Question

a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final volume v2. a total of 1.7 kj of heat is added during the expansion process. what is v2? let the ideal-gas constant r = 8.314 j/(mol • k). a- 41 l b- 26 l c- 32 l d- 35 l

Answer 1

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) = $e^{0.3098}$


v₂  =  19* $e^{0.3098}$

v₂ = 25.8999

v₂ ≈ 26 L        Option b.