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NEET Practice Questions (MCQs) with Answers & Solutions

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The first law of thermodynamics is only

The law of conservation of energy

The law of conservation of mass

The law of conservation of momentum

Both (1) and (2)

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Explanation

First low of thermodynamics is also known as Law of conservation of mass and energy.

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A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If ΔH is the enthalpy change and ΔE is the change in internal energy, then [KCET 2005]

ΔH > ΔE

ΔH < ΔE

ΔH = ΔE

The relationship depends on the capacity of the vessel

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Explanation

Formation of CO₂ from CO is an exothermic reaction; heat is evolved from the system, i.e., energy is lowered. Thus, exothermic reactions occur spontaneously on account of decrease in enthapy of system. Thus, ΔE > ΔH.

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The relation between ΔE and ΔH is [MP PET 1992; MP PMT 1996; MP PET/PMT 1998]

$ΔH = ΔE - P ΔV$

$ΔH = ΔE + P ΔV$

$ΔE = ΔV + ΔH$

$ΔE = ΔH + P ΔV$

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Explanation

ΔH = ΔE + PΔV.

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At constant T and P, which one of the following statements is correct for the reaction, $CO (g) + \frac{1}{2} O_{2} (g) \to {CO}_{2} (g)$ [AIIMS 1982, 83; KCET 1988; BHU 1995; MP PET 1997, 99]

ΔH is independent of the physical state of the reactants of that compound

ΔH > ΔE

ΔH < ΔE

ΔH = ΔE

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Explanation

$Δ n_{g} = 1 - \frac{3}{2} = \frac{- 1}{2}$ , As $Δ n_{g}$ is negative, thus ΔH < ΔE.

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For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, ΔU and w correspond to [AIIMS 2005]

$Δ U <0, w =0$

$Δ U = 0, w < 0$

$Δ U > 0, w = 0$

$Δ U < 0, w > 0$

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Explanation

Bomb calorimeter is commonly used to find the heat of combustion of organic substances which consists of a sealed combustion chamber, called a bomb. If a process is run in a sealed container then no expansion or compression is allowed, so w = 0 and âˆ†U = q.

âˆ†U < 0, w = 0

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If ΔH is the change in enthalpy and ΔE the change in internal energy accompanying a gaseous reaction [KCET 1989; CBSE PMT 1990]

ΔH is always greater than ΔE

ΔH < ΔE only if the number of moles of the products is greater than the number of the reactants

ΔH is always less than ΔE

ΔH < ΔE only if the number of moles of the products is less than the number of moles of the reactants

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Explanation

If Δn = –ve than ΔH < ΔE.

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Enthalpy for the reaction C + O₂ → CO₂ is [DPMT 1987, 91]

Positive

Negative

Zero

None

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Explanation

It is a combustion reaction, ΔH = –ve.

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The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 litres to 20 litres at 25°C is

$2 .303 \times 298 \times 0 .082 \log 2$

$298 \times 10^{7} \times 8 .31 \times 2 .303 \log 2$

$2 .303 \times 298 \times 0 .082 \log 0 .5$

$8 .31 \times 10^{7} \times 298 - 2 .303 \log 0 .5$

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Explanation

$W = 2 .303 n R T \log \frac{V_{2}}{V_{1}}$

$= 2 .303 \times 1 \times 8 .314 \times 10^{7} \times 298 \log \frac{20}{10}$

$= 298 \times 10^{7} \times 8 .314 \times 2 .303 \log 2$ .

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In a reversible isothermal process, the change in internal energy is

Zero

Positive

Negative

None of these

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Explanation

ΔE = 0 for reversible isothermal process.

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The enthalpy of neutralization of which of the following acids and bases is nearly –13.6 Kcal [Roorkee 1999]

HCN and NaOH

HCl and KOH

HCl and HCN

HCl and NH₄OH

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Explanation

Heat of neutralisation of a strong acid and strong base is equal to –13.7 kcal.

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Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.