Question

The value of gas constant R is 8.314 X. Here, X represents:
A. $\text{L atm }{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{1-}}$
B. $\text{cal mo}{{\text{l}}^{1-}}\text{ }{{\text{K}}^{-1}}$
C. $\text{J }{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}$
D. None of the above

Answer 1

Hint: For this problem, we have to study the gas constant in brief and also about the units of gas constant. Because for different conditions the unit as well as the value of gas constant also varies and then we can choose the correct option.

Complete Step-by-step answer:
- In the given question, we have to choose the correct unit of the gas constant from the given options.
- So, firstly we should know about the gas constant which has several names such as the ideal gas constant, universal gas constant or molar gas constant.
- The unit of a gas constant is generally expressed in the unit of energy per temperature per mole.
- Also, the gas constant is defined as the product of Avogadro's number and Boltzmann constant that is $\text{R = }{{\text{N}}_{\text{A}}}\text{K}$ and here the value of gas constant was given as $\text{8}\text{.314 J }{{\text{K}}^{1-}}\text{ mo}{{\text{l}}^{1-}}$.
 - Whereas in the ideal gas equation the gas constant is expressed as:
$\text{PV = nRT}$ in this equation P is considered as the pressure, V as the volume, n as the number of moles and T as the temperature.
- The symbol of the gas constant was kept 'R' in the honour of the scientist Regnault.
- So, according to these expressions, the unit of the gas constant will be $\text{J }{{\text{K}}^{1-}}\text{ mo}{{\text{l}}^{1-}}$.

Therefore, the value of X is $\text{J }{{\text{K}}^{1-}}\text{ mo}{{\text{l}}^{1-}}$.

Note: As explained that the unit of a gas constant is energy per temperature per mole so here J is the Joule i.e. unit of energy or it can also be written as \[\text{Kg }{{\text{m}}^{2}}\text{ K }{{\text{T}}^{-1}}\text{ mo}{{\text{l}}^{-1}}\text{ }{{\text{s}}^{-2}}\] because Joule can also be written as \[\text{Kg }{{\text{m}}^{2}}\text{ }{{\text{s}}^{-2}}\].