(a) Let final unsafe depth = h_{f}

Initial depth of diver = 10.3m

Initial Volume of lungs = 5.00 X 10^{-6} m^{3}

Final volume of lungs = (5.00 X 10^{-6} m^{3}) / 1.1

Final volume of lungs = (4.55 X 10^{-6} m^{3})

Initial pressure on Lungs = Atmospheric pressure + Pressure due to the water

Initial pressure on Lungs = (1.01 X 10^{5} Pa) + ρgh

Initial pressure on Lungs = (1.01 X 10^{5} Pa) + (1.01 X 10^{5 }Pa)

Initial pressure on Lungs = (2.02 X 10^{5} Pa)

Unsafe pressure on Lungs = Atmospheric pressure + Pressure due to the water

Unsafe pressure on Lungs =1.01 X 10^{5} Pa + ρgh

Unsafe pressure on Lungs =1.01 X 10^{5} Pa+ ρgh_{f} ---Eq1

PV = Constant if T is constant:

Let P_{i}= Pressure at 10.3 m deep

V_{i} = Volume at 10.3 m deep

P_{f} = Pressure at unsafe depth

V_{f} = Volume at unsafe depth

P_{i}V_{i} = P_{f}V_{f}

P_{f}= [(P_{i}V_{i}) / (V_{f})]

P_{f}= {[(2.02 X 10^{5} Pa) X (4.55 X 10^{-6} m^{3})] / (5.00 X 10^{-6} m^{3})}

P_{f}= 1.82 X 10^{5} Pa

Eq 1 becomes
1.01 X 10^{5} Pa + ρgh_{f} = 1.84 X 10^{5} Pa

ρgh_{f} = [(1.82 X 10^{5} Pa) - (1.01 X 10^{5} Pa)]

h_{f} = [(1.82 X 10^{5} Pa) - (1.01 X 10^{5} Pa)] / (ρg}

h_{f} = 8.26m

Assent = 10.3 m - 8.25m

The diver can safely ascend 2.05 m without exhaling

(b) Let P_{i}= Pressure at 10.3 m deep

V_{i} = Volume at 10.3 m deep

P_{f} = Pressure at surface

V_{f} = Volume at surface

V_{f}= [(P_{i}V_{i}) / (P_{f})]

V_{f}= {[(2.02 X 10^{5} Pa) X (5.00 X 10^{-6} m^{3})] / (1.01 X 10^{5} Pa)}

V_{f}= 1.0 X 10^{-5} m^{3}

(V_{f}) / (V_{1}) = 2 times.