Isothermal Work = PVln(v₂/v₁)
PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J
Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,
1.7 kJ = (5487.24)In(v₂/19)
1700 = (5487.24)In(v₂/19)
In(v₂/19) = (1700/5487.24) = 0.3098
In(v₂/19) = 0.3098
(v₂/19) =

v₂ = 19*

v₂ = 25.8999
v₂ ≈ 26 L Option b.
Complete question:
A diver is 10 m below the surface of water. Calculate the pressure the fluid exerted on the diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. Answer in units of Pa. Show your work.
Answer:
Tthe pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa
Explanation:
Given;
density of water, ρ = 1000 kg/m³
diver's position below the surface of the water, h = 10 m
acceleration due to gravity, g = 9.8 m/s²
Let the atmospheric pressure, P₀ = 101325 Pa
The pressure 10 m below the surface of the water is calculated as;
P = P₀ + ρgh
P = 101325 Pa + (1000 x 9.8 x 10)Pa
P = 199325 Pa
P = 1.99 x 10⁵ Pa.
Therefore, the pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa