Question

Calculate two specific heats of a gas from the following cp/cv=
1.51 and density of gas at NT.P. is 1.234 g/L J=42× 10^7 erg/cal​

Answer 1

Answer:

$\sf C_P=0.2119 \ cal/g \ K$

$\sf C_V=0.1404 \ cal/g \ K$

Explanation:

Given :

$\sf \dfrac{C_P}{C_V} = 1.15$

$\sf C_P = 1.5 \ C_V$

$\sf C_P -C_V=\dfrac{r}{J} =\dfrac{P_V}{TJ} =\dfrac{P(1/ \rho)}{TJ}$

Putting values here we get :

$\sf C_P -C_V=\dfrac{1.013\times10^6}{1234\times10^{-3}\times273\times4.2\times10^7}$

$\sf C_P -C_V=0.0715$

Now using values of $\sf C_P = 1.5 \ C_V$

$\sf C_V =\dfrac{0.0715}{0.51}$

$\sf C_V =0.1404 \ cal/g \ K$

$\sf C_P =1.51\times0.1404 \ cal/g \ K$

$\sf C_P= 0.2119 \ cal/g \ K$

Therefore , we get both specific heats.