Question

If $Pdv/s.P$(where $P$ denotes pressure in atm and $d$ denotes density in $gm/L$) is plotted for $He$ gas (assume ideal) at a particular temperature. If ${\left[\frac{d}{dP}\left(Pd\right)\right]}_{P=8.21atm}=5$, then the temperature will be:

• A. 160 K
• B. 320 K
• C. 80 K
• D. none of these

Answer 1

The correct option is A 160 K

$PM=dRT$
$Pd=P^2(\frac{M}{RT})$
$\frac{d\left(Pd\right)}{dP}=\frac{2PM}{RT}=5$
$\frac{2\times 8.21\times 4}{0.0821\times T}=5$
$T=160K$