Question

Calculate work done when $1$ mole of an ideal gas is expanded reversibly from $20L$ to $40L$ at a constant temperature of $300K$.

• A. $7.78kJ$
• B. $-1.72kJ$
• C. $11.73kJ$
• D. $-4.78kJ$

Answer 1

The correct option is B $-1.72kJ$

Work done $W=-2.303nRT\log \frac{V_2}{V_1}$

$W=-2.303\times 1\times 8.3\times 300\times \log \frac{40}{20}$

$W=-2.3030\times 8.3\times 300\times 0.3$

$W=-1720J$

$W=-1.72KJ$