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Prove that for an ideal gas undergoing a...

Prove that for an ideal gas undergoing an isothermal change, `DeltaH=0`.

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The change in enthalpy of an ideal gas undergoing a process, `DeltaH=DeltaU=nRDeltaT`. In an isothermal process, `DeltaT=0`. So,` DeltaH=DeltaU`. Again, in an isothermal process of an ideal gas, `DeltaU=0` and hence `DeltaH=0`.
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Knowledge Check

  • The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct ?

    A
    `T_1 = T_2`
    B
    `T_3 gt T_1`
    C
    `W_(isothermal gt W_(adiabatic)`
    D
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  • The change in entropy for 2 mol ideal gas in an isothermal reversible expansion from 10 mL to 100 mL at 27^(@)C is-

    A
    `26.79J*K^(-1)`
    B
    `38.29J*K^(-1)`
    C
    `59.07J*K^(-1)`
    D
    `46.26J*K^(-1)`

  • Two mole of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthapy change (in kJ) for the process is

    A
    11.4 kJ
    B
    -11.4 kJ
    C
    0 kJ
    D
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