Q. 6

# Can a process on an ideal gas be both adiabatic and isothermal?

Answer :

According to the first law of thermodynamics,

dQ = dU + dW = nC_{v}dT + dW … (i),

where

dQ = heat supplied

dU = change in internal energy

dW = work done on the gas

n= number of moles

C_{v} = specific heat capacity at constant volume

dT = change in temperature

For an adiabatic process, dQ(heat supplied) = 0.

An adiabatic process occurs without the transfer of heat or mass of substances between the thermodynamic system and the surrounding.

For an isothermal process, dT(change in temperature) = 0

An Isothermal process is a change of system, in which the temperature remains constant ∆T=0.

Putting these values in (i), we get

dW = 0,

which is not possible for either of the processes.

dW = 0 only in the case of a process where the volume is constant that is dV = 0,

since dW = PdV,

where P = pressure and dV = change in volume.

Hence, we conclude that a process cannot be both adiabatic and isothermal.

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