Question

Value of gas constant\[R\,\]in the ideal gas equation \[PV = nRT\]depends upon:
(A) Temperature of the gas.
(B) Pressure of the gas.
(C) Units in the \[P,\]\[V\] and \[T\] are measured.
(D) Nature of the gas.

Answer 1

Hint: The ideal gas equation expresses the quantitative relationship between the variables such as pressure, volume, temperature and number of moles that describe the state of a gas.

Complete step by step answer:
The ideal gas law was discovered in 1834. Ideal gas equation is an equation which is followed by the ideal gases. A gas that would obey Boyle’s law, Charle’s law and Avogadro’s law under all conditions of temperature and pressure
According to Avogadro’s law; \[V \propto n\left( {{\text{P }}and{\text{ T constant}}} \right)\]
According to Boyle’s law; \[V \propto \dfrac{1}{P}\left( {{\text{T and n constant}}} \right)\]
According to Charle’s law; \[V \propto T\left( {P{\text{ and n constant}}} \right)\]
Combining the three laws; we get
\[V \propto \dfrac{{nT}}{P}\]
\[ \Rightarrow V = R\dfrac{{nT}}{P}\]
is the proportionality constant.
Gas constant (R) is a physical constant that is used in the fundamental equations like Arrhenius equation, Nernst equation and ideal gas law.
\[\therefore PV = nRT\]
When \[1\] mole of a gas is considered, then
\[PV = RT\left( {\because n = 1mole} \right)\]
\[R\] (proportionality constant): It relates the energy scale to the temperature scale for a mole of the particles at the given temperature and pressure.
i) Gas constant
ii) It is same for all the gases
Therefore, it is also called universal gas constant. The value of \[R\] depends upon the units in which pressure and volume are taken. So, the value ofwhen pressure and volume are expressed in different units is as below.
(i) When pressure is in Pascal and volume is in \[{m^3}\]
\[R = \dfrac{{PV}}{{nT}} = \dfrac{{\left( {{{10}^5}Pa} \right) \times \left( {22.71 \times {{10}^{ - 3}}{m^3}} \right)}}{{\left( {1mol} \right)\left( {273.15K} \right)}}\]
\[ R = 8.314{\text{Pa}}{{\text{m}}^3}{\text{ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \\
   = 0.0831{\text{bar L }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \\
  1{\text{ Pascal}} = {\text{1N}}{{\text{m}}^{ - 2}} \\
   = 8.314{\text{Nm mo}}{{\text{l}}^{ - 1}}{K^{ - 1}} \\
  \therefore {\text{R}} = 8.314J{\text{ mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}{\text{ }}\left( {1Nm = 1J} \right) \\
\]
(ii) When Pressure in atmosphere and volume in litres.
\[ R = \dfrac{{\left( {1atm} \right) \times \left( {22.7L} \right)}}{{\left( {1mole} \right) \times \left( {273.15K} \right)}} \\
   = 0.082L - atm{\text{ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \\
\]
Hence the correct option is (C).

Note:
A gas which obeys the gas laws and the gas equation \[\left( {PV = nRT} \right)\] strictly at all temperatures and pressure is said to be an ideal gas.