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Statement 1 : For an isothermal reversib...

Statement 1 : For an isothermal reversible process Q=-Wi.e. work done by the system equals the heat absorbed by the system.
Statement 2: Enthalpy change `(Delta H)` is zero for isothermal process.

A

Statement 1 is True, statement 2 is True, Statement 2 is Correct explanation for Statement 1.

B

Statement 1 is True, Statement 2 is True, Statement 2 is NOT a correct explanation for Statement 1.

C

Statement 1 is True, Statement 2 is False.

D

Statement 1 is False, Statement 2 is True.

Text Solution

Verified by Experts

In an isothermal process change in internal energy `(Delta E)` is zero.
If system undergoes a change in which internal energy os yhe system remains constant `( iee Delta E = 0).` the `- W =Q.` This means that word done by the systm equals the heat absorbed by the system.
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In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

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Knowledge Check

  • Assertion : For an isothermal reversible process Q=-Wi.e. work done by the system equals the heat absorbed by the system. Reason : Enthalpy change (DeltaH) is zero for isothermal process.

    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.

  • T increase in internal energy of a system is equal to the work done on the system. Which process does the system underg?

    A
    Isochoric
    B
    Adiabatic
    C
    isobaric
    D
    Isothermal

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    The heat absorbed in a constant volume process is equal to the systems change in……….

    Statement 1 : An exothermic process which is non spontaneous at high temperature may become spontaneous at low temperature, Statement 2 : Spontaneous process is an irreversible process and may be reversed by some external agency.

    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 :

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