Question

If 5 moles of an ideal gas expands from $10 \, L$ to a volume of $100 \, L$ at $300 \, K$ under isothermal and reversible conditions, then work, $w$, is $-x \, J$. The value of $x$ is ______.
(Given $R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1}$)

Answer 1

For an isothermal reversible expansion, the work done $W$ is given by:

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

Given:

• $n = 5 \, \text{moles}$
• $R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1}$
• $T = 300 \, \text{K}$
• $V_i = 10 \, \text{L}$
• $V_f = 100 \, \text{L}$

Substitute into the formula:

$W = -2.303 \times 5 \times 8.314 \times 300 \times \log \left(\frac{100}{10}\right)$ $W = -2.303 \times 5 \times 8.314 \times 300 \times \log(10)$

Since $\log(10) = 1$:

$W = -2.303 \times 5 \times 8.314 \times 300$ $W = -28720.713 \, \text{J}$

Rounding to the nearest integer:

$W = -28721 \, \text{J}$

Thus, $x = 28721$.