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. $200J$
• D. $2000J$

Answer 1

Answer: (B)

Work done $W=p\times \Delta V=2\times 10^5\times (150-50)\times 10^{-3}$
$=2\times 10^{4}J$