Question

The value of gas constant R is:
a.) 0.082 L atm
b.) 0.987 cal ${mol}^{-1}{K}^{-1}$
c.) 8.3 J ${mol}^{-1}{K}^{-1}$
d.) 83 erg ${mol}^{-1}{K}^{-1}$

Answer 1

Hint: You should know that the gas constant is denoted by the symbol R. It is equivalent to the Boltzmann constant, but expressed in units of energy per temperature increment per mole. The values of R vary according to different unit conversions.

Complete step by step answer:

We know that from the ideal gas equation,
PV = nRT
So, $R={\displaystyle \frac{PV}{nT}}$
where P is the absolute pressure (SI unit pascals), V is the volume of gas (SI unit cubic meters), n is the amount of gas (SI unit moles), and T is the thermodynamic temperature (SI unit kelvins).
Repeated experiments show that at standard temperature (273 K) and pressure (1 atm), one mole (n = 1) of gas occupies 22.4 L volume. Using this experimental value, you can evaluate the gas constant R,
$R={\displaystyle \frac{1atm\times 22.4L}{1mol\times 273K}}$
R = 0.08205 L atm $K^{-1}{mol}^{-1}$

Now, if we need SI units, P = 101325 N/$m^2$ (Pa) instead of 1 atm. The volume is 0.0224 $m^3$ instead of 1 L. The numerical value and units for R are
R = 8.314 J $K^{-1}{mol}^{-1}$
The gas constant can be expressed in the following values and units.
    R = 0.08205 L atm.$K^{-1}{mol}^{-1}$
   =8.3145 L kPa.$K^{-1}{mol}^{-1}$ (1 atm = 101.32 kPa)
   =8.3145 J.$K^{-1}{mol}^{-1}$ (1 J = 1 L kPa)
   =1.987 cal.$K^{-1}{mol}^{-1}$ (1 cal = 4.182 J)
   = 62.364 L torr.$K^{-1}{mol}^{-1}$ (1 atm = 760 torr)
   = 8.3145 x ${10}^7$ erg.$K^{-1}{mol}^{-1}$ (1 J = ${10}^7$ erg)
Note: The name the symbol R the Regnault constant in honor of the French chemist Henri Victor Regnault, whose accurate experimental data were used to calculate the early value of the constant.