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1cm^{3} of water at its boiling point absorbs 540 calories of heat to become steam with a volume of 1671cm^{3}.If the atmospheric pressure = 1.013x10^{5} Nm^{-2} and the mechanical equivalent of heat = 4.19 J/calorie, the energy spent in this process in overcoming intermolecular forces is

  • Option 1)

    540 cal

  • Option 2)

    40 cal

  • Option 3)

    500 cal

  • Option 4)

    Zero

 

Answers (1)

As we discussed in concept

work done by a gas -

\omega = \int PdV
 

- wherein

P is pressure

dV is small change in volume.

 

 Energy given to water = 540 cal

                                      = 540 x 4.19 J

                                      = 2262.6 J

Energy required for doing work against atmospheric pressure = p\Delta V

                                                                                                       ext

= 1.01\times 10^{5}\times 1670\times 10^{-6}\:J

= 169 J \approx 40 cal.

\therefore\:energy\:spent\:in\:overcoming\:inter-molecular\:forces\:=540\:cal-40\:cal\:=\:500\:cal

 


Option 1)

540 cal

This option is incorrect.

Option 2)

40 cal

This option is incorrect.

Option 3)

500 cal

This option is correct.

Option 4)

Zero

This option is incorrect.

Posted by

Vakul

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