Calculate change in internal energy if ∆H = - 92.2 kJ, P =

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 


6.

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

  • Gas

  • Liquid

  • Equilibrium between gas and liquid

  • Super critical temperature


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


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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


B.

-88 kJ

Change in internal energy

H = E + PV

E = H - PV = -92.2 - 40 × (-1) × 101 × 10-3

      = -92.2 + 4.04

      = -88.16 kJ ≈ -88 kJ


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