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The gas constant has unit:
(This question has multiple correct options)
\[
  A.{\text{ }}L{\text{ }}atm{\text{ }}{K^{ - 1}}{\text{ }}mo{l^{ - 1}} \\
  B.{\text{ }}L{\text{ }}at{m^{ - 1}}{\text{ }}{K^{ - 1}}{\text{ }}mo{l^{ - 1}} \\
  C.{\text{ }}atm{\text{ }}c{m^3}{\text{ }}{K^{ - 1}}{\text{ }}mo{l^{ - 1}} \\
  D.{\text{ }}erg{\text{ }}{K^{ - 1}} \\
 \]

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Last updated date: 28th Mar 2024
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MVSAT 2024
Answer
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Hint- In order to solve this problem first we have to know about the term gas constant and also the formula for the gas constant further we will discuss its value and according to the formula using the basic units of different quantities we will write its units.

Complete step-by-step answer:
Gas constant: The gas constant is a physical constant denoted by R and is expressed in terms of units of energy per temperature change per mole. This is also known as Ideal gas constant, molar gas constant, uniform gas constant. The value of the gas constant is equal to constant Boltzmann but measured as the sum of pressure-volume instead of energy per increment of temperature per atom.
Gas Constant Value:
Thus, gas constant R value can be given as–
Gas constant $R = 8.3144598\left( {48} \right){\text{ }}J{\text{ }}mo{l^{ - 1}}{\text{ }}{K^{ - 1}}$
The digits within the parentheses are the variance on the gas constant value calculation.
Different Units of gas constant:
Inversely the gas constant is seen in different disciplines. Therefore it is expressed in various units. Many of the gas constant values are given below in different units-
\[
   \Rightarrow R = 8.3144598\left( {48} \right){\text{ }}J{\text{ }}mo{l^{ - 1}}{\text{ }}{K^{ - 1}} \\
   \Rightarrow R = 8.3144598\left( {48} \right) \times {\text{103 }}amu{\text{ }}{m^2}{\text{ }}{s^{ - 2}}{\text{ }}{K^{ - 1}} \\
   \Rightarrow R = 8.3144598\left( {48} \right) \times {\text{1}}{{\text{0}}^{ - 2}}{\text{ }}L{\text{ }}bar{\text{ }}mo{l^{ - 1}}{\text{ }}{K^{ - 1}} \\
   \Rightarrow R = 8.3144598\left( {48} \right){\text{ }}{m^3}{\text{ }}pa{\text{ }}mo{l^{ - 1}}{\text{ }}{K^{ - 1}} \\
   \Rightarrow R = 62.363577\left( {36} \right){\text{ }}L{\text{ }}Torr{\text{ }}mo{l^{ - 1}}{\text{ }}{K^{ - 1}} \\
   \Rightarrow R = 1.9872036\left( {11} \right) \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ }}Kcal{\text{ }}mo{l^{ - 1}}{\text{ }}{K^{ - 1}} \\
   \Rightarrow R = 8.2057338\left( {47} \right){\text{ }}{m^3}{\text{ }}atm{\text{ }}mo{l^{ - 1}}{\text{ }}{K^{ - 1}} \\
   \Rightarrow R = 0.082057338\left( {47} \right){\text{ }}L{\text{ }}atm{\text{ }}mo{l^{ - 1}}{\text{ }}{K^{ - 1}} \\
 \]
Hence by observing the above table the gas constant has unit \[L{\text{ }}atm{\text{ }}{K^{ - 1}}{\text{ }}mo{l^{ - 1}}\& atm{\text{ }}c{m^3}{\text{ }}{K^{ - 1}}{\text{ }}mo{l^{ - 1}}\]
So, the correct answer is option A and C.

Note- The gas constant is the constant of proportionality used to compare the amount of energy to the temperature scale when considering one mole of particles at a given temperature. The ideal gas constant is the variation of law from Boyle, number from Avogadro, law from Charles and law from Gay-Lussac. From a physical point of view, the gas constant is a constant of proportionality which relates the energy scale to the temperature scale at a given temperature for a mole of particles. Units for the constant gas differ according to certain units used in the calculation.
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