First slide

First law of thermo dynamics

Question

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

Moderate

Solution

$\Delta E=q+W$

-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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