Question

The molar volume of $C{O}_2$ is maximum at?
A.$273K$ and $\text{1atm}$
B. $546K$ and $\text{1atm}$
C. $273K$ and $\text{2atm}$
D. $546K$ and $\text{2atm}$

Answer 1

Hint: We have to know the laws, which manage ideal gases are normally called ideal gas laws and the laws are controlled by the observational work of Boyles and Charles. Boyle’s law states that for a given mass of gas held at a consistent temperature the gas pressure is contrarily relative to the gas volume. Charles law states that for a given fixed mass of gas held at a steady pressing factor the gas volume is straightforwardly relative to the gas temperature.

Complete answer:
We have to remember that the Ideal gas law is the condition of a speculative ideal gas. It is a decent guess to the conduct of numerous gases under numerous conditions, despite the fact that it has a few constraints. The ideal gas condition can be composed as,
$PV=nRT$
Where,
$P$ = Pressure, $V$ = Volume, $n$ = No. of moles, $R$ = Gas constant, $T$ = Temperature.
For all values $n=1$ , and $R=0.0821L.atm/mol.K$ ,
For option (A), $273K$ and $\text{1atm}$ ,
$V={\displaystyle \frac{nRT}{P}}$
Applying all the values in the above equation,
$V={\displaystyle \frac{1\times 0.0821\times 273}{1}}=22.4L$
The molar volume is $22.4L$ .
For option (B), $546K$ and $\text{1atm}$ ,
$V={\displaystyle \frac{nRT}{P}}$
Applying all the values in the above equation,
$V={\displaystyle \frac{1\times 0.0821\times 546}{1}}=44.8L$
The molar volume is $44.8L$ .
For option (C), $273K$ and $\text{2atm}$ ,
$V={\displaystyle \frac{nRT}{P}}$
Applying all the values in the above equation,
$V={\displaystyle \frac{1\times 0.0821\times 273}{2}}=11.21L$
The molar volume is $11.21L$ .
For option (D), $546K$and $\text{2atm}$ ,
$V={\displaystyle \frac{nRT}{P}}$
Applying all the values in the above equation,
$V={\displaystyle \frac{1\times 0.0821\times 546}{2}}=22.4L$
The molar volume is $22.4L$ .
Therefore, the molar volume of $C{O}_2$ is maximum at $44.8L$ .

Hence, option (B) $546K$ and $\text{1atm}$ is correct.

Note:
We have to know that, the point when the atomic mass of any gas is multiplied by its particular gas constant ( $R$ ), it is seen that the item $R$ is consistently something similar for all gases. This item is called general gas consistent and it is indicated as $R$.