Question

Carbon monoxide expands isothermally and reversibly at 300 K doing 4.754 kJ of work. If the initial volume changes from 10 dm³ to 20 dm³, calculate the number of moles of carbon monoxide. (R = 8.314 JK^-1 mol^-1)

Answer 1

W_max = -2.303 nRT log₁₀\(\frac{V_2}{V_1}\)

W_max = Maximum work done = -4.754 kJ

= -4754 J, n = Number of moles = ?,

= 8.314 JK^-1 mol^-1

T = 300 K, 

V₁ = Initial volume of carbon monoxide = 10 dm³

V₂ = Final volume of carbon monoxide = 20 dm³

∴ -4754 = 22.303 × n × 8.314 × 300 log₁₀\(\frac{20}{10}\)

∴ -4754 = – 2.303 × n × 8.314 × 300 × log10² 

∴ – 4754 = – 2.303 × n × 8.314 × 300 × 0.3010

∴ n = \(\frac{-4754}{-2.303\times 8.314\times 300 \times 0.3010}\)

= 2.75 mol

∴ Number of mol = 2.75 mol