A gas absorbs 500J heat and utilized QJ in doing work against an external pressure of 2 atm. If ΔE is -510 J, values of ΔV and W respectively are

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Detailed Solution

-510 = 500 J + WW = -1010 JW = - PΔV-1010 J = -2ΔV101J=101 lit.atm10 lit.atm= 2atm.∆V;∆V= 5 lit(1 lit=1 dm3)

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