Question

Two moles of an ideal gas expanded isothermally and reversibly from 1 liter to 10 liter at ${\text{300K}}$. The enthalpy change (in ${\text{kJ}}$) for the process is:
(A)- $11.4$
(B)- $ - 11.4$
(C)- $0$
(D)- $4.8$

Answer 1

Hint: For the calculation of an enthalpy change in the above given reaction, first we have to calculate the value of work done. Because formula which we used to calculate the enthalpy of the above given reaction is as follow: $\text{ }\!\!\Delta\!\!\text{ H= }\!\!\Delta\!\!\text{ U+W}$.

Complete answer:
From the formula of enthalpy it is clear that, we have to first calculate the value of work done for the expansion of two moles of an ideal gas from 1 liter to 10 liter at ${\text{300K}}$ by using following formula:
\[{\text{W = - 2}}{\text{.303nRTlog}}\dfrac{{{{\text{V}}_{\text{f}}}}}{{{{\text{V}}_{\text{i}}}}}\]
Where, ${\text{n}}$ = No. of moles which is equal to $2$ according to the given question.
${\text{T}}$ = Temperature of the reaction which is equal to ${\text{300K}}$.
${{\text{V}}_{\text{f}}}$ = Final volume of gas which is equal to 10 liter.
${{\text{V}}_{\text{i}}}$ = Initial volume of gas which is equal to 1 liter according to the question.
${\text{R}}$ = Universal gas constant whose value is equal to ${\text{8}}{\text{.314J/Kmol}}$.
Now putting these values on the above equation of the work done and we get,
\[{\text{W = - 2}}{\text{.303}} \times {\text{2}} \times {\text{8}}{\text{.314}} \times {\text{300}} \times {\text{log}}\dfrac{{10}}{1}\]
\[{\text{W = - 11488}}{\text{.285J}}\]
Or \[{\text{W = - 11}}{\text{.4kJ}}\]
As we know that enthalpy change for the given reaction is calculated as:
$\text{ }\!\!\Delta\!\!\text{ H= }\!\!\Delta\!\!\text{ U+W}$
And for an isothermal reaction change in internal energy is always zero i.e. $\text{ }\!\!\Delta\!\!\text{ U=0}$, so we get value of change in enthalpy is equal to the value of work done during the expansion.
$\text{ }\!\!\Delta\!\!\text{ H=W}$
\[\Delta \text{H=-11}\text{.4kJ}\]

Hence option (B) is correct i.e. change in enthalpy is equal to $ - 11.4$.

Note:
Here some of you may do wrong calculations if you put different values of universal gas constant in this question. Because many alternate values of universal constant are present in the chemistry on the basis of different measuring units, so always choose the value of ${\text{R}}$ on the basis of measuring units given in the question.