the phosphate group is small and the protein's mass is nearly the same.

Question 29Chemistry (CY)

Which quantum number set correctly describes the highest-energy electron of a ground-state chromium atom (Z = 24)?

  1. n = 3, l = 2, m = +2, s = +1/2
  2. n = 4, l = 1, m = -1, s = +1/2
  3. n = 4, l = 0, m = 0, s = +1/2 Correct answer.
  4. n = 3, l = 1, m = 0, s = -1/2

Explanation

Chromium has the configuration [Ar]3d5 4s1, since the half-filled d5 4s1 arrangement is extra-stable. The single 4s electron lies in the l = 0 orbital, so its quantum numbers are n = 4, l = 0, m = 0, s = +1/2.

Question 30Chemistry (CY)

For a particle in a one-dimensional box of length L, the energy separation between the n = 1 and n = 2 levels is proportional to which quantity?

  1. 1/L
  2. 1/L² Correct answer.
  3. 1/L³
  4. L²

Explanation

The energy levels are E_n = n²h²/(8mL²). The difference E_2 − E_1 = 3h²/(8mL²), which scales as 1/L². Doubling the box length therefore quarters the spacing.

Question 31Chemistry (CY)

Which molecule has a net dipole moment of zero in its equilibrium geometry?

  1. H₂O
  2. BF₃ Correct answer.
  3. NH₃
  4. CH₃Cl

Explanation

BF₃ is trigonal planar (D3h), and the three B–F bond dipoles cancel vectorially, giving zero net moment. NH₃, H₂O and CH₃Cl all lack the symmetry needed to cancel their bond dipoles.

Question 32Chemistry (CY)

According to molecular orbital theory, which species is predicted to be paramagnetic?

  1. N₂
  2. F₂
  3. C₂
  4. O₂ Correct answer.

Explanation

In O₂ the two highest-energy electrons occupy the degenerate π* orbitals singly, giving a triplet ground state with two unpaired electrons. The other diatomics listed have all their valence electrons paired.

Question 33Chemistry (CY)

Which of the following is the strongest intermolecular force present in liquid hydrogen fluoride?

  1. Dipole–dipole interactions
  2. Hydrogen bonding Correct answer.
  3. Ion–induced dipole interactions
  4. London dispersion forces

Explanation

HF has an H atom bonded to a highly electronegative F atom, allowing strong hydrogen bonding between molecules. This interaction dominates over ordinary dipole–dipole and dispersion forces and explains HF's unusually high boiling point relative to HCl.

Question 34Chemistry (CY)

A reaction has ΔH = −50 kJ mol⁻¹ and ΔS = −100 J K⁻¹ mol⁻¹. Above what approximate temperature does the reaction become non-spontaneous at standard conditions?

  1. It is spontaneous at all temperatures
  2. 50 K
  3. 500 K Correct answer.
  4. 5000 K

Explanation

Spontaneity requires ΔG = ΔH − TΔS < 0. With both terms negative, the reaction is spontaneous only while T < ΔH/ΔS = (−50,000)/(−100) = 500 K. Above 500 K the entropy term outweighs the enthalpy term.

Question 35Chemistry (CY)

Which statement about a catalyst in a reversible reaction at equilibrium is correct?

  1. It increases the enthalpy change of the reaction.
  2. It increases the equilibrium constant.
  3. It lowers the activation energy of both the forward and reverse reactions equally. Correct answer.
  4. It shifts the equilibrium toward products.

Explanation

A catalyst provides an alternative pathway with lower activation energy for the forward and reverse steps alike, so the equilibrium constant and position remain unchanged. Only the rate at which equilibrium is reached is affected.

Question 36Chemistry (CY)

For the reaction 2A + B → products, doubling [A] while keeping [B] constant quadruples the rate, and doubling [B] while keeping [A] constant doubles the rate. What is the overall order?

  1. 2
  2. 3 Correct answer.
  3. 1
  4. 4

Explanation

The rate depends on [A]² because doubling A quadruples the rate, and on [B]¹ because doubling B doubles the rate. The rate law is therefore rate = k[A]²[B], giving an overall order of 2 + 1 = 3.

Question 37Chemistry (CY)

What is the pH of a 0.01 M aqueous solution of a weak monoprotic acid with K_a = 1.0 × 10⁻⁶?

  1. 4.0 Correct answer.
  2. 6.0
  3. 3.0
  4. 2.0

Explanation

For a weak acid, [H⁺] ≈ √(K_a·C) = √(10⁻⁶ × 10⁻²) = 10⁻⁴ M. The pH is therefore 4.0, assuming the degree of dissociation is small enough for the approximation to hold.

Question 38Chemistry (CY)

A buffer is best prepared by mixing which pair of solutions?

  1. Weak base and a strong acid in equimolar amounts
  2. Strong acid and its salt with a strong base
  3. Weak acid and its conjugate base in comparable amounts Correct answer.
  4. Two strong acids of equal concentration

Explanation

A buffer requires a weak acid and its conjugate base (or a weak base and its conjugate acid) present in appreciable concentrations, so the pH is set by the Henderson–Hasselbalch relation. Such a mixture resists pH change on adding small amounts of acid or base.

Question 39Chemistry (CY)

Which of the following coordination complexes is expected to be diamagnetic?

