Question

Five moles of a gas expand isothermally from an initial volume of 5 mº to a final
volume of 10 mº at a constant temperature of 27°C. Find the work done by the
during this isothermal expansion. (R = 8.3 J mol-'K-')
​

Answer 1

W = 8627.85 J

Explanation:

Given information,

n=5; V1 = 5 m3; V2 = 10 m3; T = 27 0C = 300 K and R = 8.3 J/mole/K

In an isothermal process T = constant.

Work done can be calculated as,

dW = P.dV = nRTVdV ; (from PV = nRT)

⇒W = ∫V2V1nRTVdV = nRT ln(V2V1)

⇒W = 5×8.3×300×ln(105) = 5×8.3×300× 0.693

⇒W = 8627.85 JdW = P.dV = nRTVdV ; from PV = nRT

⇒W = ∫V1V2nRTVdV = nRT lnV2V1

⇒W = 5×8.3×300×ln105 = 5×8.3×300× 0.693

⇒W = 8627.85 J