Question

A bubble of air  at the bottom of a an ocean is 36.0 m deep has a volume of 1.00 cm3. The temperature at the bottom is 5.1°C while at the surface it is 17.5°C.

Assume the temperature at 30m depth to be 4.639atm and at the top to be 1 atm

Assume that the air bubble consists entirely of nitrogen N2. Calculate the change in the thermal energy as the air bubble rises from 36.0 m to the surface.

Answer 1

let
V1 = 1 cm^3 = 1*10^-6 m^3
T1 = 5.1 C = 5.1 + 273 = 278.1 K
P1 = 4.639 atm = 4.639*1.0134*10^5 pa

T2 = 17.5 C + 17.5 + 273 = 288.5 K
P2 = 1 atm = 1.0134*10^5 pa

let n is the number of moles of the gas
use, P1*V1 = n*R*T1

==> n = P1*V1/(R*T1)

= 4.639*1.0134*10^5*10^-6/(8.314*278.1)

= 2.03*10^-4 mole

change in internal energy of diatomic gas, delta_U = (5/2)*n*R*(T2 - T1)

= (5/2)*2.03*10^-4*8.314*(17.5 - 5.1)

= 0.0523 J <<<<<<<<<----------------Answer