Question

A balloon is inflated to its full extent by heating the air inside it. The volume of the balloon changes from 4.00 × 10⁶ L to 4.50 × 10⁶ L by the addition of 1.3 × 10⁸ J of heat. Assuming the balloon expands against a pressure of 1.0 atm, calculate the ∆E (in J) of the process. 1 L･atm=101.3 J.

Answer 1

Answer:

∆E = 7.9 × 10⁷ J

Explanation:

The air absorbs 1.3 × 10⁸ J of heat (Q), that is, Q = 1.3 × 10⁸ J.

The work (W) done by the expansion of the gas can be calculated using the following expression.

W = -P . ∆V

where,

P is the external pressure

∆V is the change in the volume

$W=-P.\Delta V=-1.0atm \times (4.50 \times 10^{6}L- 4.00 \times 10^{6}L) \times\frac{101.3J}{1L.atm} =-5.1 \times 10^{7}J$

The change in the internal energy (∆E) can be calculated using the following expression.

∆E = Q + W

∆E = 1.3 × 10⁸ J + (-5.1 × 10⁷ J) = 7.9 × 10⁷ J