Question

Calculate the value of molar gas constant ‘R’ in
a.) cc atm ${K^{ - 1}}mo{l^{ - 1}}$
b.) torr cc ${K^{ - 1}}mo{l^{ - 1}}$
c.) kPa liter ${K^{ - 1}}mo{l^{ - 1}}$

Answer 1

Hint: Before attempting this question one should have prior knowledge about the molar gas constant and also remember to use the ideal gas equation i.e. PV = nRT, using these instructions will help you to approach towards the solution to the problem.

Complete step by step solution: According to the question the first pressure which is given in atm while the volume in CC. therefore, when the pressure is expressed in atmosphere [atm] and volume in CC.
Then, the value of P & V are taken as:-
   P=1atm V=22400cc
The ideal gas equation is given as: PV=nRT
From here the value of R is \[R = \dfrac{{PV}}{{nT}}\]
Where P is pressure, V is volume, n is no. of moles, T is temperature.
So, \[R = \dfrac{{PV}}{{nT}}\] (equation 1)
P is taken as = 1atm
V is taken as = 22400cc and T is in Kelvin which means the value of T will be 273k and n=1.
Equating these value in equation 1 we get
\[R = \dfrac{{PV}}{{nT}}\]
$ \Rightarrow $$R = \dfrac{{1 \times 22400}}{{1 \times 273}}$= $\dfrac{{22400}}{{273}} = 82.1ccatm{K^{ - 1}}mo{l^{ - 1}}$
2) Hence, the pressure is given in terms of torr and volume in cc
So, the value of pressure in torr is=760 torr, and volume in cc is = 22400cc
T=273k and n=1.
Similarly, equating these values in equation 1 we get
    \[R = \dfrac{{PV}}{{nT}}\]
$ \Rightarrow $\[R = \dfrac{{760 \times 22400}}{{1 \times 273}}\]
$ \Rightarrow $$R = \dfrac{{17024000}}{{273}}$
So, R = 62360 torr cc ${K^{ - 1}}mo{l^{ - 1}}$
3) Now, the pressure is given in kilo pascal and volume in litre.
As we known, the value of pressure in kilopascal is 101.3KPa and volume is= 22.4litre, while T and h remain same as above two equations
T=273K n=1
Again, equating these values in equation 1 we get
\[R = \dfrac{{PV}}{{nT}}\]
\[R = \dfrac{{101.3 \times 22.4}}{{1 \times 273}}\]=$\dfrac{{2269.12}}{{273}}$= $ = 8.311$
 $ \Rightarrow $R= 8.31 K pa liter \[{K^{ - 1}}mo{l^{ - 1}}\]

Note: Molar gas constant or [R] is simply a physical constant that exists in the general gas law formulation. The constant, however, is the same for all gases, since one mole or one molecular weight in grams is the mass of gas being measured.
So, for one mole, $\dfrac{{PV}}{T} = R$ and the value of R is$8.314 mo{l^{ - 1}}{K^{ - 1}}$.