Question

What is the SI unit of universal gas constant?
$\text{A}\text{. }Watt\text{ }{{K}^{-1}}mo{{l}^{-1}}$
$\text{B}\text{. }N{{K}^{-1}}mo{{l}^{-1}}$
$\text{C}\text{. }J{{K}^{-1}}mo{{l}^{-1}}$
$\text{D}\text{. }erg{{K}^{-1}}mo{{l}^{-1}}$

Answer 1

Hint: First, use the ideal gas equation to obtain expression for universal gas constant. Then derive the SI unit by replacing terms in obtained expression by their fundamental SI units. Ideal gas equation relates the product of pressure (P) and volume (V) of the perfect gas with its amount (n) and temperature (T).

Formula used: $PV=nRT$, $P=\dfrac{F}{A}$

Complete step-by-step solution:
We know that the universal gas constant $R$is proportionality constant in the ideal gas equation. To derive the SI unit of the universal gas constant, we obtain an expression for $R$ by using the ideal gas equation.
Ideal gas equation is:
$PV=nRT$
Where
$P=\text{ Pressure exerted by the gas}$
$V=\text{ Volume of the container}$
$n=\text{ Number of moles of gas}$
$R=\text{Universal gas constant}$
$T=\text{ Absolute Temperature}$
Now we rearrange the ideal gas equation to obtain expression for universal gas constant $R$. On rearranging we get
$R=\dfrac{PV}{nT}$
In the above expression the number of moles ($n$) and temperature ($T$) are fundamental physical quantities. SI unit for $n$ is $mole$ and $T$ is $Kelvin$ with symbols $mol$ and $K$ respectively.
Pressure and volume are derived from physical quantities. Their SI units need to be derived from their respective expressions.
Pressure is written as:
$P=\dfrac{F}{A}$
Where
$F=\text{ Force applied perpendicular to the surface of object}$
$A=\text{ Surface area}$
Force has SI unit $kgm{{s}^{-2}}$ also known as Newton$(N)$. Area is $lengt{{h}^{2}}$, therefore has SI unit ${{m}^{2}}$. We substitute these units in place of respective quantities and obtain SI unit of pressure as $N{{m}^{-2}}$.
Volume is $lengt{{h}^{3}}$, therefore has SI unit ${{m}^{3}}$.
Now we replace the physical quantities in expression for $R$ with their respective SI units and get
$(\text{SI unit of }R)=\dfrac{(N{{m}^{-2}})({{m}^{3}})}{(mol)(K)}$
On simplifying the above expression we have
$(\text{SI unit of }R)=Nm{{K}^{-1}}mo{{l}^{-1}}$
We know that $N\text{ }m$ is a unit of energy and can also be written as joule ($J$).
Hence SI unit of $R$ is $J{{K}^{-1}}mo{{l}^{-1}}$.
Therefore option C is correct.

Note: $erg{{K}^{-1}}mo{{l}^{-1}}$ is also a unit of universal gas constant $R$ but it is not the SI unit. So students should not get confused with it.
The value of universal gas constant is $8.314J{{K}^{-1}}mo{{l}^{-1}}$.
Any other relation having $R$ can also be used to obtain its SI unit.