Question

For a gas Cp/R=3 then gas may be
Please explain your answer.

Answer 1

Answer:

this is the answer to ur question friend

Answer 2

❍$\:\: \underline{\underline{\bf{\large\mathfrak\red{~~Solution~~}}}}$

$c(v) = \frac{x}{2} R\\ c(p) = (\frac{1}{2} x + 1)R$

now...

❍ $\:\: \underline{\underline{\bf{\large\mathfrak{~~Given~~}}}}$

$\frac{c(p)}{R} = 3 \\ = > 1 + \frac{1}{2} x = 3 \\ = > x = 4 \\ so \: the \: system \: has \: 4 \: degress \: of \: freedom \\ we \: know \: that \: \\ for \: \\ monatomic \: gas \: x = 3 \\ for \: diatomic \: gas \: x = 5 \\ therefore \: \: \\ x = 4(is \: for \: the \: mixture \: of \:\\ monatomic \: and \: diatomic \: gas)$

____________________________

❍ More explanation

for a mixture of monatomic and diatomic gas..

$\frac{c(p)}{c(v)} = 1.50 \\ now \: we \: have \: x = 4 \\ therefore... \\ \frac{c(p)}{c(v)} = 1 + \frac{2}{x} = 1 + \frac{2}{4} = 1 + 0.5 = 1.5$

therefore.....

❒ Answer➨0ption(d)

$:\: \underline{\underline{\bf{\large\mathfrak{~hope ~this ~help~you~~}}}}$