Question

Calculate the work done by $10g$ of an ideal gas of molecular weight $44$ in expanding reversibly and isothermally from a volume of $5$ to $10$ litre at $300K$.

• A. $93.93cal$
• B. $-63.63cal$
• C. $-93.93cal$
• D. $63.63cal$

Answer 1

The correct option is C $-93.93cal$

$W=-2.303nRT{\log }_{10}\frac{V_2}{V_1}$
Number of moles of ideal gas $n=\frac{10g}{44g/mol}R=1.987cal/KT=300KAs,\frac{V_2}{V_1}=\frac{10}{5}=2$
On substituting the values, we get
$W=-2.303\times \frac{10}{44}\times 1.987\times 300\times {\log }_{10}2$
On solving, we get:
$W=-93.93cal$