Question

Air is expanded from $50$ liter to $150$ liter at $2$ atmospheric pressure ( $1$ atm = pressure =${10}^{5}Pa$). The external work done is:

• A. $200J$
• B. $2000J$
• C. $2\times {10}^4$J
• D. $2\times {10}^{12}$ J

Answer 1

The correct option is C $2\times {10}^4$J

Given,

Pressure, $P=2\times {10}^{5}pa$

Change in volume, $\Delta V=150-50=100litre=0.1m^3$

Work done =$P\Delta V=2\times {10}^5\times 0.1=2\times {10}^{4}kJ$

Hence, work done is $2\times {10}^{4}J$