Question

An ideal gas was isothermally compressed at 298K from initial volume of 35 litre to a final volume of 12 litre. Calculate ∆A and does it matter whether the path is reversible or reversible?

Answer 1

As we know

A = U - TS

or ΔA = ΔU - TΔS at constant T

as we known for isothermal process - (for ideal gas)

ΔU = O

∴ Δ A = -TΔS ---(I)

or ΔA = -T x \(\frac{q_{rev}}{T}\)

ΔA = -q_rev

∴ ΔA does not depends on the reversibility or non - reversibility of the process because we always calculate q_rev for both process.

ΔA = -q_rev = [RT ln (v₂/v₁)]

ΔA = -RT ln V₂/V₁

= -8.314 x 298 x ln (12/35)

= - 8.314 x 298 x (-1.07)

= 2651 J/mol

ΔA = 2.65 KJ/mol