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  4. Air is expanded from 50 L to 150 L at 2 atm pressure. The external work done is (1 atm =1×105 Nm-2)

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Q. Air is expanded from 50 L to 150 L at 2 atm pressure. The external work done is (1 atm =$1\times10^5 Nm^{-2}$)

Thermodynamics

Solution:

Work done $W=p\times\Delta V=2\times10^5\times(150-50)\times10^{-3}$
$= 2\times 10^4 J$