Chemistry MCQs for NEET — Practice Questions with Answers

Practice free Chemistry NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology
Language English हिंदी
Register free for difficulty & keyword filters
● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

ΔE = 0 for reversible isothermal process.

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

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

● ● ● ● ●

Work done during isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 300 K is (Gas constant = 2) [AIIMS 2000]

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

− W=+ 2.303 nRT  log p1p2

− W=2.303×1×2×300 log 101=1381.8 cal. 

● ● ● ● ●

Joule-Thomson expansion is [JIPMER 2000]

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Joule-Thomson expansion is isoenthalpic.

n thermodynamics, the Joule–Thomson effect (also known as the Joule–Kelvin effect, Kelvin–Joule effect) describes the temperature change of a real gas or liquid (as differentiated from an ideal gas) when it is forced through a valve or porous plug while keeping it insulated so that no heat is exchanged with the environment.[1][2][3]This procedure is called a throttling process or Joule–Thomson process.[4] At room temperature, all gases except hydrogen, helium and neon cool upon expansion by the Joule–Thomson process when being throttled through an orifice; these three gases experience the same effect but only at lower temperatures.[5][6] Most liquids such as hydraulic oils will be warmed by the Joule-Thomson throttling process.

The gas-cooling throttling process is commonly exploited in refrigeration processes such as air conditioners, heat pumps, and liquefiers.[7][8] In hydraulics, the warming effect from Joule-Thomson throttling can be used to find internally leaking valves as these will produce heat which can be detected by thermocouple or thermal-imaging camera. Throttling is a fundamentally irreversible process. The throttling due to the flow resistance in supply lines, heat exchangers, regenerators, and other components of (thermal) machines is a source of losses that limits the performance.

● ● ● ● ●

In an adiabatic expansion of an ideal gas [KCET (Med.) 2001; MH CET 2000]

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

q = ΔE – W, if q = 0 for adiabatic process, than ΔE = W.

● ● ● ● ●

For the reaction CH3COOH(l)+2O2(g)⇌2CO2(g)+2H2O(l) at 25°C and 1 atm. pressure, ΔH = –874 kJ. Then the change in internal energy (ΔE) is …. [Orissa JEE 2002]

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

For this reaction Δn = 0 than ΔE = ΔH.

● ● ● ● ●

One mole of an ideal gas is allowed to expand freely and adiabatically into vacuum until its volume has doubled. A statement which is not true concerning this expression is [Pb. PMT 1998]

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

ΔS = 0 is not true.

● ● ● ● ●

At 27°C, one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The values of ΔE and q are (R = 2) [BHU 2001]

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

W=2.303 nRT logP2P1 

=2.303×1×2×300 log102=965.84 

at constant temperature, ΔE = 0.

ΔE = q + w; q = –w = –965.84 cal.

● ● ● ● ●

ΔE° of combustion of isobutylene is –X kJ mol–1. The value of ΔH° is [DCE 2004]

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

(CH3)2C=CH2(g)+6O2(g)→4CO2(g)+4H2O(l)

Δng=4−6=−2 (i.e., negative)

we know that ΔH=ΔE+ΔngRT

=ΔE−(Δng)RT (∵ Δng = –ve)

∴ ΔH < ΔE

● ● ● ● ●

An ideal gas expands in volume from 1 × 10–3 m3 to 1 × 10–2 m3 at 300 K against a constant pressure of 1 × 105 Nm–2. The work done is [AIEEE 2004]

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

W=−PΔV=−1×105 (1×10−2−1×10−3) 

=−1×105×9×10−3=−900 J  

Ready to ace NEET?

Free access · No credit card required

Frequently Asked Questions

Yes. You can attempt every Chemistry 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.