The molar heat capacity for carbon monoxide at constant volume is

Question

The molar heat capacity for carbon monoxide at constant volume is Cv_m = 20.17 J(K-mol).

A 12.00-1. fixed volume flask contains CO(g) at a pressure of 10.00 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.

Delta x= J/K

Answer 1

Hello July,

To solve this problem you will need two equations:

ΔS=n*Cv*​ln(T2​/T1​​)

S is entropy = what we are solving for

n is the number of moles of the gas = need the second equation to figure out

Cv is the molar heat capacity= given 20.17J/(K*mol)

T1 and T2 are the initial and final temperatures in kelvin (K), respectively = given 25C and 800C we need to convert to Kelvin:

25C +273.15=298.15K

and

800C+273.15=1073.15K

and the second equation the ideal gas law

PV=nRT

rewritten as

n=(PV)/(RT)

n= solving for the first equation

P=pressure= given as 10kPa we need to convert to pascal k=*10^3; 10*10^3Pa=10,000Pa

V=volume =given as 12L we need to convert to cubic meters (1L=.001m^3),

12L *.001m^3/L=.012m^3

R= ideal gas constant =8.314J/(K*mol)

T=tempature= given as 25C we need to convert to kelvin: 25C+273.15= 298.15K

Plugging everything in for the ideal gas law:

n=(10,000Pa*.012m^3)/(8.314J/(K*mol)*298.15K)

n=.0484 mol

Plugging n into the first equation.

ΔS=n*Cv*​ln(T2​/T1​​)

n=.0484mol

Cv=20.17J/(K*mol)

T1=298.15K

T2=1073.15K

ΔS=.0484mol*20.17J/(K*mol) * ln(1073.15K/298.15K)