Question

The molar heat capacity for carbon monoxide at constant volume is CV,m = 20.17 J/(K·mol). A 13.00-L fixed-volume flask contains CO(g) at a pressure of 3.00 kPa and a temperature of 25.0 °C. Assuming that carbon monoxide acts as an ideal gas and that its heat capacity is constant over the given temperature range, calculate the change in entropy for the gas when it is heated to 800.0 °C.

Answer 1

i) Calculation of number of moles of CO

Ideal equation is as follows

PV = nRT

P = Peessure, 3.00 kPa = 0.0296atm

V = Volume ,13.00L

T = Temperature, 298.15K

R = gas constant , 0.082057(L atm /mol K)

n = number of moles

n = PV/RT

= 0.0296atm × 13.00L /(0.082057(L atm/mol K) ×298.15K)

= 0.01573mol

ii) calculate entropy change(∆S)

At constant volume ,

∆S = nC_v,m ln(T2/T1)

n = 0.01573

C_v,m= 20.17J/(K .mol)

T2 = Final temperature, 800℃ = 1073.15K

T1 = Initial temperature, 25℃ = 298.15K

∆S = 0.01573mol× 20.17J/K mol × ln(1073.15K/298.15K)

∆S = 0.01573mol× 20.17J/K mol × 2.303log(1073.15K/298.15K)

∆S = 0.406 J/K