What is the value of $R$ in the ideal gas equation $pV = nRT$ in units of $bar \ L \ mol^{-1} \ K^{-1}$? Calculate it.
(N/A) The value of $R$ is $0.08314 \ bar \ L \ mol^{-1} \ K^{-1}$.
From the ideal gas equation,$pV = nRT$,we have $R = \frac{pV}{nT}$.
At standard temperature and pressure $(STP)$,$1 \ mol$ of an ideal gas occupies $22.71 \ L$ at $1 \ bar$ pressure and $273.15 \ K$.
Substituting these values: $R = \frac{(1 \ bar)(22.71 \ L)}{(1 \ mol)(273.15 \ K)}$.
$R = 0.083141 \ bar \ L \ mol^{-1} \ K^{-1}$.
Thus,$R \approx 8.314 \times 10^{-2} \ bar \ L \ mol^{-1} \ K^{-1}$.
From the ideal gas equation,$pV = nRT$,we have $R = \frac{pV}{nT}$.
At standard temperature and pressure $(STP)$,$1 \ mol$ of an ideal gas occupies $22.71 \ L$ at $1 \ bar$ pressure and $273.15 \ K$.
Substituting these values: $R = \frac{(1 \ bar)(22.71 \ L)}{(1 \ mol)(273.15 \ K)}$.
$R = 0.083141 \ bar \ L \ mol^{-1} \ K^{-1}$.
Thus,$R \approx 8.314 \times 10^{-2} \ bar \ L \ mol^{-1} \ K^{-1}$.
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