Question

Calculate the values of the molar heat capacities Cv and Cp of a gas,if the ratio of the heat capacities is 1.33.What is the atomicity of the gas?Given R=8.31J/mol-k.

Answer 1

• `("C"_"p")/("C"_"v") = gamma`
• `"C"_"p" - "C"_"v" = "R"`

From above two equations

`"C"_"p" = ("R" gamma)/(gamma - 1)`

`color(white)("C"_"p") = "8.314 × 1.33"/(1.33 - 1)`

`color(white)("C"_"p") = "33.5 J/mol K"`

`"C"_"v" = "C"_"p"/gamma`

`color(white)("C"_"v") = "33.5 J/mol K"/"1.33"`

`color(white)("C"_"v") = "25.2 J/mol K"`

`gamma` is `1.33` therefore it’s triatomic (Atomicity`= 3`) gas.

`underline(bb"Type of gas" color(white)(............) bb(gamma)color(white)(....))`
`"Monoatomic" color(white)(000=) 1.66`
`"Diatomic" color(white)(..............) 1.40`
`"Triatomic" color(white)(......!!!!!!) 1.33`

Answer 2

See below

Explanation:

Generally:

`c_p - c_v = R`

Here:

`c_p/c_v = 4/3`

`implies c_v = 3R, qquad c_p = 4R`

We know:

• `c_p/c_v = (5 R + 2X)/(3 R + 2X)`

• `X = {(0 qquad qquad "monatomic"),(R qquad qquad "diatomic"),(3/2 R qquad qquad "polyatomic"):}`

`4/3 = 8/6 = (5 + 2 * 3/2)/(3 + 3 * 3/2)`

Polyatomic