What is the amount of work done when two moles of ideal gas is co

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 Multiple Choice QuestionsMultiple Choice Questions

381.

Kirchhoff's equation is

  • log k2k1 = Ea2.303 R 1T1 - 1T2

  • log p2p1 = Hv2.303 R T2 - T1T1 × T2

  • Cp = H2 - H1T2 - T1

  • log k2k1 = H2.303 R 1T1 - 1T2


382.

The heat of combustion of carbon is - 393.5 kJ/mol. The heat released upon the formation of 35.2 g of CO2 from carbon and oxygen gas is

  • + 315 kJ

  • - 31.5 kJ

  • - 315 kJ

  • + 31.5 kJ


383.

Find the correct equation

  • E2 - E1 - H2 + H1 = n2RT - n1RT

  • E2 - E1 - H2 - H1 = n2RT - n1RT

  • H2 - H1 - E2 + E1 = n2RT - n1RT

  • H2 - H1 - E2 - E1 = n2RT - n1RT


384.

Assuming enthalpy of combustion of hydrogen at 273 K is -286 kJ and enthalpy of fusion of ice at the same temperature to be +6.0 kJ, calculate enthalpy change during formation of 100 g of ice.

  • +1622 kJ

  • - 1622 kJ

  • +292 kJ

  • - 292 kJ


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385.

For which among the following reactions, change in entropy is less than zero?

  • Sublimation of iodine

  • Dissociation of hydrogen

  • Formation of water

  • Thermal decomposition of calcium carbonate


386.

Which among the following is a feature of adiabatic expansion?

  • V < 0

  • U < 0

  • U > 0

  • T = 0


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387.

What is the amount of work done when two moles of ideal gas is compressed from a volume of 1m3 to 10 dm3 at 300 K against a pressure of 100 kPa?

  • 99 kJ

  • -99kJ

  • 114.9 kJ

  • -114.9 kJ


A.

99 kJ

Given

Number of moles = 2 moles

V1 = 1m3 = 1000 dm3

V2 = 10 dm3

 We have, Work done (W) = -pextdV

= -100 kPa × (10 - 1000) dm3

-100 kPa × -999 dm3

= 99900 L = 99.9 kJ


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388.

The work done during combustion of 9 × 10-2 kg of ethane, C2H6 (g) at 300 K is (Given R = 8.314 J deg-1 mol-1, atomic mass C = 12, H= 1).

  • 6.236 kJ

  • -6.236 kJ

  • 18.71 kJ

  • -18.71 kJ


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389.

The first law of thermodynamics for isothermal process is

  • q = -W

  • U = W

  • U = qv

  • U = -qv


390.

Identify the invalid equation

  • H = ΣHproducts - ΣHreactants

  • H = U + pV

  • H°(reaction) = ΣH°(products bonds) - ΣH°(reactant bonds)

  • H = U + nRT


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