Chapter Chosen

Thermodynamics

Book Chosen

Physics Part II

Subject Chosen

Physics

Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12

A mass of diatomic gas left parenthesis straight gamma space equals space 1.4 right parenthesis at a pressure of 2 atm is compressed adiabatically so that its temperature rise from 27oC to 927o C. The pressure of the gas is final state is 

  • 28 atm

  • 68.7 atm

  • 256 atm

  • 8 atm


C.

256 atm

T1 = 273 + 27 = 300k
T2 = 273+927 = 1200 K
Gas equation for adiabatic process

pV to the power of straight gamma space equals space constant

straight p open parentheses straight T over straight P close parentheses to the power of straight gamma space equals space constant
space
left parenthesis therefore space pV space equals space RT right parenthesis

therefore comma
straight p subscript 2 over straight p subscript 1 space equals space open parentheses straight T subscript 2 over straight T subscript 1 close parentheses to the power of fraction numerator straight gamma over denominator straight gamma minus 1 end fraction end exponent

Or

straight p subscript 2 space equals straight p subscript 1 space open parentheses straight T subscript 2 over straight T subscript 1 close parentheses to the power of fraction numerator straight gamma over denominator straight gamma minus 1 end fraction end exponent

straight p subscript 2 space equals space 2 open parentheses 1200 over 300 close parentheses to the power of fraction numerator 1.4 over denominator 1.4 minus 1 end fraction end exponent

straight p subscript 2 space equals space 256 space atm

T1 = 273 + 27 = 300k
T2 = 273+927 = 1200 K
Gas equation for adiabatic process

pV to the power of straight gamma space equals space constant

straight p open parentheses straight T over straight P close parentheses to the power of straight gamma space equals space constant
space
left parenthesis therefore space pV space equals space RT right parenthesis

therefore comma
straight p subscript 2 over straight p subscript 1 space equals space open parentheses straight T subscript 2 over straight T subscript 1 close parentheses to the power of fraction numerator straight gamma over denominator straight gamma minus 1 end fraction end exponent

Or

straight p subscript 2 space equals straight p subscript 1 space open parentheses straight T subscript 2 over straight T subscript 1 close parentheses to the power of fraction numerator straight gamma over denominator straight gamma minus 1 end fraction end exponent

straight p subscript 2 space equals space 2 open parentheses 1200 over 300 close parentheses to the power of fraction numerator 1.4 over denominator 1.4 minus 1 end fraction end exponent

straight p subscript 2 space equals space 256 space atm

1782 Views

Explain why

(a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2)/2.

When two bodies are in thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperatures becomes equal.

The final temperature can be the mean temperature
fraction numerator straight T subscript 1 space plus space straight T subscript 2 over denominator 2 end fraction, only when thermal capacities of the two bodies are equal.
239 Views

Explain why
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.

(b) High specific heat capacity is required because the heat absorbed by a substance is directly proportional to the specific heat of the substance. Thus the coolant will work more effectively. 
577 Views

A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g? 

Water is flowing at a rate of 3.0 litre/min

The geyser heats the water, raising the temperature from 27°C to 77°C. 

Initial temperature, T1 = 27°C 

Final temperature, T2 = 77°C 

Rise in temperature, ΔT = T2 – T

                                      = 77 – 27

                                      = 50°C 

Heat of combustion = 4 × 104 J/g 

Specific heat of water, c = 4.2 J g–1 °C–1 

Mass of flowing water, m = 3.0 litre/min

                                     = 3000 g/min 

Total heat used, ΔQ = mc Δ

                             = 3000 × 4.2 × 50 

                             = 6.3 × 10J/min 

Therefore,

Rate of consumption =
open parentheses fraction numerator 6.3 space cross times space 10 to the power of 5 over denominator 4 space cross times space 10 to the power of 4 end fraction close parentheses space 

                                =  15.75 g/min.
464 Views

Explain why?
(c) Air pressure in a car tyre increases during driving.

During driving, due to motion of the car the temperature of air inside the tyre increases.

According to Charle's law, 
P ∝T. 

Therefore, air pressure inside the tyre increases.
231 Views

What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R= 8.3 J mol–1 K–1.)

Mass of nitrogen, m = 2.0 × 10–2 kg = 20 g

Rise in temperature, ΔT = 45°C

Molecular mass of N2M = 28 

Universal gas constant, R = 8.3 J mol–1 K–1 

Number of moles, n = straight m over straight M

                              = (fraction numerator 2 space cross times space 10 to the power of negative 2 end exponent space cross times space 10 cubed over denominator 28 end fraction)

                              = 0.714 

Molar specific heat at constant pressure for nitrogen,

C
p = 7 over 2 R

    = 7 over 2 × 8.3 

    = 29.05 J mol-1 K-1 

The total amount of heat to be supplied is given by the relation, 

ΔQ = nCΔ

     = 0.714 × 29.05 × 45 

     = 933.38 J 

Therefore, the amount of heat to be supplied is 933.38 J.
403 Views