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If two bodies are not in thermal equilibrium, then in which direction the heat will flow?


If two bodies are not in thermal equilibrium, then the heat flows from body at high temperature to the body at low temperature. 

The flow of heat takes place for the bodies which are not in thermal equilibrium. 
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Show that Cp – Cv = R.

Consider one mole of an ideal gas enclosed in a cylinder fitted with movable frictionless piston.

Let the gas be heated at constant volume first. Let the temperature of the gas increase by dT when dQ quantity of heat is supplied.


Consider one mole of an ideal gas enclosed in a cylinder fitted with

          = dU + PdV
          = dU               space space space space space space open square brackets because space d V equals 0 close square brackets

therefore space space space space space space space dQ apostrophe equals 1 cross times straight C subscript straight p cross times dT

By first law of thermodynamics,

           dQ = dU + dW = dU + PdV

or   CpdT = Cx dT + RdT

open square brackets because space at space constant space pressure space PdV space equals space RdT close square brackets
therefore space space space space space straight C subscript straight p equals straight C subscript straight v space plus space straight R
or space straight C subscript straight p minus straight C subscript straight v equals straight R

where  CpCand R are in same units.

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By using first law of thermodynamics, find the change in internal energy during boiling of liquids.

Let the latent heat of vaporisation of a liquid of mass m at its boiling temperature be L. 

Let V be the volume of liquid and Vv be the volume of vapours and let evaporation take place at constant pressure P.

According to the first law of thermodynamics,

space space space space space space space space space space increment straight Q equals increment straight U plus straight P increment straight V space

rightwards double arrow space space space space space space mL equals increment straight U plus straight P left parenthesis straight V subscript straight v minus straight V subscript straight ell right parenthesis space

rightwards double arrow space space space increment straight U equals mL minus straight P left parenthesis straight V subscript straight v minus straight V subscript straight ell right parenthesis space space 

This is the required expression for change in internal energy. 

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Find the internal energy of air in a room of size 3m x 4m x 6m at pressure 75cm of mercury.

The internal energy of V volume of gas at pressure P is given by, 

            space space space space space space
space space space space space space space space space space space space space U equals fraction numerator P V over denominator y minus 1 end fraction             ... (1)

we have,

Volume of the room, V=3m x 4m x 6m = 72m

Pressure, P=75cm of mercury

                =75 x 13.6 x 980 dyne/cm2 

                = 9.996 x 10N/m

Air is mixture of diatomic gases.

Therefore,
                 y = 1.4

Putting values in equation (1), we have

So, internal energy is given by, 


space straight U space equals fraction numerator 9.996 cross times 10 to the power of 4 cross times 72 over denominator 1.4 minus 1 end fraction space space space

space space space space space equals space 1.8 cross times 10 to the power of 7 straight J          
            
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When two bodies will be said to be in thermal equilibrium?

When the temperature of two bodies is same they are said to be in thermal equilibrium. 
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