Question

Which are the values of $R$?
A. $1.99calde{g^{ - 1}}mo{l^{ - 1}}$
B. $0.0821litre - atm{g^{ - 1}}mo{l^{ - 1}}$
C. $9.8kcalde{g^{ - 1}}mo{l^{ - 1}}$
D. $8.3joulede{g^{ - 1}}mo{l^{ - 1}}$

Answer 1

$R$ is known as universal gas constant which has different types of values in various units. It is derived from the ideal gas equation.

Complete step by step answer:
In order to obtain the value of $R$ in different units at first we need to derive a gas equation. The ideal gas equation is based on four gas laws named as Boyle’s law, Charle’s law, Guy Lussac’s law and Avogadro’s law.
Boyle's law states that at constant temperature the volume of a gas is inversely proportional to the pressure of the gas.
$V \propto \dfrac{1}{P}{\text{ or }}PV = constant$
The Charle’s law states that at constant pressure the volume of a gas is directly proportional to the temperature of the gas.
$V \propto T{\text{ or }}\dfrac{V}{T} = constant$
The Guy –Lussac’s law states that at constant volume the pressure of a gas is directly proportional to the temperature of the gas.
$P \propto T{\text{ or }}\dfrac{P}{T} = constant$
Te Avogadro’s law states that the volume of the gas is directly proportional to the number of particles present in a container.
$V \propto n{\text{ or }}\dfrac{{\text{V}}}{n} = constant$
Combining all the above gas laws the ideal gas equation is written as
$PV = nRT$ where $P$ is pressure, $V$ is volume, $n$ is the number of moles of gas, $T$ is temperature and $R$ is the gas constant. From the equation the pressure and volume is directly proportional to the number of moles and the temperature and the pressure is inversely proportional to the volume.
The value of $R$ (gas constant) depends on the other parameters of the equation which changes with the unit system.
At S.T.P means at Standard Temperature and Pressure, the $P = 1atm$ , $T = 273K$ or $0^\circ C$ , if $n = 1mole$ and volume of gas is $V = 22.4lit$.
$R = \dfrac{{PV}}{{nT}}$ ---(A)
Thus inserting the values of $P$, $V$ , $n$ and $T$ in the equation (A),
$R = \dfrac{{1atm \times 22.4lit}}{{1mol \times 273K}} = 0.0821litatmmo{l^{ - 1}}{K^{ - 1}}$
In C.G.S system, $P = {\text{ }}1atm = 1.013 \times {10^6}dyne/c{m^2}$ , $V = 22400mL$, inserting the values in the equation (A)
$R = \dfrac{{1.013 \times {{10}^6}dyne/c{m^2} \times 22400mL}}{{1mol \times 273K}}$
$R = 8.31 \times {10^7}dynec{m^2}mo{l^{ - 1}}{K^{ - 1}}$
$R = 8.31 \times {10^7}ergsmo{l^{ - 1}}{K^{ - 1}}$, $(1ergs = 1dynec{m^2})$
In SI system$1J = {10^7}ergs$ , so $R = {\text{ }}8.31Jmo{l^{ - 1}}{K^{ - 1}}$
Also $1cal = 4.18J$ . Thus dividing the value of $R$ with$4.18$:
$R = \dfrac{{8.31}}{{4.18}} = 1.98calmo{l^{ - 1}}{K^{ - 1}}$

So, the correct answer is Option A, B,D.

Note: The choice of the correct value of $R$ is important for determining the unknown parameters of the gas. The choice of R depends on the units used for describing other parameters.