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.

2 Answers
Jun 8, 2018
  • #("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#

Jun 8, 2018

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