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,

... (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/cm

= 9.996 x 10

Air is mixture of diatomic gases.

Therefore,

y = 1.4

Putting values in equation (1), we have

So, internal energy is given by,

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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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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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Show that C_{p} – C_{v} = 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.

= dU + PdV

= dU

By first law of thermodynamics,

dQ = dU + dW = dU + PdV

or C_{p}dT = C_{v }x dT + RdT

where C_{p}_{, }C_{v }and 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_{ℓ} be the volume of liquid and V_{v} be the volume of vapours and let evaporation take place at constant pressure P.

According to the first law of thermodynamics,

This is the required expression for change in internal energy.

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