Question

Ten moles of an ideal gas at constant temperature $600 \,K$ is compressed from $100 \,L$ to $10 \,L$. The work done in the process is

• A. $ 4.11\times 10^4\,J $
• B. $ -4.11\times 10^4\,J $
• C. $ 11.4\times 10^4\,J $
• D. $ -11.4\times 10^4\,J $

Answer 1

Answer: (D)

Work done in an isothermal process
$W =2.3026 \,n R T \log _{10}\left(\frac{V_{2}}{V_{1}}\right) $
$=2.3026 \times 10 \times 8.3 \times 600 \log _{10}\left(\frac{10}{100}\right)$
$=-11.4 \times 10^{4} J$