Question

The molar volume of $C{O}_2$ is maximum at:

• A. $NTP$
• B. $0^{\circ }C$ and $2.0$ atm
• C. $127^{\circ }C$ and $1$ atm
• D. $273^{\circ }C$ and $2$ atm

Answer 1

Answer: (C)

From ideal gas equation,
$pV=nRT$
$\Rightarrow V=\frac{nRT}{p}$
For $n=1mole$
$V=\frac{RT}{p}$
$\Rightarrow V\propto T$
$\therefore$ The molar volume of $C{O}_2$ at
a)NTP
$V=\frac{0.0821\times 273}{1}=22.4L$
b) $0^{\circ }C$ and $2.0atm$
$V=\frac{0.0821\times 273}{2.0}=11.2L$
c) $127^{\circ }C$ and $1atm$
$V=\frac{0.0821\times 400}{1}=32.84L$
d) $273^{\circ }C$ and $2atm$
$V=\frac{0.0821\times 546}{2}=22.4L$
Therefore the molar volume of $C{O}_2$ is maximum at
$127^{\circ }C$ and $1atm$.