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 Multiple Choice QuestionsMultiple Choice Questions

181.

Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T2. The final pressure pf is:

  • 2 straight p subscript straight i open parentheses fraction numerator straight T subscript 1 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses
  • 2 straight p subscript straight i open parentheses fraction numerator straight T subscript 2 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses
  • 2 straight p subscript straight i space open parentheses fraction numerator straight T subscript 1 straight T subscript 2 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses
  • 2 straight p subscript straight i space open parentheses fraction numerator straight T subscript 1 straight T subscript 2 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses
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182.

The intermolecular interaction that is dependent on the inverse cube of the distance between the molecule is:

  • ion-ion interaction

  • ion-dipole interaction

  • London force

  • London force

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183.

If Z is a compressibility factor, Vander Waal's equation at low pressure can be written as 

  • straight Z space equals 1 plus RT over pb
  • straight Z space equals space 1 minus straight a over VRT
  • straight Z space equals space 1 minus pb over RT
  • straight Z space equals space 1 minus pb over RT
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184.

For the gaseous state, if most probable speed is denoted by C*, average speed by C and mean square speed by C, then for a large number of molecules the ratios of these speeds are:

  • straight C asterisk times colon stack space straight C with bar on top space colon thin space straight C space equals space 1.225 space colon 1.128 space space colon 1
  • straight C to the power of asterisk times space colon thin space straight C with bar on top space colon space straight C space equals space 1.128 space colon thin space 1.225 space colon space 1
  • straight C to the power of asterisk times space colon thin space straight C with bar on top space colon space straight C space equals space 1.28 space colon thin space 1.225 space colon space
  • straight C to the power of asterisk times space colon thin space straight C with bar on top space colon space straight C space equals space 1.28 space colon thin space 1.225 space colon space
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185.

The compressibility factor for a real gas at high pressure is

  • 1 plus RT over pb
  • 1
  • 1 plus pb over RT
  • 1 plus pb over RT
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186.

'a' and 'b' are van der Waals constants for gases. Chlorine is more easily liquefied than ethane because :

  • a and b for Cl2 > a and b for C2H6

  • a and b for Cl2 < a and b for C2H6

  • a and Cl2 < a for C2H6 but b for Cl2 > b for C2H6

  • a and Cl2 < a for C2H6 but b for Cl2 > b for C2H6

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187.

If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established?(Given: Vapour pressure of H2O at 300 K is 3170 Pa; R = 8.314 J K–1 mol–1)

  • 5.56 x 10-3 mol

  • 1.53 x 10-2 mol

  • 4.46 x 10-2 mol

  • 4.46 x 10-2 mol

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188.

Assuming that water vapour is an ideal gas, the internal energy change(∆U) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (Given: Molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol-1 and R = 8.3 J mol–1K–1 will be) –

  • 4.100 kJ mol–1

  • 3.7904 kJ mol–1

  • 37.904 kJ mol–1

  • 37.904 kJ mol–1


C.

37.904 kJ mol–1


∆ng =1−0 =1
∆H =∆U +∆ngRT
∆U =∆H −∆ngRT
= 41 – 8.3 × 10-3 × 373
= 37.9 kJ mol-1

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189.

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel,
PCl5 (g) ⇌ PCl3(g) + cl2(g)
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be

  • open parentheses fraction numerator straight x over denominator straight x plus 1 end fraction close parentheses straight P
  • open parentheses fraction numerator 2 straight x over denominator straight x plus 1 end fraction close parentheses straight P
  • open parentheses fraction numerator straight x over denominator straight x minus 1 end fraction close parentheses straight P
  • open parentheses fraction numerator straight x over denominator straight x minus 1 end fraction close parentheses straight P
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190.

The formation of the oxide ion O2-(g) requires first an exothermic and then an endothermic step as shown below
straight O space left parenthesis straight g right parenthesis space space plus straight e to the power of minus space straight O to the power of minus left parenthesis straight g right parenthesis space increment straight H to the power of straight o space equals space minus space 142 space kJ space mol to the power of negative 1 end exponent
straight O to the power of minus space left parenthesis straight g right parenthesis space space plus straight e to the power of minus straight O to the power of 2 minus end exponent space left parenthesis straight g right parenthesis increment straight H to the power of straight o space equals space 844 space kJ space mol to the power of minus space

  • Oxygen is more electronegative

  • O- ion has comparatively larger size than oxygen atom

  • O- ion will tend to resist the addition of another electron

  • O- ion will tend to resist the addition of another electron

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