Question

Question #9a548

Answer 1

The change in internal energy is of 1061 J.

Explanation:

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`.