Question
Deduce values of gas constant $'R\ ’$ in different units.
✓
Answer
$i. R$ in $SI$ Unit $($in Joules$):$ Value of $R$ can be calculated by using the $SI$ units of $P, V$ and $T$.
Pressure $P$ is measured in $N m ^{-2}$ or $Pa$, volume $V$ in meter cube $\left( m ^3\right)$ and temperature $T$ in Kelvin $(K)$.
$R =\frac{ PV }{ nT }$
$\therefore R =\frac{10^5 Pa \times 22.71 \times 10^{-3} m ^3}{1 mol \times 273.15 K }$
$\therefore R =8.314 J K ^{-1} mol ^{-1} \text { }$
$ii. R$ in litre atmosphere: If pressure $(P)$ is expressed in atmosphere $(atm)$ and volume in litre $(L)$ or decimeter cube $\left( dm ^3\right)$ and Temperature in kelvin $(K), ($that is, old $STP$ conditions$)$,
then value of $R$ is, $R =\frac{1 atm \times 22.414 L }{1 mol \times 273.15 K }$
$\therefore R =0.0821 \ L \ atm K ^{-1} mol ^{-1}$
$OR$
$R=0.0821 \ dm ^3 \ atm \ K ^{-1} \ mol ^{-1}$
$iii$. $R$ in calories: We know, $1$ calorie $=4.184$ Joules
$\therefore R =\frac{8.314}{4.184}=1.987 \cong 2 cal K ^{-1} mol ^{-1}$
Pressure $P$ is measured in $N m ^{-2}$ or $Pa$, volume $V$ in meter cube $\left( m ^3\right)$ and temperature $T$ in Kelvin $(K)$.
$R =\frac{ PV }{ nT }$
$\therefore R =\frac{10^5 Pa \times 22.71 \times 10^{-3} m ^3}{1 mol \times 273.15 K }$
$\therefore R =8.314 J K ^{-1} mol ^{-1} \text { }$
$ii. R$ in litre atmosphere: If pressure $(P)$ is expressed in atmosphere $(atm)$ and volume in litre $(L)$ or decimeter cube $\left( dm ^3\right)$ and Temperature in kelvin $(K), ($that is, old $STP$ conditions$)$,
then value of $R$ is, $R =\frac{1 atm \times 22.414 L }{1 mol \times 273.15 K }$
$\therefore R =0.0821 \ L \ atm K ^{-1} mol ^{-1}$
$OR$
$R=0.0821 \ dm ^3 \ atm \ K ^{-1} \ mol ^{-1}$
$iii$. $R$ in calories: We know, $1$ calorie $=4.184$ Joules
$\therefore R =\frac{8.314}{4.184}=1.987 \cong 2 cal K ^{-1} mol ^{-1}$
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