In P versus V graph, the horizontal line is found in which exists

Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The pH of the solution obtained on neutralisation of 40 mL 0.1 M NaOH with 40 mL 0.1 M CH3COOH is

  • 7

  • 8

  • 6

  • 3


2.

Inert gases are mixed in iodine vapours. Then there are _____ between them.

  • H-bonding

  • van der Waals forces

  • electrostatic forces

  • metallic bonds


3.

The order of bond length is

  • O2 < O3 < O22-

  • O2 < O22- < O3

  • O22- < O3 < O2

  • O2 = O22- > O3


4.

Largest difference in radii is found in case of the pair

  • Li, Na

  • Na, K

  • K, Rb

  • Rb, Cs


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5.

X-rays are emitted during

  • α, n reaction

  • K-capture

  • n, α-reaction

  • β- emission 


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6.

In P versus V graph, the horizontal line is found in which exists

  • Gas

  • Liquid

  • Equilibrium between gas and liquid

  • Super critical temperature


C.

Equilibrium between gas and liquid

In the graph, at point A, CO2 exists as a gas, as pressure is increased, the volume of the gas decreases along the curve AB. At B liquefaction of the gas starts. Hence, volume decreases rapidly along BC because liquid has much less volume than the gas. At point C, liquefaction is complete. Thus, along horizontal line, gas converts into liquid.


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7.

During estimation of nickel, we prepare nickel dimethylglyoxime, a scarlet red solid. This compound is

  • ionic

  • covalent

  • metallic

  • non-ionic complex


8.

Critical temperatures for A, B, C and D gases are 25°C, 10°C, -80°C and 15°C respectively. Which gas will be liquefied more easily?

  • A

  • B

  • C

  • D


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9.

During titration of acetic acid with aqueous NaOH solution, the neutralisation graph has a vertical line. This line indicates

  • alkaline nature of equivalence

  • acidic nature of equivalence

  • neutral nature of equivalence

  • depends on experimental proceeding


10.

Calculate change in internal energy if H = - 92.2 kJ, P = 40 atm and V = -1L.

  • -42 kJ

  • -88 kJ

  • +88 kJ

  • +42 kJ


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