States Of Matter

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Question
CBSEENCH11008094

'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 for Cl2 > a for C2H6 but b for Cl2 < b for C2H6

Solution

D.

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

Vander Waals, constant is due to force of attraction and b due to the infinite size of molecules. Thus, greater the value a and smaller the value b, larger the liquefaction.

  a b
Cl2  6.579 L2 bar mol-2 0.05622 L bar mol-2
C2H5 5.562 L2 bar mol-2 0.06380 L mol-1

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Question
CBSEENCH11008207

An ideal gas expands in volume from 1×10-3 m3 to 1×10-2 m3 at 300 K against a constant pressure of 1×105 Nm-2. The work done is

  • -900 J

  • 900 J

  • 2780 J

  • -2780 J

Solution

A.

-900 J

Work done due to change in volume against constant pressure is 
W = - p(V2-V1)
 = - 1 x 105 Nm-2 (1 x 10-2- 1 x 10-3)m3

Question
CBSEENCH11008203

As the temperature is raised from 20°C to 40°C, the average kinetic energy of neon atoms changes by a factor of which of the following?

  • 1/2

  • 2

  • 313/293

  • square root of 313 over 293 end root

Solution

C.

313/293

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Question
CBSEENCH11008142

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

  • 41.00 kJ mol–1

Solution

C.

37.904 kJ mol–1

straight H subscript 2 straight O space left parenthesis straight l right parenthesis space rightwards arrow with vaporisation on top space straight H subscript 2 straight O space left parenthesis straight g right parenthesis
∆ng =1−0 =1
∆H =∆U +∆ngRT
∆U =∆H −∆ngRT
= 41 – 8.3 × 10-3 × 373
= 37.9 kJ mol-1

Question
CBSEENCH11008067

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 1.225 space colon space 1. space 128 space

Solution

C.

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 WiredFaculty