For a reversible process at $T = 300K$, the volume of $2\;mol$ of ideal gas is increased from $1L$ to $10L$, the $\Delta H$ for isothermal change is:
A) $11.47J$
B) $4.98kJ$
C) $0$
D) $ - 11.47J$

Hint:We are aware of the reversible process which occurs slowly and take infinite time for their completion. We also know that isothermal process is the one where the temperature remains constant and amount of heat is equivalent to the negative value of work done.


Complete solution:
As we know that for a reversible process the intermediate stages are defined and each intermediate stage is in equilibrium with its previous and next one stage. Also, the same path is traced while a process is reversible.
Now we are given that the process is isothermal for which we have to calculate the change in enthalpy. We know that the formula of change in enthalpy is simply the sum of change in initial energy and temperature along with universal gas constant. We can show this with the help of the reaction as given below:
$\Delta H = \Delta U + R\Delta T$
Now, we are aware of the fact that for an isothermal process, the temperature is always constant, so in the above reaction $\Delta T = 0$. The initial energy is also zero for an isothermal process, hence the $\Delta U = 0$ which results into the enthalpy as zero, $\Delta H = 0$.
So the given data for reversible processes will not at all be used.

So, the correct answer will be Option C.


Note:Remember that if the work is done by the system in an isothermal process then value of work is negative and heat is positive which depicts that heat energy is absorbed by the system and if the work is done on the system then the value of work is positive and heat energy is negative suggesting that the heat energy is released from the system.