Thermal Properties of Matter

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Question 1

‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:

  • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction
  • fraction numerator 3 straight p subscript straight o straight V subscript straight o over denominator 2 nR end fraction
  • 9 over 2 fraction numerator straight p subscript straight o straight V subscript straight o over denominator nR end fraction
  • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator nR end fraction

Solution

A.

fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction

As, T will be maximum temperature where product of pV is maximum

Equation of line AB, we have
straight y minus straight y subscript 1 space equals space fraction numerator straight y subscript 2 minus straight y subscript 1 over denominator straight x subscript 2 minus straight x subscript 1 end fraction space left parenthesis straight x minus straight x subscript 1 right parenthesis
rightwards double arrow space straight p minus straight p subscript straight o space equals fraction numerator 2 straight p subscript straight o minus straight p subscript straight o over denominator straight V subscript straight o minus 2 straight V subscript straight o end fraction space left parenthesis straight V minus 2 straight V subscript straight o right parenthesis
straight p minus straight p subscript straight o space equals space fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction left parenthesis straight V minus 2 straight V subscript straight o right parenthesis
straight p space equals space fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction straight V squared space plus 3 straight p subscript straight o straight V
nRT space equals space fraction numerator negative straight p subscript straight o over denominator straight v subscript straight o end fraction straight V squared space plus space 3 straight p subscript straight o straight V
straight T space equals space 1 over nR open parentheses fraction numerator negative straight p subscript straight o straight V squared over denominator straight v subscript straight o end fraction space plus 3 straight p subscript straight o straight V close parentheses
For space maximum space temperature
fraction numerator partial differential straight T over denominator partial differential straight V end fraction space equals space 0
fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction left parenthesis 2 straight V right parenthesis space equals space minus space 3 straight p subscript straight o
straight V space equals space 3 divided by 2 straight V subscript straight o
left parenthesis condition space for space maximum space temperature right parenthesis
Thus comma space the space maximum space temperature space of space the space gas space during space the space process space will space be
straight T subscript max space equals space 1 over nR open parentheses fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction space straight x 9 over 4 straight V subscript 0 superscript 2 space plus space 3 straight p subscript straight o space straight x space 3 over 2 straight V subscript straight o close parentheses
equals space 1 over nR open parentheses negative 9 over 4 straight p subscript straight o straight V subscript straight o space plus space 9 over 2 straight p subscript straight o straight V subscript straight o close parentheses space equals space 9 over 4 fraction numerator straight p subscript straight o straight V subscript straight o over denominator nR end fraction

Question 2

100g of water is heated from 30°C to 50°C ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/Kg/K)

  • 8.4 kJ

  • 84 kJ

  • 2.1 kJ

  • 4.2 kJ

Solution

A.

8.4 kJ

ΔQ = M,S,ΔT
= 100 × 10-3 × 4.184 × 20 = 8.4 × 103
ΔQ = 84 kJ, ΔW = 0
ΔQ = ΔV + ΔW
ΔV = 8.4 kJ

Question 3

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by:(Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C

  • 1250°C

  • 825°C

  • 800°C

  • 885° C

Solution

D.

885° C

Heat given = Heat taken
(100) (0.1)(T – 75) = (100)(0.1)(45) + (170)(1)(45)
10(T – 75) = 450 + 7650 = 8100
T – 75 = 810
T = 885 °C

Question 4

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

  • doubled

  • four times

  • one fourth

  • halved

Solution

A.

doubled

straight H space equals space fraction numerator straight V squared increment straight t over denominator straight R end fraction
straight H apostrophe space equals space space fraction numerator straight V squared over denominator straight R apostrophe end fraction. increment straight t
Given space straight R apostrophe space equals space straight R divided by 2
Question 5

A light ray is incident perpendicular to one face of a 90° prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45°, we conclude that the refractive index n

  • n<2

  • straight n greater than square root of 2
  • straight n space greater than space fraction numerator 1 over denominator square root of 2 end fraction
  • n<1/2

Solution

B.

straight n greater than square root of 2

Angle of incidence i > C for total internal reflection.
Here i = 45° inside the medium. ∴ 45° > sin−1 (1/n)
⇒ n > √2.

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