A bubble of air is underwater at temperature 15\ ^ ^ C and the...

Let the initial volume of the bubble be $V_1$.

Initial conditions (underwater):
$P_1 = 1.5\ \text{bar}$
T_1 = 15\ ^{}^{\circ}\text{C} = 15 + 273 = 288\ \text{K}

Final conditions (at the surface):
$P_2 = 1.0\ \text{bar}$
T_2 = 25\ ^{}^{\circ}\text{C} = 25 + 273 = 298\ \text{K}

Using the combined gas law equation:
$\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$

Substituting the given values:
$\dfrac{1.5 \times V_1}{288} = \dfrac{1.0 \times V_2}{298}$

$V_2 = \dfrac{1.5 \times 298}{288} V_1$

$V_2 = 1.5 \times 1.0347 V_1$

$V_2 \approx 1.552 V_1 \approx 1.6 V_1$

The volume will become greater by a factor of $1.6$.

Answer: Volume will become greater by a factor of $1.6$.