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For an isolated system, Delta U=0. What ...

For an isolated system, `Delta U=0`. What will be `DeltaS`?

A

`0`

B

`gt 0`

C

`lt 0`

D

`ge 0`

Text Solution

Verified by Experts

For an isolated system `Delta U =0` and for a spontaneous process, total entropy change must be position Mereover `Delta S = ( q _(rev))/( T) = ( Delta H )/( T) = ( Delta U + p Delta V)/(T) = ( p Delta V)/(T) (therefore Delta U = 0) i.e., T Delta S gt 0`
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The change in internal energy (U) can be brought about in two ways: (i) Either by allowing the heat of flow into the system or out of the system (ii) By doing work on the system or the work done by the system Using the symbol q to represent heat transferred to system and using work done by the system - w, we can represent the internal energy change of a system Delta U, as : q = Delta U + (-w) (First law of thermodynamics) If the reaction is caried out in a closed container with constant volume so that Delta V = 0 Hence, q _(1) = Delta U On the other hand, if a reaction carried out in open vessel that keeps the pressure contans and allows the volume of the system to change freely. In such case, Delta V ne 0 and - w = P. Delta V. Hence, q _(B) = Delta U + p Delta V Also q _(P) = q _(v) + Delta _(g) RT As reactions carried out at constant pressure are so common, the heat change for such preocess is given a special symbol Delta H, called the enhalpy change of the reaction . The enthalpy (H) of the system is the name given to the quantity (U+ PV). In which of the following cases Delta H and DeltaU are not equal to each other?

The change in internal energy (U) can be brought about in two ways: (i) Either by allowing the heat of flow into the system or out of the system (ii) By doing work on the system or the work done by the system Using the symbol q to represent heat transferred to system and using work done by the system - w, we can represent the internal energy change of a system Delta U, as : q = Delta U + (-w) (First law of thermodynamics) If the reaction is caried out in a closed container with constant volume so that Delta V = 0 Hence, q _(1) = Delta U On the other hand, if a reaction carried out in open vessel that keeps the pressure contans and allows the volume of the system to change freely. In such case, Delta V ne 0 and - w = P. Delta V. Hence, q _(B) = Delta U + p Delta V Also q _(P) = q _(v) + Delta_(g) RT As reactions carried out at constant pressure are so common, the heat change for such preocess is given a special symbol Delta H, called the enhalpy change of the reaction . The enthalpy (H) of the system is the name given to the quantity (U+ PV). One mole of a gas is allowed to expand freely in vacuum at 300 K. The work done during the process is :

Knowledge Check

  • Assertion : A process for which Delta _("Ssystem") gt0 as well as Delta H gt 0, passes from non-spontaneous to spontaneous state as temperature is increased. Reason : At higher temperature T Delta S exceeds DeltaH.

    A
    and (R) are correct and (R) is the correct explanation of(A).
    B
    If both (A) and (R) are correct, but (R) is not the correct explanation of(A).
    C
    If(A) is correct, but (R) is incorrect.
    D
    If both (A) and (R) are incorrect.

  • For -reaction, N_2(g)+3 H_2(g) hArr 2 NH_3(g) which of the following is valid? Delta H=Delta U , Delta HgtDelta U , Delta HltDelta U , None of the above

    A
    Delta H=Delta U'
    B
    Delta HgtDelta U'
    C
    Delta HltDelta U'
    D
    None of the above

  • The enthalpy of a reaction at 273K is-3.57 kJ: What will be the enthalpy of reaction at 373 K if Delta C = U

    A
    `-3.57`
    B
    `0`
    C
    `- 3.57 xx (373)/(273)`
    D
    `- 375`

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    For which process does Delta U =0 holds true?

    A square is drawn on the altitude of an equilateral triangle of side 4 cm. a) What is the altitude of the triangle? b) What is the area and perimeter of the square?

    The change in internal energy (U) can be brought about in two ways: (i) Either by allowing the heat of flow into the system or out of the system (ii) By doing work on the system or the work done by the system Using the symbol q to represent heat transferred to system and using work done by the system - w, we can represent the internal energy change of a system Delta U, as : q = Delta U + (-w) (First law of thermodynamics) If the reaction is caried out in a closed container with constant volume so that Delta V = 0 Hence, q _(1) = Delta U On the other hand, if a reaction carried out in open vessel that keeps the pressure contans and allows the volume of the system to change freely. In such case, Delta V ne 0 and - w = P. Delta V. Hence, q _(B) = Delta U + p Delta V Also q _(P) = q _(v) + Delta _(g) RT As reactions carried out at constant pressure are so common, the heat change for such preocess is given a special symbol Delta H, called the enhalpy change of the reaction . The enthalpy (H) of the system is the name given to the quantity (U+ PV). The latent heat of vaporization of liquid at 500 K and 1 atmospheric pressure is 10.0 kcal/mol. What will be. the change in internal energy of 3 mole of the liquid at the same temperature and pressure?

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