Question

Air is expanded from $50\, L$ to $150\, L$ at $2 atm$ pressure. The external work done is $(1\, atm =1 \times 10^{5} N / m ^{2})$

• A. $2 \times 10^{-8} J$
• B. $2 \times 10^{4} J$
• C. $200\, J$
• D. $2000\, J$

Answer 1

Answer: (B)

Work done
$W=$ pressure $(p) \times$ volume-change $(\Delta V)$
Given, $p=2\, atm =2 \times 10^{5} N / m ^{2}$
$\Delta V =V_{2}-V_{1}$
$=150-50=100\, L$
$=100 \times 10^{-3} m ^{3}$
$\therefore W =p \Delta V$
$\therefore W =2 \times 10^{5} \times 100 \times 10^{-3}$
$=2 \times 10^{4} J$