  1. [Fe(CN)₆]⁴⁻ Correct answer.
  2. [Co(H₂O)₆]²⁺
  3. [Fe(CN)₆]³⁻
  4. [CoF₆]³⁻

Explanation

[Fe(CN)₆]⁴⁻ contains Fe(II), a d⁶ ion, and cyanide is a strong-field ligand, so the t2g level is completely filled with paired electrons (t2g⁶ eg⁰). The complex is low-spin and diamagnetic.

Question 40Chemistry (CY)

Which complex can exhibit optical isomerism?

  1. [PtCl₄]²⁻
  2. cis-[Co(en)₂Cl₂]⁺ Correct answer.
  3. trans-[Co(en)₂Cl₂]⁺
  4. [Co(NH₃)₆]³⁺

Explanation

The cis isomer of [Co(en)₂Cl₂]⁺ lacks an internal plane or centre of symmetry and exists as a pair of non-superimposable mirror images (Δ and Λ). The trans isomer has a plane of symmetry and is achiral.

Question 41Chemistry (CY)

Which reagent converts a primary alcohol directly into a carboxylic acid under aqueous acidic conditions?

  1. Dess–Martin periodinane
  2. LiAlH₄ in ether
  3. PCC in CH₂Cl₂
  4. Acidified KMnO₄ Correct answer.

Explanation

Hot acidified permanganate is a strong oxidant that takes a primary alcohol through the aldehyde to the carboxylic acid. PCC and Dess–Martin periodinane stop at the aldehyde stage.

Question 42Chemistry (CY)

Which product is formed when propene is treated with HBr in the presence of peroxides?

  1. 2-bromopropane
  2. Propan-2-ol
  3. 1,2-dibromopropane
  4. 1-bromopropane Correct answer.

Explanation

Peroxides initiate a radical chain, and the bromine radical adds to the terminal carbon to give the more stable secondary radical. Hydrogen abstraction then yields 1-bromopropane, the anti-Markovnikov product.

Question 43Chemistry (CY)

Which compound gives a positive iodoform test?

  1. Propan-1-ol
  2. Ethanol Correct answer.
  3. Methanol
  4. Diethyl ether

Explanation

The iodoform test is positive for compounds containing the CH₃CO– group or a CH₃CH(OH)– unit that can be oxidised to it. Ethanol is oxidised to acetaldehyde, which then forms the yellow precipitate of CHI₃.

Question 44Chemistry (CY)

In the electrophilic aromatic substitution of nitrobenzene, the nitro group is best described as:

  1. Activating and ortho/para directing
  2. Activating and meta directing
  3. Deactivating and meta directing Correct answer.
  4. Deactivating and ortho/para directing

Explanation

The nitro group withdraws electron density inductively and by resonance, deactivating the ring toward electrophiles. Its electron withdrawal destabilises the ortho and para arenium intermediates more than the meta one, so substitution occurs mainly at the meta position.

Question 45Chemistry (CY)

Which of the following is the rate-determining step in an SN1 reaction?

  1. Protonation of the substrate
  2. Loss of the leaving group to form a carbocation Correct answer.
  3. Rearrangement of the carbocation
  4. Attack of the nucleophile on the carbocation

Explanation

In an SN1 mechanism the slow, rate-determining step is the heterolysis of the C–X bond to give a carbocation and a leaving group. Nucleophilic attack on the carbocation then follows rapidly.

Question 46Chemistry (CY)

How many signals are observed in the ¹H NMR spectrum of 1-bromopropane?

  1. 3 Correct answer.
  2. 4
  3. 2
  4. 5

Explanation

1-Bromopropane has three chemically distinct sets of protons: the CH₃ group, the CH₂ adjacent to it, and the CH₂ bonded to bromine. Each set gives one signal, so three signals are seen.

Question 47Chemistry (CY)

Which compound shows a strong absorption band near 1700 cm⁻¹ in its IR spectrum?

  1. Diethyl ether
  2. Hexane
  3. Ethanol
  4. Acetone Correct answer.

Explanation

The C=O stretching vibration of a saturated ketone appears near 1715 cm⁻¹ and is typically the strongest band in the spectrum. Acetone is the only listed compound containing a carbonyl group.

Question 48Chemistry (CY)

Which of the following polymers is a condensation polymer?

  1. Polystyrene
  2. Nylon-6,6 Correct answer.
  3. Polyethylene
  4. Polypropylene

Explanation

Nylon-6,6 is formed by step-growth condensation of adipic acid and hexamethylenediamine with elimination of water. The other three are addition polymers of alkene monomers.

Question 49Chemistry (CY)

Which of the following statements about the entropy change accompanying the isothermal expansion of an ideal gas into a vacuum is correct?

  1. ΔS cannot be defined for an irreversible process
  2. ΔS < 0 because the gas expands
  3. ΔS = 0 because q = 0
  4. ΔS > 0 because the gas occupies a larger volume Correct answer.

Explanation

For an ideal gas expanding into a vacuum, ΔU = 0 and q = 0, but entropy is a state function. Since ΔS = nR ln(V₂/V₁) with V₂ > V₁, the entropy change is positive, reflecting the greater number of accessible microstates.

Question 50Chemistry (CY)

Which quantum-mechanical principle requires that the wavefunction be antisymmetric with respect to interchange of two electrons?

  1. Hund's rule
  2. Aufbau principle
  3. Pauli exclusion principle Correct answer.
  4. Heisenberg uncertainty principle

Explanation

The Pauli exclusion principle follows from the requirement that a many-electron wavefunction be antisymmetric under exchange of any two electrons. This antisymmetry forbids two electrons from occupying the same spin-orbital.