Question

Two moles of an ideal gas expended isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is:

• A. 11.4
• B. -11.4
• C. 0
• D. 4.8

Answer 1

Answer: (B)

-11.4

Work done in a reversible isothermal process is:

$W=-2.303nRT\log \frac{V_f}{V_i}.....\left(1\right)$

Given:-

$n=2\text{moles}$

$T=300K$

$V_f=10L$

$V_i=1L$

$R=$ Gas constant $=8.314J/K-mol$

Substituting these values in ${eq}^n\left(1\right)$, we have

$W=-2.303\times 2\times 8.314\times 300\times \log \frac{10}{1}$

$\Rightarrow W=-11488.285J=-11.4kJ$

Now as we know that,

$\Delta H=\Delta U+W$

For an isothermal process,

$\Delta U=0$

$\therefore \Delta H=W=-11.4kJ$

Hence the enthalpy change for the given process is $-11.4kJ$.

Hence, the correct option is $\text{B}$