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For an isothermal expansion of an ideal ...

For an isothermal expansion of an ideal gas, the correct combination of the thermodynamic parameters will be-

A

`DeltaU=0,Q=0,wne0 and DeltaH ne0`

B

`DeltaU ne0,Qne0,wne0 and DeltaH=0`.

C

`DeltaU=0,Qne0,w=0 and DeltaHne0`

D

`DeltaU=0,Qne0,wne0 and DeltaH=0`

Text Solution

Verified by Experts

The correct Answer is:
D
For an ideal gas undergoing an isothermal process, `DeltaD=0`. When a gas undergoes expansion it absorbs heat from its surroundings and performs external work. So, in such a process, `qne0 and wne0`.
We know, `DeltaH=DeltaU+DeltaU(PV)`. For ideal gas, `PV=nRT`.
`therefore DeltaH=DeltaU+Delta(nRT)=DeltaU+nRDeltaT`
In an isothermal proces of an ideal gas, `DeltaU=0,DeltaT=0 and` hence `DeltaH=0`.
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Knowledge Check

  • In the isothermal expansion of an ideal gas

    A
    there is no change in the temperature of the gas
    B
    there is no change in the internal energy of the gas
    C
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    B
    there is no change in the internal energy of the gas
    C
    the work done by the gas is equal to the heat supplied to the gas
    D
    the work done by the gas is equal to the change in its internal energy

  • For an ideal gas

    A
    the change in internal energy in a constant pressure process from temperature `The_1 to The_2` is equal to ``nC_v(T_2-T_1)` where `c_v` is the molar heat capacity at constant volume and n is the number of moles of the gas
    B
    the change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process
    C
    the internal energy does not change in an isothermal process
    D
    no heat is added or removed in an adiabatic process

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