Question

Five moles of an ideal gas expands isothermally at 300 K from an initial volume of 100 L to a final volume of 500 L. Calculate: (a) the maximum work the gas can deliver, (b) the heat accompanying the process, (c) ∆S for the gas. (Please explain why did you use the equation, what conditions did you see from the question, etc)

Answer 1

Given,

n = 5

Isothermal process, T = 300 K

Initial Volume = 100 liters

Final Volume = 500 liters

Maximum work is done in a reversible process

Therefore

Maximum work = w_rev = -nRTln(V_final/V_initial) = - 5 * 8.314 * 300 * ln(500/100) = -20071.30 J

W_rev = -20.071 kJ

In an isothermal process, $\Delta$T = 0 (since temperature remains constant)

Therefore $\Delta$U = C_v$\Delta$T = 0 (i.e internal energy change = 0)

From first law of thermodynamics,

Heat, q = $\Delta$U - w

=> q = 0 - (-20.071 kJ)

=> q = 20.071 kJ

$\Delta$S = q_rev/T = 20071.30/300 = 66.90 J/K