Question

The graph shows logarithmic readings of pressure and volume for two ideal gases $A$ and $B$ undergoing adiabatic process. It can be concluded that (assume one of the gases $A$ and $B$ is monatomic and other is diatomic)

• A. $A$ is diatomic
• B. $B$ is diatomic
• C. $B$ is monatomic
• D. Both $\left(a\right)$ and $\left(b\right)$

Answer 1

The correct option is D Both $\left(a\right)$ and $\left(b\right)$

For an adiabatic process
$P{V}^{\gamma }=K$
Taking $lo{g}_e$on both sides,
$ln\left(P{V}^{\gamma }\right)=ln\left(k\right)$
$lnP+\gamma lnV=lnk$
$lnP=lnk-\gamma lnV$
Above equation is of the form $y=C-mx$
$\Rightarrow m=\gamma$
Since slope of $m_B>m_A\Rightarrow {\gamma }_B>{\gamma }_A$
If $B$ is monatomic, then $A$ must be diatomtic.