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Calculateq, W, DeltaU and DeltaH for the...

Calculate`q, W, DeltaU` and `DeltaH` for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of `1.0` bar to a final pressure of `0.1` bar at a constant temperature of `273K.`

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For isothermal reversible expansion:
`W=-2.303nRT"log"(P_(1))/(P_(2))`
`=-2.303xx1xx8.314xx273 "log"(1.0)/(0.1)`
`W=-5227.2 J`
At constant temperature, for expansion `DeltaT=0` Therefore, `DeltaU=0`
Also, `DeltaU=1xxC_(v)xxDeltaT=0... `(for `1` mole)
and `DeltaU=q+W`
`:. q=-W=5227.2J`
Also when temperature is constant `P_(1)V_(1)=P_(2)V_(2)` or `PV` is constant.
Thus, `DeltaH=DeltaU+Delta(PV) :. DeltaH=0`
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Knowledge Check

  • Calculate Q and w for the isothermal reversible expansion of one mole an ideal gas from an initial pressure of 1.0 bar to a final pressure of 0.1 bar at a constant temperature of 273 K respectively.

    A
    5.22kJ, -5.22 kJ
    B
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    C
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    D
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    The pressure-volume of varies thermodynamic process is shown in graphs: Work is the mole of transference of energy. It has been observed that reversible work done by the system is the maximum obtainable work. w_(rev) gt w_(irr) The works of isothermal and adiabatic processes are different from each other. w_("isothermal reversible") = 2.303 nRT log_(10) ((V_(2))/(V_(1))) = 2.303 nRT log_(10)((P_(2))/(P_(1))) w_("adiabatic reversible") = C_(V) (T_(1)-T_(2)) The q value and work done in isothermal reversible expansion of one mole of an ideal gas from initial pressure of 1 bar to final pressure of 0.1 bar at constant temperature 273K are:

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