The work done during the expansion of a gas from a volume of 4 dm

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 Multiple Choice QuestionsMultiple Choice Questions

221.

Enthalpy of solution of NaOH (solid) in water is -41.6 kJ mol-1. When NaOH is dissolved in water, the temperature of water :

  • increases

  • decreases

  • does not change

  • fluctuates indefinitely


222.

Hess's law is used to calculate :

  • enthalpy of reaction

  • entropy of reaction

  • work done in reaction

  • All of the above


223.

The f for CO2 (g), CO (g) and H2O (g) are -393.5, -110.5 and -241.8 kJ/ mol respectively. The standard enthalpy change (in kJ) for the reaction-

CO2 (g) + H2 (g) → CO (g) + H2O (g) is :

  • 524.1

  • 41.2

  • -262.5

  • -41.2


224.

If at 298 K the bond energies of C-H, C-C, C=C and H-H bonds are respectively 414, 347, 615 and 435 kJ mol-1, the value of enthalpy change for the reaction

H2C=CH2 (g) + H2 (g) → H3C - CH3 (g) at 298 K will be

  • +250 kJ

  • -250 kJ

  • +125 kJ

  • -125 kJ


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225.

In an isochoric process, ΔH for a system is equal to:

  • p.ΔV

  • pV

  • E+pΔV

  • ΔE


226.

Consider the reaction,

N2 + 3H2 → 2NH3

carried out at constant temperature and pressure. If H and U are the enthalpy and internal energy changes for the reaction, which of the following expression is true?

  • H > U

  • H <U

  • H = U

  • H = 0


227.

The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol respectively. The enthalpy of formation of carbon monoxide per mole is

  • 110.5 kJ

  • 676.5 kJ

  • -676.5 kJ

  • -110.5 kJ


228.

For the reaction

A(g) + 2B(g) → 2C (g) + 3D (g)

the change of enthalpy at 27° is 19 kcal. The value of ΔE is:

  • 21.2 kcal

  • 17.8 kcal

  • 18.4 kcal

  • 20.6 kcal


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229.

The maximum number of molecule is present in

  • 15 L of H2 gas at STP

  • 5 L of N2 gas at STP

  • 0.5 g of H2 gas

  • 10 g of O2 gas


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230.

The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm, is

  • -6 J

  • -608 J

  • +304 J

  • -304 J


B.

-608 J

Work done (W) = -Pext (V2 - V1)

                      = -3 × (6 - 4) = -6 L-atm

                      = -6 × 101.32 J ( 1 L-atm = 101.32 J)

                      = -607.92 ≈ 608 J


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