Question

One-half mole of helium is expanded adiabatically and quasi-statically from an initial pressure of 4.00 atm and temperature of 540 K to a final pressure of 1.00 atm. Find the following values for the gas.

a) the work done by the gas KJ

b) the change in the internal energy KJ

Answer 1

f = 3

gamma= 5/3

2 1

P1 = 4 atm =405300 Pa

pV1 = nRT

405300 * V1 = 0.5 * 8.3144598 * 540

V1= 0.00553887033 m^3

4*(0.00553887033)^(5/3) = 1* V2^(5/3)

V2 = 0.01272498236 m^3

W = 405300 *(0.00553887033)^(5/3) ((0.01272498236)^(-2/3) - (0.00553887033)^(-2/3)) /(-2/3)

=1433.317932 J = 1.43 kJ

to find T2

(405300/4) * 0.01272498236 = 0.5 * 8.3144598 * T

T = 310.148552917 K

change in internal energy change = 3/2 * 0.5* 8.3144598 * (310.148552917-540)

= - 1433.31796256 J = -1.433 kJ