Question #9a548

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Answer 1 · 2018-02-08T20:19:48

The change in internal energy is of 1061 J.

The internal energy variation $DeltaU$ is related to the system's heat $Q$ by:

$DeltaU = Q - W$ , where $W$ is the work done by/on the system.

Since the pressure does not vary, $W$ can be simply calculated as

$W = pDeltaV$ ; where $p$ and $DeltaV$ stand for the pressure and the system's volume variation, respectively.

Putting the numerical values into our expression for $W$ gives

$W = 1.5 * 10^5 Pa * (2.2 - 7.5) * 10^-3 m^3$ ;

$W = -7.95 * 10^2 J$ .

Since we have now both $Q$ and $W$ , we can calculate $DeltaU$ :

$DeltaU = Q - W$ ;

$DeltaU = 266 J - (-795)J$ ;

$DeltaU = 1061 J$ .