Question

air expands from $5$ liters to $10$ liters at $2$ atm pressure. External workdone is

• A. $10J$
• B. $1000J$
• C. $3000J$
• D. $300J$

Answer 1

The correct option is B $1000J$

Given,

$V_1=5liter=5\times {10}^{-3}{m}^3$

$V_2=10liter=10\times {10}^{-3}{m}^3$

$P=2atm=2\times {10}^{5}Pa$

Work done, $W=P(V_2-V_1)$

$W=2\times {10}^5(10-5)\times {10}^{-3}$

$W=2\times 100\times 5$

$W=1000J$

The correct option is B.