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**Hint:**An ideal gas obeys ideal gas equation \[{\text{PV = nRT}}\]where ‘R’ is the universal gas constant whose units depend upon pressure(P), volume(V) and temperature(T).

**Complete step by step answer:**The unit of pressure in an ideal gas equation can be found from the gas constant values. The standard unit for pressure is Pascal (Pa) which is equal to 1 Newton (N) per square meter. Pressure can be expressed in any of the following units:

A. Atmospheric(atm)

B. Torr

C. Pounds per square inch (Psi)

D. mmHg

E. Pascal (Pa)

F. Bar

Since at various pressure and temperature units the gas constant value varies. The various values of gas constant with their corresponding units are given below:

\[

{\text{R = 0}}{\text{.0821 L atm }}{{\text{K}}^{{\text{ - 1}}}} \\

{\text{R = 8}}{\text{.314 J }}{{\text{K}}^{{\text{ - 1}}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}} \\

{\text{R = 1}}{\text{.987 cal }}{{\text{K}}^{{\text{ - 1}}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}} \\

\]

The given ideal gas constant is 0.0821 which has the unit of \[{\text{L atm }}{{\text{K}}^{{\text{ - 1}}}}\]thus we get the unit of pressure is atm that is expressed in atmospheric.

**So, the correct option is A.**

**Additional Information:**ideal gas obeys the following obeys ideal gas equation

\[{\text{PV = nRT}}\]

where ‘P’ is the pressure of the gas, ‘V’ is the volume of the gas, ‘n’ is the number of moles , ‘T’ is the temperature and ‘R’ is the universal gas

At S.T.P (Standard Temperature and Pressure)

P = 1 atm, T = 273 K or 0° C and volume of gas is 22.4 L,

Let’s assume n = 1. Therefore, by calculation we get R = 0.0821 \[{\text{L atm }}{{\text{K}}^{{\text{ - 1}}}}\]

In C.G.S system, n and T remains same (no unit change), ${\text{P = 1atm = 1}}{{\text{0}}^{\text{5}}}{{ \times 101325dyn/}}{{\text{m}}^{\text{2}}}$, V= 22400ml, then by calculation we get ${\text{R = 8}}{\text{.314 \times 1}}{{\text{0}}^{\text{7}}}{\text{ergs mo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}$

In M.K.S system or SI system

$

{\text{P = 101325N/}}{{\text{m}}^{\text{2}}}{\text{ and V = 22}}{{.4 \times 1}}{{\text{0}}^{\text{ - }}}{\text{3}}{{\text{m}}^{\text{3}}} \\

\Rightarrow {\text{R = 8}}{\text{.314J/Mol}}{\text{.kel}} \\

$

**Note:**Useful unit conversions:

- Joule (J) to Kilojoule (kJ)

${\text{1kJ = 1}}{{\text{0}}^{\text{3}}}{\text{J}}$

- Joule(J) to Calories (cal)

1 Cal = 4.184 J

- Atmospheric(atm) to Torr

${\text{1 atm = 760 torr}}$

- Pounds per square inch (Psi) to mmHg

${\text{1Psi = 51}}{\text{.7151\;mmHg}}$

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