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A Mixture Contains 1 Mole of Helium (Cp = 2.5 R, Cv = 1.5 R) and 1 Mole of Hydrogen (Cp = 3.5 R, Cv = 2.5 R). Calculate the Values of Cp, Cv and γ for the Mixture. - Physics

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प्रश्न

A mixture  contains 1 mole of helium (Cp = 2.5 R, Cv = 1.5 R) and 1 mole of hydrogen (Cp= 3.5 R, Cv = 2.5 R). Calculate the values of Cp, Cv and γ for the mixture.

संक्षेप में उत्तर
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उत्तर

Specific heat at constant pressure of helium, Cp' = 2.5 R
Specific heat at constant pressure of hydrogen, Cp" = 3.5 R
Specific heat at constant volume of helium, Cv' = 1.5 R
Specific heat at constant volume of hydrogen, Cv" = 2.5 R

n1 = n2 = 1 mol

dU = nCvdT

For the mixture of two gases,

dU1 +dU2 = 1 mol

[n1 + n2] CvdT = n1C'vdT + n2C"vdT,

where Cv  is the heat capacity of the mixture

`=> "C"_"v" =("n"_1"C"'_"v" + "n"_2"C"''_"v")/("n"_1+"n"_2)`

`= (1.5"R" +2.5"R")/2 =2"R"`

Cp = Cv + R = 2R + R = 3R

`gamma  = "C"_"p"/"C"_"v" = (3"R")/(2"R") = 1.5`

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अध्याय 5: Specific Heat Capacities of Gases - Exercises [पृष्ठ ७८]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 5 Specific Heat Capacities of Gases
Exercises | Q 12 | पृष्ठ ७८

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