Express $1$ atm pressure into $N/m^{2}$ and bar.
(N/A) The standard atmospheric pressure is defined as the pressure exerted by a mercury column of $760 \,mm$ height at $0 \,^{\circ}C$ under standard gravity.
$1 \,atm = 1.01325 \times 10^{5} \,N/m^{2}$ (or $Pa$).
Since $1 \,bar = 10^{5} \,Pa$, we can write:
$1 \,atm = 1.01325 \,bar$.
$1 \,atm = 1.01325 \times 10^{5} \,N/m^{2}$ (or $Pa$).
Since $1 \,bar = 10^{5} \,Pa$, we can write:
$1 \,atm = 1.01325 \,bar$.
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