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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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10 cm³ ; 500J

10³dm³ ; -510J

10dm³ ; 1010J

5 dm³; – 1010J

answer is D.

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Detailed Solution

-510 = 500 J + W
W = -1010 J
W = - PΔV
-1010 J = -2ΔV
101J=101 lit.atm

10 lit.atm= 2atm. $∆ V$ ;

$∆ V$ = 5 lit( $1 lit = 1 {dm}^{3}$ )

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