260121

Question : 2 moles of an ideal gas is isothermally expanded to 3 times its original volume at 300K. Calculate the work done and heat absorbed by the gas?
(Given R=8.31 J/molK and log_e3=4.771)

Doubt by Jayant

Solution : 

n=2
V₁=V
V₂=3V
T = 300 K
In Isothermal Process
PV = constant
P₁V₁=P₂V₂
P₁/P₂ = V₂/V₁
P₁/P₂ = 3V/V
P₁/P₂ = 3

According to First Law of Thermodynamics
ΔQ=ΔU+ΔW
During Isothermal Process
ΔU=0
ΔQ=ΔW
ΔW=nRTln[V₂/V₁]
ΔW=nRTln[3v/v]
ΔW=2×8.31×300×ln3
ΔW=4986×4.771
ΔW=23788.206 J