
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
414k+ views
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 ${\text{1atm}}$ ,
$V = \dfrac{{nRT}}{P}$
Applying all the values in the above equation,
$V = \dfrac{{1 \times 0.0821 \times 273}}{1} = 22.4L$
The molar volume is $22.4L$ .
For option (B), $546K$ and ${\text{1atm}}$ ,
$V = \dfrac{{nRT}}{P}$
Applying all the values in the above equation,
$V = \dfrac{{1 \times 0.0821 \times 546}}{1} = 44.8L$
The molar volume is $44.8L$ .
For option (C), $273K$ and ${\text{2atm}}$ ,
$V = \dfrac{{nRT}}{P}$
Applying all the values in the above equation,
$V = \dfrac{{1 \times 0.0821 \times 273}}{2} = 11.21L$
The molar volume is $11.21L$ .
For option (D), $546K$and ${\text{2atm}}$ ,
$V = \dfrac{{nRT}}{P}$
Applying all the values in the above equation,
$V = \dfrac{{1 \times 0.0821 \times 546}}{2} = 22.4L$
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 $R$.
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 ${\text{1atm}}$ ,
$V = \dfrac{{nRT}}{P}$
Applying all the values in the above equation,
$V = \dfrac{{1 \times 0.0821 \times 273}}{1} = 22.4L$
The molar volume is $22.4L$ .
For option (B), $546K$ and ${\text{1atm}}$ ,
$V = \dfrac{{nRT}}{P}$
Applying all the values in the above equation,
$V = \dfrac{{1 \times 0.0821 \times 546}}{1} = 44.8L$
The molar volume is $44.8L$ .
For option (C), $273K$ and ${\text{2atm}}$ ,
$V = \dfrac{{nRT}}{P}$
Applying all the values in the above equation,
$V = \dfrac{{1 \times 0.0821 \times 273}}{2} = 11.21L$
The molar volume is $11.21L$ .
For option (D), $546K$and ${\text{2atm}}$ ,
$V = \dfrac{{nRT}}{P}$
Applying all the values in the above equation,
$V = \dfrac{{1 \times 0.0821 \times 546}}{2} = 22.4L$
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 $R$.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
